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:
p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))
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:
P(a(x0), p(a(b(x1)), x2)) → P(a(a(x1)), x2)
P(a(x0), p(a(b(x1)), x2)) → P(a(b(a(x2))), p(a(a(x1)), x2))
The TRS R consists of the following rules:
p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
P(a(x0), p(a(b(x1)), x2)) → P(a(a(x1)), x2)
P(a(x0), p(a(b(x1)), x2)) → P(a(b(a(x2))), p(a(a(x1)), x2))
The TRS R consists of the following rules:
p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))
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.
P(a(x0), p(a(b(x1)), x2)) → P(a(a(x1)), x2)
P(a(x0), p(a(b(x1)), x2)) → P(a(b(a(x2))), p(a(a(x1)), x2))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
P(x1, x2) = P(x1, x2)
a(x1) = x1
p(x1, x2) = p(x1, x2)
b(x1) = b(x1)
Recursive Path Order [2].
Precedence:
p2 > b1 > P2
The following usable rules [14] were oriented:
p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
p(a(x0), p(a(b(x1)), x2)) → p(a(b(a(x2))), p(a(a(x1)), x2))
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.