(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

H(x, y) → F(x, y, x)
F(0, 1, x) → H(x, 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.

(3) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

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:
  • Matcher: [ ]
  • Semiunifier: [y / 1, x / 0, x' / g(0, 1)]




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 [x / 0, y' / 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 [x / 0, y / 1]

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

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



(4) NO