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(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(y))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F1(s1(0)) -> F1(0)
F1(+2(x, y)) -> F1(y)
F1(+2(x, y)) -> F1(x)
F1(+2(x, s1(0))) -> F1(x)
The TRS R consists of the following rules:
f1(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(y))
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(s1(0)) -> F1(0)
F1(+2(x, y)) -> F1(y)
F1(+2(x, y)) -> F1(x)
F1(+2(x, s1(0))) -> F1(x)
The TRS R consists of the following rules:
f1(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
F1(+2(x, y)) -> F1(y)
F1(+2(x, y)) -> F1(x)
F1(+2(x, s1(0))) -> F1(x)
The TRS R consists of the following rules:
f1(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
F1(+2(x, y)) -> F1(y)
F1(+2(x, y)) -> F1(x)
F1(+2(x, s1(0))) -> F1(x)
Used argument filtering: F1(x1) = x1
+2(x1, x2) = +2(x1, x2)
s1(x1) = s
0 = 0
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(0) -> s1(0)
f1(s1(0)) -> s1(s1(0))
f1(s1(0)) -> *2(s1(s1(0)), f1(0))
f1(+2(x, s1(0))) -> +2(s1(s1(0)), f1(x))
f1(+2(x, y)) -> *2(f1(x), f1(y))
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.