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(0, 1, X) → f(X, X, X)
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(0, 1, X) → f(X, X, X)
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(0, 1, X) → F(X, X, X)
The TRS R consists of the following rules:
f(0, 1, X) → f(X, X, X)
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(0, 1, X) → F(X, X, X)
The TRS R consists of the following rules:
f(0, 1, X) → f(X, X, X)
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(0, 1, X) → F(X, X, X)
The TRS R consists of the following rules:
f(0, 1, X) → f(X, X, X)
g(X, Y) → X
g(X, Y) → Y
s = F(g(0, Y), g(X', 1), X) evaluates to t =F(X, X, X)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [X' / 0, Y / 1, X / g(0, 1)]
- Matcher: [ ]
Rewriting sequence
F(g(0, 1), g(0, 1), g(0, 1)) → F(g(0, 1), 1, g(0, 1))
with rule g(X', Y') → Y' at position [1] and matcher [Y' / 1, X' / 0]
F(g(0, 1), 1, g(0, 1)) → F(0, 1, g(0, 1))
with rule g(X', Y) → X' at position [0] and matcher [X' / 0, Y / 1]
F(0, 1, g(0, 1)) → F(g(0, 1), g(0, 1), g(0, 1))
with rule F(0, 1, X) → F(X, X, X)
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.