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:
f3(0, 1, x) -> f3(s1(x), x, x)
f3(x, y, s1(z)) -> s1(f3(0, 1, z))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f3(0, 1, x) -> f3(s1(x), x, x)
f3(x, y, s1(z)) -> s1(f3(0, 1, z))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F3(x, y, s1(z)) -> F3(0, 1, z)
F3(0, 1, x) -> F3(s1(x), x, x)
The TRS R consists of the following rules:
f3(0, 1, x) -> f3(s1(x), x, x)
f3(x, y, s1(z)) -> s1(f3(0, 1, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
F3(x, y, s1(z)) -> F3(0, 1, z)
F3(0, 1, x) -> F3(s1(x), x, x)
The TRS R consists of the following rules:
f3(0, 1, x) -> f3(s1(x), x, x)
f3(x, y, s1(z)) -> s1(f3(0, 1, z))
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.
F3(x, y, s1(z)) -> F3(0, 1, z)
Used argument filtering: F3(x1, x2, x3) = x3
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F3(0, 1, x) -> F3(s1(x), x, x)
The TRS R consists of the following rules:
f3(0, 1, x) -> f3(s1(x), x, x)
f3(x, y, s1(z)) -> s1(f3(0, 1, z))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 1 less node.