(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:
- Semiunifier: [x' / g(0, 1), y / 1, x / 0]
- Matcher: [ ]
Rewriting sequenceF(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) FALSE