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(b(x1)) → b(a(x1))
b(a(x1)) → a(c(b(x1)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x1)) → b(a(x1))
b(a(x1)) → a(c(b(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(b(x1)) → B(a(x1))
B(a(x1)) → A(c(b(x1)))
B(a(x1)) → B(x1)
A(b(x1)) → A(x1)
The TRS R consists of the following rules:
a(b(x1)) → b(a(x1))
b(a(x1)) → a(c(b(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(b(x1)) → B(a(x1))
B(a(x1)) → A(c(b(x1)))
B(a(x1)) → B(x1)
A(b(x1)) → A(x1)
The TRS R consists of the following rules:
a(b(x1)) → b(a(x1))
b(a(x1)) → a(c(b(x1)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
B(a(x1)) → B(x1)
The TRS R consists of the following rules:
a(b(x1)) → b(a(x1))
b(a(x1)) → a(c(b(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.
B(a(x1)) → B(x1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:
POL(B(x1)) = (1/4)x_1
POL(a(x1)) = 1/4 + (2)x_1
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(b(x1)) → b(a(x1))
b(a(x1)) → a(c(b(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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
A(b(x1)) → A(x1)
The TRS R consists of the following rules:
a(b(x1)) → b(a(x1))
b(a(x1)) → a(c(b(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(b(x1)) → A(x1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:
POL(A(x1)) = (1/4)x_1
POL(b(x1)) = 1/4 + (2)x_1
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(b(x1)) → b(a(x1))
b(a(x1)) → a(c(b(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.