YES Termination w.r.t. Q proof of MNZ_10_8.ari

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 TransformationProof (⇔, 3 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 TransformationProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 DependencyGraphProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 TransformationProof (⇔, 0 ms)
        ↳15 QDP
        ↳16 DependencyGraphProof (⇔, 0 ms)
        ↳17 QDP
        ↳18 TransformationProof (⇔, 0 ms)
        ↳19 QDP
        ↳20 DependencyGraphProof (⇔, 0 ms)
        ↳21 QDP
        ↳22 TransformationProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 DependencyGraphProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 TransformationProof (⇔, 0 ms)
        ↳27 QDP
        ↳28 DependencyGraphProof (⇔, 0 ms)
        ↳29 QDP
        ↳30 TransformationProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 DependencyGraphProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 TransformationProof (⇔, 0 ms)
        ↳35 QDP
        ↳36 DependencyGraphProof (⇔, 0 ms)
        ↳37 QDP
        ↳38 TransformationProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 DependencyGraphProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 TransformationProof (⇔, 1 ms)
        ↳43 QDP
        ↳44 DependencyGraphProof (⇔, 0 ms)
        ↳45 QDP
        ↳46 TransformationProof (⇔, 0 ms)
        ↳47 QDP
        ↳48 DependencyGraphProof (⇔, 0 ms)
        ↳49 QDP
        ↳50 QDPOrderProof (⇔, 27 ms)
        ↳51 QDP
        ↳52 QDPOrderProof (⇔, 1490 ms)
        ↳53 QDP
        ↳54 QDPOrderProof (⇔, 1901 ms)
        ↳55 QDP
        ↳56 QDPOrderProof (⇔, 1723 ms)
        ↳57 QDP
        ↳58 QDPOrderProof (⇔, 1201 ms)
        ↳59 QDP
        ↳60 QDPOrderProof (⇔, 1513 ms)
        ↳61 QDP
        ↳62 QDPOrderProof (⇔, 1031 ms)
        ↳63 QDP
        ↳64 QDPOrderProof (⇔, 619 ms)
        ↳65 QDP
        ↳66 QDPOrderProof (⇔, 727 ms)
        ↳67 QDP
        ↳68 QDPOrderProof (⇔, 83 ms)
        ↳69 QDP
        ↳70 PisEmptyProof (⇔, 0 ms)
        ↳71 YES
    ↳72 QDP
        ↳73 QDPOrderProof (⇔, 79 ms)
        ↳74 QDP
        ↳75 DependencyGraphProof (⇔, 0 ms)
        ↳76 QDP
        ↳77 UsableRulesProof (⇔, 0 ms)
        ↳78 QDP
        ↳79 QDPSizeChangeProof (⇔, 0 ms)
        ↳80 YES

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(0)) → F(s(0))
G(x) → H(x, x)
S(x) → H(x, 0)
S(x) → H(0, x)
F(g(x)) → G(g(f(x)))
F(g(x)) → G(f(x))
F(g(x)) → F(x)
G(s(x)) → S(s(g(x)))
G(s(x)) → S(g(x))
G(s(x)) → G(x)
H(f(x), g(x)) → F(s(x))
H(f(x), g(x)) → S(x)
S(s(0)) → K(0)
S(s(s(0))) → K(s(0))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(0))))))
K(s(s(0))) → S(s(s(s(s(0)))))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))
H(k(x), g(x)) → K(s(x))
H(k(x), g(x)) → S(x)

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(g(x)) → G(g(f(x)))
G(x) → H(x, x)
H(f(x), g(x)) → F(s(x))
F(g(x)) → G(f(x))
G(s(x)) → S(s(g(x)))
S(s(0)) → F(s(0))
F(g(x)) → F(x)
S(s(s(0))) → K(s(0))
K(s(s(0))) → S(s(s(s(s(s(s(s(0))))))))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(0))))))
K(s(s(0))) → S(s(s(s(s(0)))))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))
G(s(x)) → S(g(x))
G(s(x)) → G(x)
H(f(x), g(x)) → S(x)
H(k(x), g(x)) → K(s(x))
H(k(x), g(x)) → S(x)

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule S(s(0)) → F(s(0)) at position [0] we obtained the following new rules [LPAR04]:

S(s(0)) → F(h(0, 0)) → S(s(0)) → F(h(0, 0))

(6) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(g(x)) → G(g(f(x)))
G(x) → H(x, x)
H(f(x), g(x)) → F(s(x))
F(g(x)) → G(f(x))
G(s(x)) → S(s(g(x)))
F(g(x)) → F(x)
S(s(s(0))) → K(s(0))
K(s(s(0))) → S(s(s(s(s(s(s(s(0))))))))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(0))))))
K(s(s(0))) → S(s(s(s(s(0)))))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))
G(s(x)) → S(g(x))
G(s(x)) → G(x)
H(f(x), g(x)) → S(x)
H(k(x), g(x)) → K(s(x))
H(k(x), g(x)) → S(x)
S(s(0)) → F(h(0, 0))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 6 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

K(s(s(0))) → S(s(s(s(s(s(s(s(0))))))))
S(s(s(0))) → K(s(0))
K(s(s(0))) → S(s(s(s(s(s(s(0)))))))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(s(0))))))
K(s(s(0))) → S(s(s(s(s(0)))))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(10) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule S(s(s(0))) → K(s(0)) at position [0] we obtained the following new rules [LPAR04]:

S(s(s(0))) → K(h(0, 0)) → S(s(s(0))) → K(h(0, 0))

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

K(s(s(0))) → S(s(s(s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(s(0)))))))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(s(0))))))
K(s(s(0))) → S(s(s(s(s(0)))))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))
S(s(s(0))) → K(h(0, 0))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(13) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(0))))))
K(s(s(0))) → S(s(s(s(s(0)))))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(14) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule K(s(s(0))) → S(s(s(s(s(s(s(s(0)))))))) at position [0] we obtained the following new rules [LPAR04]:

K(s(s(0))) → S(h(s(s(s(s(s(s(0)))))), 0)) → K(s(s(0))) → S(h(s(s(s(s(s(s(0)))))), 0))
K(s(s(0))) → S(h(0, s(s(s(s(s(s(0)))))))) → K(s(s(0))) → S(h(0, s(s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0))) → K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0)))))))) → K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0)))) → K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0)))))))) → K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0))))) → K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0)))))))) → K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0)))))) → K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0)))))))) → K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0))))))) → K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0)))))))) → K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0))))))) → K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0)))))))) → K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0))))))) → K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0)))))))) → K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))

(15) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(0))))))
K(s(s(0))) → S(s(s(s(s(0)))))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))
K(s(s(0))) → S(h(s(s(s(s(s(s(0)))))), 0))
K(s(s(0))) → S(h(0, s(s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(16) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(17) Obligation:

Q DP problem:
The TRS P consists of the following rules:

K(s(s(0))) → S(s(s(s(s(s(s(0)))))))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(s(0))))))
K(s(s(0))) → S(s(s(s(s(0)))))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(18) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule K(s(s(0))) → S(s(s(s(s(s(s(0))))))) at position [0] we obtained the following new rules [LPAR04]:

K(s(s(0))) → S(h(s(s(s(s(s(0))))), 0)) → K(s(s(0))) → S(h(s(s(s(s(s(0))))), 0))
K(s(s(0))) → S(h(0, s(s(s(s(s(0))))))) → K(s(s(0))) → S(h(0, s(s(s(s(s(0)))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0))) → K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0))))))) → K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0)))) → K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0))))))) → K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0))))) → K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0))))))) → K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0)))))) → K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0))))))) → K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0)))))) → K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0))))))) → K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0)))))) → K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0))))))) → K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))

(19) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(s(0))))))
K(s(s(0))) → S(s(s(s(s(0)))))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(h(s(s(s(s(s(0))))), 0))
K(s(s(0))) → S(h(0, s(s(s(s(s(0)))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(20) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

K(s(s(0))) → S(s(s(s(s(s(0))))))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(0)))))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(22) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule K(s(s(0))) → S(s(s(s(s(s(0)))))) at position [0] we obtained the following new rules [LPAR04]:

K(s(s(0))) → S(h(s(s(s(s(0)))), 0)) → K(s(s(0))) → S(h(s(s(s(s(0)))), 0))
K(s(s(0))) → S(h(0, s(s(s(s(0)))))) → K(s(s(0))) → S(h(0, s(s(s(s(0))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0))) → K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0)))))) → K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0)))) → K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0)))))) → K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0))))) → K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0)))))) → K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0))))) → K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0)))))) → K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0))))) → K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0)))))) → K(s(s(0))) → S(s(s(s(s(h(0, 0))))))

(23) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(0)))))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(h(s(s(s(s(0)))), 0))
K(s(s(0))) → S(h(0, s(s(s(s(0))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(24) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

K(s(s(0))) → S(s(s(s(s(0)))))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(26) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule K(s(s(0))) → S(s(s(s(s(0))))) at position [0] we obtained the following new rules [LPAR04]:

K(s(s(0))) → S(h(s(s(s(0))), 0)) → K(s(s(0))) → S(h(s(s(s(0))), 0))
K(s(s(0))) → S(h(0, s(s(s(0))))) → K(s(s(0))) → S(h(0, s(s(s(0)))))
K(s(s(0))) → S(s(h(s(s(0)), 0))) → K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0))))) → K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0)))) → K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0))))) → K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0)))) → K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0))))) → K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0)))) → K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0))))) → K(s(s(0))) → S(s(s(s(h(0, 0)))))

(27) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(0))))
K(s(s(0))) → S(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(h(s(s(s(0))), 0))
K(s(s(0))) → S(h(0, s(s(s(0)))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(28) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(29) Obligation:

Q DP problem:
The TRS P consists of the following rules:

K(s(s(0))) → S(s(s(s(0))))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(30) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule K(s(s(0))) → S(s(s(s(0)))) at position [0] we obtained the following new rules [LPAR04]:

K(s(s(0))) → S(h(s(s(0)), 0)) → K(s(s(0))) → S(h(s(s(0)), 0))
K(s(s(0))) → S(h(0, s(s(0)))) → K(s(s(0))) → S(h(0, s(s(0))))
K(s(s(0))) → S(k(s(0))) → K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → S(s(f(s(0)))) → K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(s(0), 0))) → K(s(s(0))) → S(s(h(s(0), 0)))
K(s(s(0))) → S(s(h(0, s(0)))) → K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(k(0))) → K(s(s(0))) → S(s(k(0)))
K(s(s(0))) → S(s(s(h(0, 0)))) → K(s(s(0))) → S(s(s(h(0, 0))))

(31) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))
K(s(s(0))) → S(h(s(s(0)), 0))
K(s(s(0))) → S(h(0, s(s(0))))
K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(s(0), 0)))
K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(k(0)))
K(s(s(0))) → S(s(s(h(0, 0))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(32) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(33) Obligation:

Q DP problem:
The TRS P consists of the following rules:

K(s(s(0))) → S(s(s(0)))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))
K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(s(0), 0)))
K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(k(0)))
K(s(s(0))) → S(s(s(h(0, 0))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(34) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule K(s(s(0))) → S(s(s(0))) at position [0] we obtained the following new rules [LPAR04]:

K(s(s(0))) → S(f(s(0))) → K(s(s(0))) → S(f(s(0)))
K(s(s(0))) → S(h(s(0), 0)) → K(s(s(0))) → S(h(s(0), 0))
K(s(s(0))) → S(h(0, s(0))) → K(s(s(0))) → S(h(0, s(0)))
K(s(s(0))) → S(k(0)) → K(s(s(0))) → S(k(0))
K(s(s(0))) → S(s(h(0, 0))) → K(s(s(0))) → S(s(h(0, 0)))

(35) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))
K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(s(0), 0)))
K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(k(0)))
K(s(s(0))) → S(s(s(h(0, 0))))
K(s(s(0))) → S(f(s(0)))
K(s(s(0))) → S(h(s(0), 0))
K(s(s(0))) → S(h(0, s(0)))
K(s(s(0))) → S(k(0))
K(s(s(0))) → S(s(h(0, 0)))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(36) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(37) Obligation:

Q DP problem:
The TRS P consists of the following rules:

K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))
K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(s(0), 0)))
K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(k(0)))
K(s(s(0))) → S(s(s(h(0, 0))))
K(s(s(0))) → S(f(s(0)))
K(s(s(0))) → S(k(0))
K(s(s(0))) → S(s(h(0, 0)))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(38) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule K(s(s(0))) → S(f(s(0))) at position [0] we obtained the following new rules [LPAR04]:

K(s(s(0))) → S(f(h(0, 0))) → K(s(s(0))) → S(f(h(0, 0)))

(39) Obligation:

Q DP problem:
The TRS P consists of the following rules:

K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))
K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(s(0), 0)))
K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(k(0)))
K(s(s(0))) → S(s(s(h(0, 0))))
K(s(s(0))) → S(k(0))
K(s(s(0))) → S(s(h(0, 0)))
K(s(s(0))) → S(f(h(0, 0)))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(40) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(41) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))
K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(s(0), 0)))
K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(k(0)))
K(s(s(0))) → S(s(s(h(0, 0))))
K(s(s(0))) → S(k(0))
K(s(s(0))) → S(s(h(0, 0)))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(42) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule K(s(s(0))) → S(k(0)) at position [0] we obtained the following new rules [LPAR04]:

K(s(s(0))) → S(0) → K(s(s(0))) → S(0)

(43) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))
K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(s(0), 0)))
K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(k(0)))
K(s(s(0))) → S(s(s(h(0, 0))))
K(s(s(0))) → S(s(h(0, 0)))
K(s(s(0))) → S(0)

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(44) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(45) Obligation:

Q DP problem:
The TRS P consists of the following rules:

K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))
K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(s(0), 0)))
K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(k(0)))
K(s(s(0))) → S(s(s(h(0, 0))))
K(s(s(0))) → S(s(h(0, 0)))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(46) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule K(s(s(0))) → S(s(h(0, 0))) at position [0] we obtained the following new rules [LPAR04]:

K(s(s(0))) → S(h(h(0, 0), 0)) → K(s(s(0))) → S(h(h(0, 0), 0))
K(s(s(0))) → S(h(0, h(0, 0))) → K(s(s(0))) → S(h(0, h(0, 0)))

(47) Obligation:

Q DP problem:
The TRS P consists of the following rules:

K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))
K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(s(0), 0)))
K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(k(0)))
K(s(s(0))) → S(s(s(h(0, 0))))
K(s(s(0))) → S(h(h(0, 0), 0))
K(s(s(0))) → S(h(0, h(0, 0)))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(48) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(49) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))
K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(s(0), 0)))
K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(k(0)))
K(s(s(0))) → S(s(s(h(0, 0))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(50) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → S(s(k(0)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

POL(S(x1)) = 1 +
[1,1]
·x1

POL(s(x1)) =
/1\
\1/
 +
/00\
\10/
·x1

POL(0) =
/0\
\1/

POL(K(x1)) = 1 +
[1,1]
·x1

POL(h(x1, x2)) =
/1\
\1/
 +
/00\
\10/
·x1 +
/00\
\00/
·x2

POL(k(x1)) =
/0\
\0/
 +
/10\
\01/
·x1

POL(f(x1)) =
/1\
\1/
 +
/00\
\10/
·x1

POL(g(x1)) =
/1\
\1/
 +
/00\
\10/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
h(k(x), g(x)) → k(s(x))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(0)) → k(0)
k(0) → 0
k(s(0)) → s(0)

(51) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))
K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(s(0), 0)))
K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(s(h(0, 0))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(52) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


K(s(s(0))) → S(s(h(0, s(s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(f(s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(s(h(0, 0))))))))
K(s(s(0))) → S(s(h(0, s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(s(f(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, 0)))))))
K(s(s(0))) → S(s(h(0, s(s(s(0))))))
K(s(s(0))) → S(s(s(s(f(s(0))))))
K(s(s(0))) → S(s(s(s(s(h(0, 0))))))
K(s(s(0))) → S(s(h(0, s(s(0)))))
K(s(s(0))) → S(s(s(f(s(0)))))
K(s(s(0))) → S(s(s(s(h(0, 0)))))
K(s(s(0))) → S(s(f(s(0))))
K(s(s(0))) → S(s(h(0, s(0))))
K(s(s(0))) → S(s(s(h(0, 0))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x1)) = 2A +
[0A,-I,-I]
·x1

POL(s(x1)) =
/1A\
|-I|
\-I/
 +
/-I2A1A\
|-I0A0A|
\-I0A0A/
·x1

POL(0) =
/1A\
|-I|
\1A/

POL(K(x1)) = 3A +
[-I,0A,-I]
·x1

POL(h(x1, x2)) =
/1A\
|-I|
\-I/
 +
/-I2A-I\
|-I0A-I|
\-I0A-I/
·x1 +
/-I2A-I\
|-I-I-I|
\-I0A-I/
·x2

POL(k(x1)) =
/0A\
|1A|
\1A/
 +
/0A-I-I\
|-I0A0A|
\-I-I-I/
·x1

POL(f(x1)) =
/2A\
|0A|
\0A/
 +
/-I-I-I\
|-I-I-I|
\-I-I-I/
·x1

POL(g(x1)) =
/2A\
|0A|
\0A/
 +
/0A2A-I\
|-I0A-I|
\-I0A0A/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
h(k(x), g(x)) → k(s(x))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(0)) → k(0)
k(0) → 0
k(s(0)) → s(0)

(53) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(s(h(s(0), 0)))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(54) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


K(s(s(0))) → S(s(h(s(s(s(s(s(0))))), 0)))
K(s(s(0))) → S(s(h(s(s(s(s(0)))), 0)))
K(s(s(0))) → S(s(h(s(s(s(0))), 0)))
K(s(s(0))) → S(s(h(s(s(0)), 0)))
K(s(s(0))) → S(s(k(s(0))))
K(s(s(0))) → S(s(h(s(0), 0)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x1)) = 1A +
[1A,2A,2A]
·x1

POL(s(x1)) =
/0A\
|3A|
\3A/
 +
/-I-I1A\
|3A0A3A|
\-I-I0A/
·x1

POL(0) =
/1A\
|3A|
\0A/

POL(K(x1)) = 1A +
[1A,3A,-I]
·x1

POL(h(x1, x2)) =
/0A\
|-I|
\-I/
 +
/-I-I-I\
|-I-I0A|
\-I-I0A/
·x1 +
/-I-I0A\
|-I0A2A|
\-I-I0A/
·x2

POL(k(x1)) =
/0A\
|3A|
\3A/
 +
/0A-I-I\
|3A-I0A|
\-I-I0A/
·x1

POL(f(x1)) =
/0A\
|-I|
\-I/
 +
/-I-I-I\
|-I-I-I|
\-I-I-I/
·x1

POL(g(x1)) =
/0A\
|-I|
\-I/
 +
/0A-I3A\
|-I3A2A|
\-I-I2A/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
h(k(x), g(x)) → k(s(x))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(0)) → k(0)
k(0) → 0
k(s(0)) → s(0)

(55) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
K(s(s(0))) → S(s(s(k(0))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(56) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


K(s(s(0))) → S(s(s(s(s(s(h(0, s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(0)))))))
K(s(s(0))) → S(s(s(s(h(0, s(0))))))
K(s(s(0))) → S(s(s(h(0, s(0)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x1)) = 0A +
[0A,-I,0A]
·x1

POL(s(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|0A0A-I|
\0A0A-I/
·x1

POL(0) =
/0A\
|-I|
\1A/

POL(K(x1)) = 1A +
[-I,0A,0A]
·x1

POL(h(x1, x2)) =
/0A\
|-I|
\0A/
 +
/0A-I-I\
|-I0A-I|
\0A-I-I/
·x1 +
/-I0A-I\
|-I0A-I|
\-I-I-I/
·x2

POL(k(x1)) =
/1A\
|0A|
\0A/
 +
/0A0A0A\
|-I0A-I|
\-I0A0A/
·x1

POL(f(x1)) =
/0A\
|0A|
\0A/
 +
/0A0A0A\
|-I0A0A|
\0A0A-I/
·x1

POL(g(x1)) =
/1A\
|1A|
\1A/
 +
/0A0A0A\
|0A0A-I|
\0A-I0A/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
h(k(x), g(x)) → k(s(x))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(0)) → k(0)
k(0) → 0
k(s(0)) → s(0)

(57) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(k(0))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(58) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


K(s(s(0))) → S(s(s(h(0, s(s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(h(0, s(s(s(0))))))))
K(s(s(0))) → S(s(s(s(s(h(0, s(s(0))))))))
K(s(s(0))) → S(s(s(h(0, s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(h(0, s(s(0)))))))
K(s(s(0))) → S(s(s(h(0, s(s(0))))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x1)) = 0A +
[-I,1A,-I]
·x1

POL(s(x1)) =
/0A\
|-I|
\-I/
 +
/0A-I-I\
|-I-I0A|
\-I0A0A/
·x1

POL(0) =
/0A\
|0A|
\-I/

POL(K(x1)) = 1A +
[0A,-I,-I]
·x1

POL(h(x1, x2)) =
/0A\
|-I|
\-I/
 +
/-I-I-I\
|-I-I0A|
\-I-I0A/
·x1 +
/-I-I-I\
|-I-I-I|
\-I-I-I/
·x2

POL(k(x1)) =
/0A\
|0A|
\0A/
 +
/-I-I-I\
|0A-I-I|
\0A-I-I/
·x1

POL(f(x1)) =
/0A\
|0A|
\0A/
 +
/-I-I-I\
|-I-I-I|
\-I-I-I/
·x1

POL(g(x1)) =
/0A\
|0A|
\0A/
 +
/0A-I0A\
|-I0A0A|
\-I-I0A/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
h(k(x), g(x)) → k(s(x))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(0)) → k(0)
k(0) → 0
k(s(0)) → s(0)

(59) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
K(s(s(0))) → S(s(s(k(0))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(60) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


K(s(s(0))) → S(s(s(h(s(s(s(s(0)))), 0))))
K(s(s(0))) → S(s(s(s(h(s(s(s(0))), 0)))))
K(s(s(0))) → S(s(s(s(s(h(s(s(0)), 0))))))
K(s(s(0))) → S(s(s(s(s(s(h(s(0), 0)))))))
K(s(s(0))) → S(s(s(h(s(s(s(0))), 0))))
K(s(s(0))) → S(s(s(s(h(s(s(0)), 0)))))
K(s(s(0))) → S(s(s(s(s(h(s(0), 0))))))
K(s(s(0))) → S(s(s(h(s(s(0)), 0))))
K(s(s(0))) → S(s(s(s(h(s(0), 0)))))
K(s(s(0))) → S(s(s(h(s(0), 0))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x1)) = 1A +
[0A,1A,1A]
·x1

POL(s(x1)) =
/0A\
|-I|
\-I/
 +
/0A-I-I\
|0A0A-I|
\0A0A0A/
·x1

POL(0) =
/0A\
|-I|
\1A/

POL(K(x1)) = 2A +
[1A,-I,-I]
·x1

POL(h(x1, x2)) =
/0A\
|-I|
\-I/
 +
/-I-I-I\
|-I0A-I|
\-I0A-I/
·x1 +
/-I-I-I\
|-I-I-I|
\-I0A-I/
·x2

POL(k(x1)) =
/0A\
|0A|
\-I/
 +
/-I-I-I\
|0A-I-I|
\0A-I0A/
·x1

POL(f(x1)) =
/0A\
|0A|
\-I/
 +
/-I-I-I\
|0A-I-I|
\0A-I-I/
·x1

POL(g(x1)) =
/0A\
|-I|
\0A/
 +
/0A-I-I\
|0A0A0A|
\-I0A0A/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
h(k(x), g(x)) → k(s(x))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(0)) → k(0)
k(0) → 0
k(s(0)) → s(0)

(61) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(k(0)))))
K(s(s(0))) → S(s(s(k(0))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(62) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


K(s(s(0))) → S(s(s(k(0))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x1)) = 1A +
[1A,-I,0A]
·x1

POL(s(x1)) =
/1A\
|1A|
\-I/
 +
/0A-I-I\
|1A0A-I|
\0A1A-I/
·x1

POL(0) =
/0A\
|-I|
\1A/

POL(K(x1)) = 1A +
[2A,-I,-I]
·x1

POL(k(x1)) =
/0A\
|1A|
\1A/
 +
/0A-I-I\
|0A-I0A|
\0A-I1A/
·x1

POL(h(x1, x2)) =
/1A\
|-I|
\-I/
 +
/0A-I-I\
|-I0A-I|
\-I1A-I/
·x1 +
/-I-I-I\
|-I0A-I|
\-I1A-I/
·x2

POL(f(x1)) =
/1A\
|-I|
\-I/
 +
/0A-I-I\
|-I-I-I|
\-I-I-I/
·x1

POL(g(x1)) =
/1A\
|-I|
\-I/
 +
/0A-I-I\
|-I1A-I|
\-I2A1A/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
h(k(x), g(x)) → k(s(x))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(0)) → k(0)
k(0) → 0
k(s(0)) → s(0)

(63) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(k(0)))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(64) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


K(s(s(0))) → S(s(s(s(k(s(0))))))
K(s(s(0))) → S(s(s(s(s(k(0))))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x1)) = 0A +
[2A,-I,0A]
·x1

POL(s(x1)) =
/0A\
|-I|
\0A/
 +
/-I0A0A\
|-I-I-I|
\0A-I-I/
·x1

POL(0) =
/0A\
|-I|
\1A/

POL(K(x1)) = 2A +
[3A,-I,2A]
·x1

POL(k(x1)) =
/0A\
|-I|
\-I/
 +
/0A1A-I\
|-I-I-I|
\-I-I0A/
·x1

POL(h(x1, x2)) =
/0A\
|-I|
\-I/
 +
/-I-I-I\
|-I-I-I|
\-I-I-I/
·x1 +
/-I0A-I\
|-I-I-I|
\0A-I-I/
·x2

POL(f(x1)) =
/0A\
|-I|
\-I/
 +
/-I0A-I\
|-I0A-I|
\-I0A-I/
·x1

POL(g(x1)) =
/0A\
|0A|
\-I/
 +
/0A0A0A\
|-I0A0A|
\0A0A0A/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
h(k(x), g(x)) → k(s(x))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(0)) → k(0)
k(0) → 0
k(s(0)) → s(0)

(65) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(k(0)))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(66) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


K(s(s(0))) → S(s(s(k(s(0)))))
K(s(s(0))) → S(s(s(s(k(0)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x1)) = 0A +
[0A,-I,-I]
·x1

POL(s(x1)) =
/0A\
|-I|
\-I/
 +
/-I-I0A\
|-I0A0A|
\0A0A-I/
·x1

POL(0) =
/1A\
|-I|
\-I/

POL(K(x1)) = 1A +
[-I,-I,-I]
·x1

POL(k(x1)) =
/0A\
|-I|
\0A/
 +
/0A-I-I\
|-I0A-I|
\-I0A0A/
·x1

POL(h(x1, x2)) =
/0A\
|-I|
\-I/
 +
/-I-I-I\
|-I-I-I|
\-I-I-I/
·x1 +
/-I-I0A\
|-I-I0A|
\-I0A-I/
·x2

POL(f(x1)) =
/0A\
|-I|
\-I/
 +
/-I0A0A\
|-I0A0A|
\0A0A-I/
·x1

POL(g(x1)) =
/0A\
|0A|
\-I/
 +
/0A-I0A\
|0A0A0A|
\0A0A0A/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
h(k(x), g(x)) → k(s(x))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(0)) → k(0)
k(0) → 0
k(s(0)) → s(0)

(67) Obligation:

Q DP problem:
The TRS P consists of the following rules:

S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(68) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


S(s(s(s(s(s(s(s(s(s(0)))))))))) → K(s(s(0)))
K(s(s(0))) → S(s(s(s(s(k(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(k(0)))))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

POL(S(x1)) = 2 +
[0,1]
·x1

POL(s(x1)) =
/2\
\0/
 +
/01\
\10/
·x1

POL(0) =
/0\
\0/

POL(K(x1)) = 3 +
[2,1]
·x1

POL(k(x1)) =
/2\
\2/
 +
/03\
\03/
·x1

POL(h(x1, x2)) =
/0\
\0/
 +
/01\
\10/
·x1 +
/01\
\10/
·x2

POL(f(x1)) =
/0\
\0/
 +
/00\
\00/
·x1

POL(g(x1)) =
/0\
\0/
 +
/33\
\33/
·x1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
h(k(x), g(x)) → k(s(x))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(0)) → k(0)
k(0) → 0
k(s(0)) → s(0)

(69) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(70) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(71) YES

(72) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(x) → H(x, x)
H(f(x), g(x)) → F(s(x))
F(g(x)) → G(g(f(x)))
G(s(x)) → G(x)
F(g(x)) → G(f(x))
F(g(x)) → F(x)

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(73) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


F(g(x)) → G(g(f(x)))
F(g(x)) → G(f(x))
F(g(x)) → F(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( F(x1) ) = 2x1 + 2

POL( G(x1) ) = x1 + 2

POL( f(x1) ) = 2x1

POL( g(x1) ) = x1 + 2

POL( h(x1, x2) ) = x1

POL( s(x1) ) = x1

POL( k(x1) ) = max{0, -2}

POL( 0 ) = 0

POL( H(x1, x2) ) = x1 + 2


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
h(k(x), g(x)) → k(s(x))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(x) → h(x, 0)
s(x) → h(0, x)
s(s(0)) → k(0)
k(0) → 0
k(s(0)) → s(0)

(74) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(x) → H(x, x)
H(f(x), g(x)) → F(s(x))
G(s(x)) → G(x)

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(75) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(76) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(s(x)) → G(x)

The TRS R consists of the following rules:

s(s(0)) → f(s(0))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))
s(s(0)) → k(0)
k(0) → 0
s(s(s(0))) → k(s(0))
k(s(0)) → s(0)
s(s(s(s(s(s(s(s(s(s(0)))))))))) → k(s(s(0)))
k(s(s(0))) → s(s(s(s(s(s(s(s(0))))))))
h(k(x), g(x)) → k(s(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(77) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(78) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(s(x)) → G(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(79) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • G(s(x)) → G(x)
    The graph contains the following edges 1 > 1

(80) YES