(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)
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(a, f(a, f(a, f(x, b)))) → F(f(a, f(a, f(a, x))), b)
F(a, f(a, f(a, f(x, b)))) → F(a, f(a, f(a, x)))
F(a, f(a, f(a, f(x, b)))) → F(a, f(a, x))
F(a, f(a, f(a, f(x, b)))) → F(a, x)
F(f(f(a, x), b), b) → F(f(a, f(f(x, b), b)), b)
F(f(f(a, x), b), b) → F(a, f(f(x, b), b))
F(f(f(a, x), b), b) → F(f(x, b), b)
F(f(f(a, x), b), b) → F(x, b)
The TRS R consists of the following rules:
f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(f(f(a, x), b), b) → F(f(x, b), b)
F(f(f(a, x), b), b) → F(f(a, f(f(x, b), b)), b)
F(f(f(a, x), b), b) → F(x, b)
The TRS R consists of the following rules:
f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
F(f(f(a, x), b), b) → F(f(x, b), b)
F(f(f(a, x), b), b) → F(x, b)
Used ordering: Polynomial interpretation [POLO]:
POL(F(x1, x2)) = x1 + x2
POL(a) = 1
POL(b) = 0
POL(f(x1, x2)) = x1 + x2
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(f(f(a, x), b), b) → F(f(a, f(f(x, b), b)), b)
The TRS R consists of the following rules:
f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
F(f(f(a, x), b), b) → F(f(a, f(f(x, b), b)), b)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:
POL(F(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
POL(f(x1, x2)) = | | + | | · | x1 | + | | · | x2 |
The following usable rules [FROCOS05] were oriented:
f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)
(9) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(11) TRUE
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a, f(a, f(a, f(x, b)))) → F(a, f(a, x))
F(a, f(a, f(a, f(x, b)))) → F(a, f(a, f(a, x)))
F(a, f(a, f(a, f(x, b)))) → F(a, x)
The TRS R consists of the following rules:
f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
F(a, f(a, f(a, f(x, b)))) → F(a, f(a, x))
F(a, f(a, f(a, f(x, b)))) → F(a, x)
Used ordering: Polynomial interpretation [POLO]:
POL(F(x1, x2)) = x1 + x2
POL(a) = 1
POL(b) = 0
POL(f(x1, x2)) = x1 + x2
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a, f(a, f(a, f(x, b)))) → F(a, f(a, f(a, x)))
The TRS R consists of the following rules:
f(a, f(a, f(a, f(x, b)))) → f(f(a, f(a, f(a, x))), b)
f(f(f(a, x), b), b) → f(f(a, f(f(x, b), b)), b)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) RootLabelingFC2Proof (EQUIVALENT transformation)
We used root labeling (second transformation) [ROOTLAB] with the following heuristic:
LabelAll: All function symbols get labeled
As Q is empty the root labeling was sound AND complete.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{f,b}(x, b)))) → F_{a,f}(a, f_{a,f}(a, f_{a,f}(a, x)))
F_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{a,b}(x, b)))) → F_{a,f}(a, f_{a,f}(a, f_{a,a}(a, x)))
F_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{b,b}(x, b)))) → F_{a,f}(a, f_{a,f}(a, f_{a,b}(a, x)))
The TRS R consists of the following rules:
f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{f,b}(x, b)))) → f_{f,b}(f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, x))), b)
f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{a,b}(x, b)))) → f_{f,b}(f_{a,f}(a, f_{a,f}(a, f_{a,a}(a, x))), b)
f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{b,b}(x, b)))) → f_{f,b}(f_{a,f}(a, f_{a,f}(a, f_{a,b}(a, x))), b)
f_{f,b}(f_{f,b}(f_{a,f}(a, x), b), b) → f_{f,b}(f_{a,f}(a, f_{f,b}(f_{f,b}(x, b), b)), b)
f_{f,b}(f_{f,b}(f_{a,a}(a, x), b), b) → f_{f,b}(f_{a,f}(a, f_{f,b}(f_{a,b}(x, b), b)), b)
f_{f,b}(f_{f,b}(f_{a,b}(a, x), b), b) → f_{f,b}(f_{a,f}(a, f_{f,b}(f_{b,b}(x, b), b)), b)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{f,b}(x, b)))) → F_{a,f}(a, f_{a,f}(a, f_{a,f}(a, x)))
The TRS R consists of the following rules:
f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{f,b}(x, b)))) → f_{f,b}(f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, x))), b)
f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{a,b}(x, b)))) → f_{f,b}(f_{a,f}(a, f_{a,f}(a, f_{a,a}(a, x))), b)
f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{b,b}(x, b)))) → f_{f,b}(f_{a,f}(a, f_{a,f}(a, f_{a,b}(a, x))), b)
f_{f,b}(f_{f,b}(f_{a,f}(a, x), b), b) → f_{f,b}(f_{a,f}(a, f_{f,b}(f_{f,b}(x, b), b)), b)
f_{f,b}(f_{f,b}(f_{a,a}(a, x), b), b) → f_{f,b}(f_{a,f}(a, f_{f,b}(f_{a,b}(x, b), b)), b)
f_{f,b}(f_{f,b}(f_{a,b}(a, x), b), b) → f_{f,b}(f_{a,f}(a, f_{f,b}(f_{b,b}(x, b), b)), b)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04].
The following pairs can be oriented strictly and are deleted.
F_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{f,b}(x, b)))) → F_{a,f}(a, f_{a,f}(a, f_{a,f}(a, x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
POL(F_{a,f}(x1, x2)) = | | + | / | -I | 0A | -I | \ |
| | -I | -I | -I | | |
\ | -I | -I | 0A | / |
| · | x1 | + | / | 0A | -I | -I | \ |
| | 2A | 1A | -I | | |
\ | 2A | 1A | -I | / |
| · | x2 |
POL(f_{a,f}(x1, x2)) = | | + | / | -I | -I | -I | \ |
| | -I | 0A | -I | | |
\ | -I | 0A | -I | / |
| · | x1 | + | / | -I | 0A | -I | \ |
| | -I | -I | 0A | | |
\ | 0A | -I | -I | / |
| · | x2 |
POL(f_{f,b}(x1, x2)) = | | + | / | -I | -I | -I | \ |
| | 0A | -I | -I | | |
\ | 0A | -I | 2A | / |
| · | x1 | + | / | -I | 0A | -I | \ |
| | -I | 0A | 0A | | |
\ | -I | -I | -I | / |
| · | x2 |
POL(f_{b,b}(x1, x2)) = | | + | / | -I | -I | -I | \ |
| | 3A | 0A | 0A | | |
\ | 3A | 2A | 2A | / |
| · | x1 | + | / | -I | -I | -I | \ |
| | 0A | 0A | -I | | |
\ | 0A | -I | -I | / |
| · | x2 |
POL(f_{a,b}(x1, x2)) = | | + | / | -I | -I | -I | \ |
| | 0A | 0A | 0A | | |
\ | 0A | 2A | -I | / |
| · | x1 | + | / | -I | 0A | -I | \ |
| | 0A | 0A | 0A | | |
\ | 3A | 0A | -I | / |
| · | x2 |
POL(f_{a,a}(x1, x2)) = | | + | / | -I | -I | -I | \ |
| | -I | -I | 0A | | |
\ | -I | -I | 0A | / |
| · | x1 | + | / | 0A | 0A | 0A | \ |
| | -I | 0A | -I | | |
\ | 0A | -I | -I | / |
| · | x2 |
The following usable rules [FROCOS05] were oriented:
f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{f,b}(x, b)))) → f_{f,b}(f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, x))), b)
f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{b,b}(x, b)))) → f_{f,b}(f_{a,f}(a, f_{a,f}(a, f_{a,b}(a, x))), b)
f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{a,b}(x, b)))) → f_{f,b}(f_{a,f}(a, f_{a,f}(a, f_{a,a}(a, x))), b)
f_{f,b}(f_{f,b}(f_{a,f}(a, x), b), b) → f_{f,b}(f_{a,f}(a, f_{f,b}(f_{f,b}(x, b), b)), b)
f_{f,b}(f_{f,b}(f_{a,a}(a, x), b), b) → f_{f,b}(f_{a,f}(a, f_{f,b}(f_{a,b}(x, b), b)), b)
f_{f,b}(f_{f,b}(f_{a,b}(a, x), b), b) → f_{f,b}(f_{a,f}(a, f_{f,b}(f_{b,b}(x, b), b)), b)
(20) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{f,b}(x, b)))) → f_{f,b}(f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, x))), b)
f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{a,b}(x, b)))) → f_{f,b}(f_{a,f}(a, f_{a,f}(a, f_{a,a}(a, x))), b)
f_{a,f}(a, f_{a,f}(a, f_{a,f}(a, f_{b,b}(x, b)))) → f_{f,b}(f_{a,f}(a, f_{a,f}(a, f_{a,b}(a, x))), b)
f_{f,b}(f_{f,b}(f_{a,f}(a, x), b), b) → f_{f,b}(f_{a,f}(a, f_{f,b}(f_{f,b}(x, b), b)), b)
f_{f,b}(f_{f,b}(f_{a,a}(a, x), b), b) → f_{f,b}(f_{a,f}(a, f_{f,b}(f_{a,b}(x, b), b)), b)
f_{f,b}(f_{f,b}(f_{a,b}(a, x), b), b) → f_{f,b}(f_{a,f}(a, f_{f,b}(f_{b,b}(x, b), b)), b)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(22) TRUE