Termination w.r.t. Q of the following Term Rewriting System could be disproven:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(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:
F(g(x), x, y) → F(y, y, g(y))
F(g(x), x, y) → G(y)
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(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:
F(g(x), x, y) → F(y, y, g(y))
F(g(x), x, y) → G(y)
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
F(g(x), x, y) → F(y, y, g(y))
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
The TRS P consists of the following rules:
F(g(x), x, y) → F(y, y, g(y))
The TRS R consists of the following rules:
f(g(x), x, y) → f(y, y, g(y))
g(g(x)) → g(x)
s = F(g(g(x')), g(g(x')), y) evaluates to t =F(y, y, g(y))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [y / g(g(x''))]
- Matcher: [x' / x'', x'' / g(x'')]
Rewriting sequence
F(g(g(x')), g(g(x')), g(g(x''))) → F(g(g(x')), g(x'), g(g(x'')))
with rule g(g(x''')) → g(x''') at position [1] and matcher [x''' / x']
F(g(g(x')), g(x'), g(g(x''))) → F(g(g(x'')), g(g(x'')), g(g(g(x''))))
with rule F(g(x), x, y) → F(y, y, g(y))
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.