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(a(b(x1))) → c(x1)
a(c(x1)) → b(c(a(a(x1))))
b(c(x1)) → x1
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(a(b(x1))) → c(x1)
a(c(x1)) → b(c(a(a(x1))))
b(c(x1)) → x1
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
A(c(x1)) → B(c(a(a(x1))))
A(c(x1)) → A(x1)
A(c(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(a(b(x1))) → c(x1)
a(c(x1)) → b(c(a(a(x1))))
b(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A(c(x1)) → B(c(a(a(x1))))
A(c(x1)) → A(x1)
A(c(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(a(b(x1))) → c(x1)
a(c(x1)) → b(c(a(a(x1))))
b(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
A(c(x1)) → A(x1)
A(c(x1)) → A(a(x1))
The TRS R consists of the following rules:
a(a(b(x1))) → c(x1)
a(c(x1)) → b(c(a(a(x1))))
b(c(x1)) → x1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
A(c(x1)) → A(x1)
A(c(x1)) → A(a(x1))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:
POL(c(x1)) = 1/4 + (4)x_1
POL(a(x1)) = (4)x_1
POL(A(x1)) = (4)x_1
POL(b(x1)) = 1/4 + (1/4)x_1
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:
a(a(b(x1))) → c(x1)
a(c(x1)) → b(c(a(a(x1))))
b(c(x1)) → x1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(a(b(x1))) → c(x1)
a(c(x1)) → b(c(a(a(x1))))
b(c(x1)) → x1
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.