(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0, 1, x) → f(h(x), h(x), x)
h(0) → 0
h(g(x, y)) → y
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is
h(0) → 0
h(g(x, y)) → y
The TRS R 2 is
f(0, 1, x) → f(h(x), h(x), x)
The signature Sigma is {
f}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(0, 1, x) → f(h(x), h(x), x)
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
f(0, 1, x0)
h(0)
h(g(x0, x1))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, x) → F(h(x), h(x), x)
F(0, 1, x) → H(x)
The TRS R consists of the following rules:
f(0, 1, x) → f(h(x), h(x), x)
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
f(0, 1, x0)
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, x) → F(h(x), h(x), x)
The TRS R consists of the following rules:
f(0, 1, x) → f(h(x), h(x), x)
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
f(0, 1, x0)
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(7) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, x) → F(h(x), h(x), x)
The TRS R consists of the following rules:
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
f(0, 1, x0)
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(9) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
f(0, 1, x0)
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, x) → F(h(x), h(x), x)
The TRS R consists of the following rules:
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(11) Narrowing (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
F(
0,
1,
x) →
F(
h(
x),
h(
x),
x) at position [0] we obtained the following new rules [LPAR04]:
F(0, 1, 0) → F(0, h(0), 0)
F(0, 1, g(x0, x1)) → F(x1, h(g(x0, x1)), g(x0, x1))
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, 0) → F(0, h(0), 0)
F(0, 1, g(x0, x1)) → F(x1, h(g(x0, x1)), g(x0, x1))
The TRS R consists of the following rules:
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(13) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, g(x0, x1)) → F(x1, h(g(x0, x1)), g(x0, x1))
The TRS R consists of the following rules:
h(0) → 0
h(g(x, y)) → y
The set Q consists of the following terms:
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(15) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, g(x0, x1)) → F(x1, h(g(x0, x1)), g(x0, x1))
The TRS R consists of the following rules:
h(g(x, y)) → y
The set Q consists of the following terms:
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(17) Rewriting (EQUIVALENT transformation)
By rewriting [LPAR04] the rule
F(
0,
1,
g(
x0,
x1)) →
F(
x1,
h(
g(
x0,
x1)),
g(
x0,
x1)) at position [1] we obtained the following new rules [LPAR04]:
F(0, 1, g(x0, x1)) → F(x1, x1, g(x0, x1))
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(0, 1, g(x0, x1)) → F(x1, x1, g(x0, x1))
The TRS R consists of the following rules:
h(g(x, y)) → y
The set Q consists of the following terms:
h(0)
h(g(x0, x1))
We have to consider all minimal (P,Q,R)-chains.
(19) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(20) TRUE