(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(0, 1, x) → f(x, x, x)
f(x, y, z) → 2
02
12
g(x, x, y) → y
g(x, y, y) → x

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:

F(0, 1, x) → F(x, x, x)

The TRS R consists of the following rules:

f(0, 1, x) → f(x, x, x)
f(x, y, z) → 2
02
12
g(x, x, y) → y
g(x, y, y) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) NonTerminationLoopProof (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(1, 0, 0), g(1, y, y), 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 / 0, x / g(1, 0, 0)]




Rewriting sequence

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

F(g(1, 0, 0), 1, g(1, 0, 0))F(g(1, 2, 0), 1, g(1, 0, 0))
with rule 02 at position [0,1] and matcher [ ]

F(g(1, 2, 0), 1, g(1, 0, 0))F(g(2, 2, 0), 1, g(1, 0, 0))
with rule 12 at position [0,0] and matcher [ ]

F(g(2, 2, 0), 1, g(1, 0, 0))F(0, 1, g(1, 0, 0))
with rule g(x', x', y) → y at position [0] and matcher [x' / 2, y / 0]

F(0, 1, g(1, 0, 0))F(g(1, 0, 0), g(1, 0, 0), g(1, 0, 0))
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.



(4) NO