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(x1) → b(b(x1))
b(b(b(x1))) → a(x1)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
a(x1) → b(b(x1))
b(b(b(x1))) → a(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(x1) → B(b(x1))
A(x1) → B(x1)
B(b(b(x1))) → A(x1)
The TRS R consists of the following rules:
a(x1) → b(b(x1))
b(b(b(x1))) → a(x1)
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:
A(x1) → B(b(x1))
A(x1) → B(x1)
B(b(b(x1))) → A(x1)
The TRS R consists of the following rules:
a(x1) → b(b(x1))
b(b(b(x1))) → a(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(x1) → B(x1)
B(b(b(x1))) → A(x1)
The remaining pairs can at least be oriented weakly.
A(x1) → B(b(x1))
Used ordering: Polynomial interpretation [25,35]:
POL(B(x1)) = x_1
POL(a(x1)) = 4 + x_1
POL(A(x1)) = 2 + x_1
POL(b(x1)) = 2 + x_1
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:
a(x1) → b(b(x1))
b(b(b(x1))) → a(x1)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
A(x1) → B(b(x1))
The TRS R consists of the following rules:
a(x1) → b(b(x1))
b(b(b(x1))) → a(x1)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.