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:



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.