Termination w.r.t. Q proof of MNZ_10

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 (⇔, 0 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 (⇔, 10 ms)

        ↳39 QDP

        ↳40 DependencyGraphProof (⇔, 0 ms)

        ↳41 QDP

        ↳42 TransformationProof (⇔, 0 ms)

        ↳43 QDP

        ↳44 DependencyGraphProof (⇔, 0 ms)

        ↳45 QDP

        ↳46 QDPOrderProof (⇔, 13 ms)

        ↳47 QDP

        ↳48 QDPOrderProof (⇔, 20 ms)

        ↳49 QDP

        ↳50 QDPOrderProof (⇔, 5 ms)

        ↳51 QDP

        ↳52 QDPOrderProof (⇔, 619 ms)

        ↳53 QDP

        ↳54 QDPOrderProof (⇔, 578 ms)

        ↳55 QDP

        ↳56 QDPOrderProof (⇔, 424 ms)

        ↳57 QDP

        ↳58 QDPOrderProof (⇔, 442 ms)

        ↳59 QDP

        ↳60 DependencyGraphProof (⇔, 0 ms)

        ↳61 TRUE

    ↳62 QDP

        ↳63 QDPOrderProof (⇔, 74 ms)

        ↳64 QDP

        ↳65 DependencyGraphProof (⇔, 0 ms)

        ↳66 QDP

        ↳67 UsableRulesProof (⇔, 0 ms)

        ↳68 QDP

        ↳69 QDPSizeChangeProof (⇔, 0 ms)

        ↳70 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(s(0))) → k(0)
k(0) → s(0)
s(s(s(s(0)))) → k(s(0))
k(s(0)) → s(s(0))
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)))))))
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(s(0))) → K(0)
K(0) → S(0)
S(s(s(s(0)))) → K(s(0))
K(s(0)) → S(s(0))
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)))
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(s(0))) → k(0)
k(0) → s(0)
s(s(s(s(0)))) → k(s(0))
k(s(0)) → s(s(0))
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)))))))
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 4 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(s(0)))) → K(s(0))
K(s(0)) → S(s(0))
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(s(0))) → k(0)
k(0) → s(0)
s(s(s(s(0)))) → k(s(0))
k(s(0)) → s(s(0))
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)))))))
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(s(0)))) → K(s(0))
K(s(0)) → S(s(0))
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(s(0))) → k(0)
k(0) → s(0)
s(s(s(s(0)))) → k(s(0))
k(s(0)) → s(s(0))
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)))))))
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(0)) → S(s(0))
S(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(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(s(0))) → k(0)
k(0) → s(0)
s(s(s(s(0)))) → k(s(0))
k(s(0)) → s(s(0))
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)))))))
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 K(s(0)) → S(s(0)) at position [0] we obtained the following new rules [LPAR04]:

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

(11) Obligation:

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

S(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(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(0)) → 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(s(0))) → k(0)
k(0) → s(0)
s(s(s(s(0)))) → k(s(0))
k(s(0)) → s(s(0))
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)))))))
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:

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

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

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

(15) Obligation:

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

K(s(s(0))) → S(s(s(s(s(s(s(0)))))))
K(s(s(0))) → S(s(s(s(s(s(0))))))
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)))
S(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(s(0))) → k(0)
k(0) → s(0)
s(s(s(s(0)))) → k(s(0))
k(s(0)) → s(s(0))
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)))))))
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 1 less node.

(17) Obligation:

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

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)))

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

(43) Obligation:

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

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(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(k(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(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(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(k(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(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(s(h(0, 0))))))
K(s(s(0))) → S(k(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(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(s(h(0, 0)))))
K(s(s(0))) → S(k(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))))
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(s(0))) → k(0)
k(0) → s(0)
s(s(s(s(0)))) → k(s(0))
k(s(0)) → s(s(0))
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)))))))
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 2 less nodes.

(45) Obligation:

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

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

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

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

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

POL(K(x₁)) = 1 +
[ 0, 0 ]

· x₁

POL(s(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 0 0 /

· x₁

POL(0) =
/ 0 \

\ 1 /

POL(S(x₁)) = 0 +
[ 1, 0 ]

· x₁

POL(h(x₁, x₂)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁ +
/ 1 0 \

\ 0 0 /

· x₂

POL(k(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 0 0 /

· x₁

POL(f(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(g(x₁)) =
/ 0 \

\ 0 /

+
/ 1 1 \

\ 0 0 /

· 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(0)) → s(s(0))
s(s(0)) → f(s(0))
s(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(0))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(s(0))) → k(0)
k(0) → s(0)

(47) Obligation:

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

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

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

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

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

POL(S(x₁)) = 1 +
[ 0, 1 ]

· x₁

POL(s(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 1 1 /

· x₁

POL(0) =
/ 1 \

\ 0 /

POL(K(x₁)) = 1 +
[ 1, 1 ]

· x₁

POL(h(x₁, x₂)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 1 /

· x₁ +
/ 0 0 \

\ 0 0 /

· x₂

POL(k(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 1 1 /

· x₁

POL(f(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁

POL(g(x₁)) =
/ 0 \

\ 0 /

+
/ 0 0 \

\ 1 1 /

· 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(0)) → s(s(0))
s(s(0)) → f(s(0))
s(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(0))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(s(0))) → k(0)
k(0) → s(0)

(49) Obligation:

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

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

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

POL(S(x₁)) = 0 +
[ 0, 1 ]

· x₁

POL(s(x₁)) =
/ 1 \

\ 1 /

+
/ 0 0 \

\ 1 0 /

· x₁

POL(0) =
/ 0 \

\ 1 /

POL(K(x₁)) = 0 +
[ 0, 1 ]

· x₁

POL(k(x₁)) =
/ 1 \

\ 1 /

+
/ 0 0 \

\ 1 0 /

· x₁

POL(h(x₁, x₂)) =
/ 1 \

\ 1 /

+
/ 0 0 \

\ 0 0 /

· x₁ +
/ 0 0 \

\ 1 0 /

· x₂

POL(f(x₁)) =
/ 0 \

\ 1 /

+
/ 1 0 \

\ 1 0 /

· x₁

POL(g(x₁)) =
/ 1 \

\ 1 /

+
/ 0 0 \

\ 1 0 /

· 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(0)) → s(s(0))
s(s(0)) → f(s(0))
s(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(0))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(s(0))) → k(0)
k(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(0))))))))) → K(s(s(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(k(s(0))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → S(k(s(0)))
K(s(s(0))) → 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(s(0))) → k(0)
k(0) → s(0)
s(s(s(s(0)))) → k(s(0))
k(s(0)) → s(s(0))
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)))))))
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(k(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, -I, 2A ]

· x₁

POL(s(x₁)) =
/ 1A \

| 1A |

\ 1A /

+
/ -I -I 0A \

| 0A -I 0A |

\ 0A 0A -I /

· x₁

POL(0) =
/ 1A \

| 2A |

\ 0A /

POL(K(x₁)) = 1A +
[ -I, -I, 3A ]

· x₁

POL(k(x₁)) =
/ 1A \

| 1A |

\ -I /

+
/ 0A -I 0A \

| 0A 0A 0A |

\ 0A 0A -I /

· x₁

POL(h(x₁, x₂)) =
/ 1A \

| 1A |

\ 0A /

+
/ -I -I 0A \

| -I -I 0A |

\ 0A -I -I /

· x₁ +
/ -I -I 0A \

| -I -I 0A |

\ -I -I -I /

· x₂

POL(f(x₁)) =
/ 1A \

| -I |

\ -I /

+
/ 0A 0A 0A \

| -I 0A 0A |

\ 0A 0A -I /

· x₁

POL(g(x₁)) =
/ 1A \

| 1A |

\ 0A /

+
/ 0A -I 0A \

| -I 0A 0A |

\ 0A -I 0A /

· 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(0)) → s(s(0))
s(s(0)) → f(s(0))
s(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(0))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(s(0))) → k(0)
k(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(0))))))))) → K(s(s(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(k(s(0))))
K(s(s(0))) → S(s(s(k(0))))
K(s(s(0))) → 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(s(0))) → k(0)
k(0) → s(0)
s(s(s(s(0)))) → k(s(0))
k(s(0)) → s(s(0))
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)))))))
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(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 +
[ 0A, -I, 3A ]

· x₁

POL(s(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ 0A -I 0A \

| -I -I -I |

\ 0A 1A -I /

· x₁

POL(0) =
/ 0A \

| 0A |

\ -I /

POL(K(x₁)) = 1A +
[ 3A, 3A, 3A ]

· x₁

POL(k(x₁)) =
/ 0A \

| -I |

\ -I /

+
/ 0A -I 0A \

| -I -I -I |

\ 0A 1A 0A /

· x₁

POL(h(x₁, x₂)) =
/ -I \

| -I |

\ 0A /

+
/ 0A -I -I \

| -I -I -I |

\ -I -I -I /

· x₁ +
/ 0A -I -I \

| -I -I -I |

\ 0A -I -I /

· x₂

POL(f(x₁)) =
/ 0A \

| -I |

\ 0A /

+
/ 0A 1A -I \

| -I -I -I |

\ 0A -I -I /

· x₁

POL(g(x₁)) =
/ -I \

| -I |

\ 0A /

+
/ 0A 1A 0A \

| -I -I -I |

\ 0A -I 0A /

· 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(0)) → s(s(0))
s(s(0)) → f(s(0))
s(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(0))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(s(0))) → k(0)
k(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(0))))))))) → K(s(s(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(k(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(s(0))) → k(0)
k(0) → s(0)
s(s(s(s(0)))) → k(s(0))
k(s(0)) → s(s(0))
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)))))))
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(k(s(0))))
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)) = 0A +
[ 0A, 1A, -I ]

· x₁

POL(s(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I 0A 0A \

| -I 0A -I |

\ 0A 0A -I /

· x₁

POL(0) =
/ 2A \

| 0A |

\ 0A /

POL(K(x₁)) = 1A +
[ 0A, -I, -I ]

· x₁

POL(k(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I 0A 0A \

| -I -I -I |

\ 0A -I -I /

· x₁

POL(h(x₁, x₂)) =
/ 0A \

| 0A |

\ -I /

+
/ -I -I 0A \

| -I 0A -I |

\ -I 0A -I /

· x₁ +
/ -I -I 0A \

| -I -I -I |

\ -I 0A -I /

· x₂

POL(f(x₁)) =
/ 0A \

| 0A |

\ -I /

+
/ -I 0A -I \

| -I 0A -I |

\ 0A 0A -I /

· x₁

POL(g(x₁)) =
/ 0A \

| 0A |

\ -I /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ 0A 0A 0A /

· 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(0)) → s(s(0))
s(s(0)) → f(s(0))
s(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(0))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(s(0))) → k(0)
k(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(0))))))))) → K(s(s(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(s(0))) → k(0)
k(0) → s(0)
s(s(s(s(0)))) → k(s(0))
k(s(0)) → s(s(0))
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)))))))
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.

S(s(s(s(s(s(s(s(s(0))))))))) → K(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, -I, 0A ]

· x₁

POL(s(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I 0A -I \

| 0A -I 0A |

\ 0A -I -I /

· x₁

POL(0) =
/ 0A \

| 0A |

\ 1A /

POL(K(x₁)) = 0A +
[ -I, -I, 0A ]

· x₁

POL(k(x₁)) =
/ 0A \

| 0A |

\ 0A /

+
/ 0A 0A -I \

| 0A -I 0A |

\ 0A -I -I /

· x₁

POL(h(x₁, x₂)) =
/ 0A \

| -I |

\ 0A /

+
/ -I -I -I \

| 0A -I -I |

\ -I -I -I /

· x₁ +
/ -I 0A -I \

| 0A -I -I |

\ 0A -I -I /

· x₂

POL(f(x₁)) =
/ 0A \

| 0A |

\ -I /

+
/ -I -I 0A \

| -I -I 0A |

\ -I -I 0A /

· x₁

POL(g(x₁)) =
/ 1A \

| 1A |

\ 1A /

+
/ 0A 0A 0A \

| 0A 0A 0A |

\ 0A 0A 0A /

· 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(0)) → s(s(0))
s(s(0)) → f(s(0))
s(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(0))))))))) → k(s(s(0)))
g(s(x)) → s(s(g(x)))
s(s(s(0))) → k(0)
k(0) → s(0)

(59) Obligation:

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

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

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

(60) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(61) TRUE

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

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

(63) 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, -1}

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

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

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

(65) DependencyGraphProof (EQUIVALENT transformation)

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

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

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

(67) 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.

(68) 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.

(69) 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

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Obligation:

(5) TransformationProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Complex Obligation (AND)

(9) Obligation:

(10) TransformationProof (EQUIVALENT transformation)

(11) Obligation:

(12) DependencyGraphProof (EQUIVALENT transformation)

(13) Obligation:

(14) TransformationProof (EQUIVALENT transformation)

(15) Obligation:

(16) DependencyGraphProof (EQUIVALENT transformation)

(17) Obligation:

(18) TransformationProof (EQUIVALENT transformation)

(19) Obligation:

(20) DependencyGraphProof (EQUIVALENT transformation)

(21) Obligation:

(22) TransformationProof (EQUIVALENT transformation)

(23) Obligation:

(24) DependencyGraphProof (EQUIVALENT transformation)

(25) Obligation:

(26) TransformationProof (EQUIVALENT transformation)

(27) Obligation:

(28) DependencyGraphProof (EQUIVALENT transformation)

(29) Obligation:

(30) TransformationProof (EQUIVALENT transformation)

(31) Obligation:

(32) DependencyGraphProof (EQUIVALENT transformation)

(33) Obligation:

(34) TransformationProof (EQUIVALENT transformation)

(35) Obligation:

(36) DependencyGraphProof (EQUIVALENT transformation)

(37) Obligation:

(38) TransformationProof (EQUIVALENT transformation)

(39) Obligation:

(40) DependencyGraphProof (EQUIVALENT transformation)

(41) Obligation:

(42) TransformationProof (EQUIVALENT transformation)

(43) Obligation:

(44) DependencyGraphProof (EQUIVALENT transformation)

(45) Obligation:

(46) QDPOrderProof (EQUIVALENT transformation)

(47) Obligation:

(48) QDPOrderProof (EQUIVALENT transformation)

(49) Obligation:

(50) QDPOrderProof (EQUIVALENT transformation)

(51) Obligation:

(52) QDPOrderProof (EQUIVALENT transformation)

(53) Obligation:

(54) QDPOrderProof (EQUIVALENT transformation)

(55) Obligation:

(56) QDPOrderProof (EQUIVALENT transformation)

(57) Obligation:

(58) QDPOrderProof (EQUIVALENT transformation)

(59) Obligation:

(60) DependencyGraphProof (EQUIVALENT transformation)

(61) TRUE

(62) Obligation:

(63) QDPOrderProof (EQUIVALENT transformation)

(64) Obligation:

(65) DependencyGraphProof (EQUIVALENT transformation)

(66) Obligation:

(67) UsableRulesProof (EQUIVALENT transformation)

(68) Obligation:

(69) QDPSizeChangeProof (EQUIVALENT transformation)

(70) YES