Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))
Q is empty.
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))
Q is empty.
We have applied [15,7] to switch to innermost. The TRS R 1 is
f(h(x)) → h(g(x))
f(g(x)) → g(f(f(x)))
The TRS R 2 is
f'(s(x), y, y) → f'(y, x, s(x))
The signature Sigma is {f'}
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))
The set Q consists of the following terms:
f(g(x0))
f(h(x0))
f'(s(x0), x1, x1)
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(g(x)) → F(f(x))
F'(s(x), y, y) → F'(y, x, s(x))
F(g(x)) → F(x)
The TRS R consists of the following rules:
f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))
The set Q consists of the following terms:
f(g(x0))
f(h(x0))
f'(s(x0), x1, x1)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
F(g(x)) → F(f(x))
F'(s(x), y, y) → F'(y, x, s(x))
F(g(x)) → F(x)
The TRS R consists of the following rules:
f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))
The set Q consists of the following terms:
f(g(x0))
f(h(x0))
f'(s(x0), x1, x1)
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(g(x)) → F(f(x))
F'(s(x), y, y) → F'(y, x, s(x))
F(g(x)) → F(x)
The TRS R consists of the following rules:
f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))
The set Q consists of the following terms:
f(g(x0))
f(h(x0))
f'(s(x0), x1, x1)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F(g(x)) → F(f(x))
F(g(x)) → F(x)
The TRS R consists of the following rules:
f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))
The set Q consists of the following terms:
f(g(x0))
f(h(x0))
f'(s(x0), x1, x1)
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].
The following pairs can be oriented strictly and are deleted.
F(g(x)) → F(f(x))
F(g(x)) → F(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1) = x1
g(x1) = g(x1)
f(x1) = x1
h(x1) = h
Recursive path order with status [2].
Precedence:
trivial
Status:
trivial
The following usable rules [14] were oriented:
f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(g(x)) → g(f(f(x)))
f(h(x)) → h(g(x))
f'(s(x), y, y) → f'(y, x, s(x))
The set Q consists of the following terms:
f(g(x0))
f(h(x0))
f'(s(x0), x1, x1)
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.