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:
g(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))
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:
G(x, s(y)) → G(f(x, y), 0)
G(f(x, y), 0) → G(y, 0)
G(s(x), y) → G(f(x, y), 0)
G(f(x, y), 0) → G(x, 0)
The TRS R consists of the following rules:
g(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
G(x, s(y)) → G(f(x, y), 0)
G(f(x, y), 0) → G(y, 0)
G(s(x), y) → G(f(x, y), 0)
G(f(x, y), 0) → G(x, 0)
The TRS R consists of the following rules:
g(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
G(x, s(y)) → G(f(x, y), 0)
G(f(x, y), 0) → G(y, 0)
G(s(x), y) → G(f(x, y), 0)
G(f(x, y), 0) → G(x, 0)
The TRS R consists of the following rules:
g(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G(f(x, y), 0) → G(y, 0)
G(s(x), y) → G(f(x, y), 0)
G(f(x, y), 0) → G(x, 0)
The TRS R consists of the following rules:
g(0, f(x, x)) → x
g(x, s(y)) → g(f(x, y), 0)
g(s(x), y) → g(f(x, y), 0)
g(f(x, y), 0) → f(g(x, 0), g(y, 0))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.