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:

h(x, y) → f(x, y, x)
f(0, 1, x) → h(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:

h(x, y) → f(x, y, x)
f(0, 1, x) → h(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) → H(x, x)
H(x, y) → F(x, y, x)

The TRS R consists of the following rules:

h(x, y) → f(x, y, x)
f(0, 1, x) → h(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) → H(x, x)
H(x, y) → F(x, y, x)

The TRS R consists of the following rules:

h(x, y) → f(x, y, x)
f(0, 1, x) → h(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) → H(x, x)
H(x, y) → F(x, y, x)

The TRS R consists of the following rules:

h(x, y) → f(x, y, x)
f(0, 1, x) → h(x, x)
g(x, y) → x
g(x, y) → y


s = H(g(0, y), g(x, 1)) evaluates to t =H(g(0, y), g(0, y))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

H(g(0, 1), g(0, 1)) → H(g(0, 1), 1)
with rule g(x, y') → y' at position [1] and matcher [y' / 1, x / 0]

H(g(0, 1), 1) → F(g(0, 1), 1, g(0, 1))
with rule H(x', y') → F(x', y', x') at position [] and matcher [y' / 1, x' / g(0, 1)]

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 [y / 1, x' / 0]

F(0, 1, g(0, 1)) → H(g(0, 1), g(0, 1))
with rule F(0, 1, x) → H(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.