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:
f(x, 0) → s(0)
f(s(x), s(y)) → s(f(x, y))
g(0, x) → g(f(x, x), x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, 0) → s(0)
f(s(x), s(y)) → s(f(x, y))
g(0, x) → g(f(x, x), x)
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:
G(0, x) → F(x, x)
G(0, x) → G(f(x, x), x)
F(s(x), s(y)) → F(x, y)
The TRS R consists of the following rules:
f(x, 0) → s(0)
f(s(x), s(y)) → s(f(x, y))
g(0, x) → g(f(x, x), x)
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:
G(0, x) → F(x, x)
G(0, x) → G(f(x, x), x)
F(s(x), s(y)) → F(x, y)
The TRS R consists of the following rules:
f(x, 0) → s(0)
f(s(x), s(y)) → s(f(x, y))
g(0, x) → g(f(x, x), x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F(s(x), s(y)) → F(x, y)
The TRS R consists of the following rules:
f(x, 0) → s(0)
f(s(x), s(y)) → s(f(x, y))
g(0, x) → g(f(x, x), x)
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.
F(s(x), s(y)) → F(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:
POL(s(x1)) = 1/2 + (13/4)x_1
POL(F(x1, x2)) = (15/4)x_2
The value of delta used in the strict ordering is 15/8.
The following usable rules [17] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(x, 0) → s(0)
f(s(x), s(y)) → s(f(x, y))
g(0, x) → g(f(x, x), x)
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.