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