YES Termination w.r.t. Q proof of MNZ_10_2.ari

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

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

(15) Obligation:

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

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(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(0)))))))) → k(s(s(0)))
k(s(s(0))) → 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(0)))))
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(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(0)))))))) → k(s(s(0)))
k(s(s(0))) → 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(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)))))

(19) Obligation:

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

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(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(0)))))))) → k(s(s(0)))
k(s(s(0))) → 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(0))))
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(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(0)))))))) → k(s(s(0)))
k(s(s(0))) → 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(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))))

(23) Obligation:

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

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

(27) Obligation:

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

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

(31) Obligation:

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

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

(33) Obligation:

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

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

K(s(s(0))) → S(0) → K(s(s(0))) → S(0)

(35) Obligation:

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

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

(37) Obligation:

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

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

(39) Obligation:

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

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

(41) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → K(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))))))
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(0)))))))) → k(s(s(0)))
k(s(s(0))) → 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) 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 +
[0,1]
·x1

POL(s(x1)) =
/1\
\1/
+
/00\
\10/
·x1

POL(0) =
/0\
\0/

POL(K(x1)) = 1 +
[0,1]
·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(0))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
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)

(43) Obligation:

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

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

POL(S(x1)) = -I +
[0A,-I,-I]
·x1

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

POL(0) =
/2A\
|-I|
\-I/

POL(K(x1)) = 1A +
[0A,-I,-I]
·x1

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

POL(k(x1)) =
/2A\
|2A|
\-I/
+
/-I-I0A\
|-I-I-I|
\-I-I-I/
·x1

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

POL(g(x1)) =
/3A\
|3A|
\2A/
+
/0A-I0A\
|1A-I0A|
\-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(0))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
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)

(45) Obligation:

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

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(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(0)))))))) → k(s(s(0)))
k(s(s(0))) → 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(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)) = 3A +
[1A,-I,0A]
·x1

POL(s(x1)) =
/0A\
|-I|
\1A/
+
/0A2A-I\
|-I-I1A|
\-I-I0A/
·x1

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

POL(K(x1)) = 1A +
[3A,3A,-I]
·x1

POL(k(x1)) =
/0A\
|-I|
\1A/
+
/0A2A-I\
|-I0A-I|
\-I-I0A/
·x1

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

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

POL(g(x1)) =
/0A\
|1A|
\0A/
+
/0A2A3A\
|-I-I1A|
\-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(0))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
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)

(47) Obligation:

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

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(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(0)))))))) → k(s(s(0)))
k(s(s(0))) → 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.


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(s(k(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\
\2/
+
/01\
\10/
·x1

POL(0) =
/0\
\0/

POL(K(x1)) = 2 +
[0,3]
·x1

POL(k(x1)) =
/2\
\2/
+
/20\
\30/
·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\
\22/
·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(0))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
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)

(49) 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(0)))))))) → k(s(s(0)))
k(s(s(0))) → 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) PisEmptyProof (EQUIVALENT transformation)

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

(51) YES

(52) 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(0)))))))) → k(s(s(0)))
k(s(s(0))) → 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.

(53) 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) ) = x1

POL( G(x1) ) = 0

POL( f(x1) ) = 2x1

POL( g(x1) ) = x1 + 2

POL( h(x1, x2) ) = 0

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

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

POL( 0 ) = 0

POL( H(x1, x2) ) = 0


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(0))))))
s(s(0)) → f(s(0))
s(s(s(0))) → k(s(0))
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)

(54) 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(0)))))))) → k(s(s(0)))
k(s(s(0))) → 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.

(55) DependencyGraphProof (EQUIVALENT transformation)

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

(56) 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(0)))))))) → k(s(s(0)))
k(s(s(0))) → 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.

(57) 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.

(58) 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.

(59) 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

(60) YES