Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)
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:
F(t, x, y) → G(x, y)
G(s(x), s(y)) → G(x, y)
F(t, x, y) → F(g(x, y), x, s(y))
The TRS R consists of the following rules:
f(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, 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:
F(t, x, y) → G(x, y)
G(s(x), s(y)) → G(x, y)
F(t, x, y) → F(g(x, y), x, s(y))
The TRS R consists of the following rules:
f(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G(s(x), s(y)) → G(x, y)
The TRS R consists of the following rules:
f(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)
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.
G(s(x), s(y)) → G(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:
POL(s(x1)) = 1 + x_1
POL(G(x1, x2)) = x_2
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, 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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(t, x, y) → F(g(x, y), x, s(y))
The TRS R consists of the following rules:
f(t, x, y) → f(g(x, y), x, s(y))
g(s(x), 0) → t
g(s(x), s(y)) → g(x, y)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.