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(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → x
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
G(h(x)) → G(x)
F(a, h(x)) → G(x)
H(g(x)) → H(a)
F(a, h(x)) → F(g(x), h(x))
The TRS R consists of the following rules:
f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
G(h(x)) → G(x)
F(a, h(x)) → G(x)
H(g(x)) → H(a)
F(a, h(x)) → F(g(x), h(x))
The TRS R consists of the following rules:
f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(a, h(x)) → G(x)
G(h(x)) → G(x)
F(a, h(x)) → F(g(x), h(x))
H(g(x)) → H(a)
The TRS R consists of the following rules:
f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G(h(x)) → G(x)
The TRS R consists of the following rules:
f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → x
Q is empty.
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.
G(h(x)) → G(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
G(x1) = G(x1)
h(x1) = h(x1)
Lexicographic Path Order [19].
Precedence:
h1 > G1
The following usable rules [14] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F(a, h(x)) → F(g(x), h(x))
The TRS R consists of the following rules:
f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → x
Q is empty.
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(a, h(x)) → F(g(x), h(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2) = x1
a = a
h(x1) = x1
g(x1) = g
Lexicographic Path Order [19].
Precedence:
a > g
The following usable rules [14] were oriented:
g(h(x)) → g(x)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(a, h(x)) → f(g(x), h(x))
h(g(x)) → h(a)
g(h(x)) → g(x)
h(h(x)) → x
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.