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:
f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(0)) -> g1(d1(1))
g1(c1(1)) -> g1(d1(0))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(0)) -> g1(d1(1))
g1(c1(1)) -> g1(d1(0))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F1(f1(x)) -> F1(c1(f1(x)))
G1(c1(1)) -> G1(d1(0))
F1(f1(x)) -> F1(d1(f1(x)))
G1(c1(0)) -> G1(d1(1))
The TRS R consists of the following rules:
f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(0)) -> g1(d1(1))
g1(c1(1)) -> g1(d1(0))
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:
F1(f1(x)) -> F1(c1(f1(x)))
G1(c1(1)) -> G1(d1(0))
F1(f1(x)) -> F1(d1(f1(x)))
G1(c1(0)) -> G1(d1(1))
The TRS R consists of the following rules:
f1(f1(x)) -> f1(c1(f1(x)))
f1(f1(x)) -> f1(d1(f1(x)))
g1(c1(x)) -> x
g1(d1(x)) -> x
g1(c1(0)) -> g1(d1(1))
g1(c1(1)) -> g1(d1(0))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 4 less nodes.