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.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
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
↳ QDPOrderProof
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.
We use the reduction pair processor [13].
The following pairs can be strictly oriented and are deleted.
F3(x, y, s1(z)) -> F3(0, 1, z)
The remaining pairs can at least by weakly be oriented.
F3(0, 1, x) -> F3(s1(x), x, x)
Used ordering: Combined order from the following AFS and order.
F3(x1, x2, x3) = x3
s1(x1) = s1(x1)
0 = 0
1 = 1
Lexicographic Path Order [19].
Precedence:
0 > [s1, 1]
The following usable rules [14] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ 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 [13,14,18] contains 0 SCCs with 1 less node.