NO
Termination w.r.t. Q proof of ../tpdb/TRS/nontermin/AG01/#4.3.trs
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, y), x, z) → f(z, z, z)
g(x, y) → x
g(x, y) → y
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(x, y), x, z) → f(z, z, z)
g(x, y) → x
g(x, y) → 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(g(x, y), x, z) → F(z, z, z)
The TRS R consists of the following rules:
f(g(x, y), x, z) → f(z, z, z)
g(x, y) → x
g(x, y) → y
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
F(g(x, y), x, z) → F(z, z, z)
The TRS R consists of the following rules:
f(g(x, y), x, z) → f(z, z, z)
g(x, y) → x
g(x, y) → y
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, y), x, z) → F(z, z, z)
The TRS R consists of the following rules:
f(g(x, y), x, z) → f(z, z, z)
g(x, y) → x
g(x, y) → y
s = F(g(x', y), g(x', y'), z) evaluates to t =F(z, z, z)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [y' / y, z / g(x', y)]
- Matcher: [ ]
Rewriting sequence
F(g(x', y), g(x', y), g(x', y)) → F(g(x', y), x', g(x', y))
with rule g(x'', y') → x'' at position [1] and matcher [y' / y, x'' / x']
F(g(x', y), x', g(x', y)) → F(g(x', y), g(x', y), g(x', y))
with rule F(g(x, y), x, z) → F(z, z, z)
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.