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:
a(f, a(f, x)) → a(x, x)
a(h, x) → a(f, a(g, a(f, x)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(f, a(f, x)) → a(x, x)
a(h, x) → a(f, a(g, a(f, 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:
A(f, a(f, x)) → A(x, x)
A(h, x) → A(f, x)
A(h, x) → A(f, a(g, a(f, x)))
A(h, x) → A(g, a(f, x))
The TRS R consists of the following rules:
a(f, a(f, x)) → a(x, x)
a(h, x) → a(f, a(g, a(f, 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:
A(f, a(f, x)) → A(x, x)
A(h, x) → A(f, x)
A(h, x) → A(f, a(g, a(f, x)))
A(h, x) → A(g, a(f, x))
The TRS R consists of the following rules:
a(f, a(f, x)) → a(x, x)
a(h, x) → a(f, a(g, a(f, 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:
A(f, a(f, x)) → A(x, x)
A(h, x) → A(f, a(g, a(f, x)))
A(h, x) → A(f, x)
A(h, x) → A(g, a(f, x))
The TRS R consists of the following rules:
a(f, a(f, x)) → a(x, x)
a(h, x) → a(f, a(g, a(f, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
A(f, a(f, x)) → A(x, x)
A(h, x) → A(f, x)
The TRS R consists of the following rules:
a(f, a(f, x)) → a(x, x)
a(h, x) → a(f, a(g, a(f, 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.
A(f, a(f, x)) → A(x, x)
A(h, x) → A(f, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
A(x1, x2) = A(x1, x2)
f = f
a(x1, x2) = a(x1, x2)
h = h
Lexicographic Path Order [19].
Precedence:
a2 > A2
h > f > A2
The following usable rules [14] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(f, a(f, x)) → a(x, x)
a(h, x) → a(f, a(g, a(f, 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.