(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, f(y, x)) → f(f(x, x), f(a, 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:
F(x, f(y, x)) → F(f(x, x), f(a, y))
F(x, f(y, x)) → F(x, x)
F(x, f(y, x)) → F(a, y)
The TRS R consists of the following rules:
f(x, f(y, x)) → f(f(x, x), f(a, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
F(x, f(y, x)) → F(x, x)
F(x, f(y, x)) → F(a, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(F(x1, x2)) = | 4A | + | -I | · | x1 | + | 0A | · | x2 |
| POL(f(x1, x2)) = | 5A | + | 1A | · | x1 | + | 1A | · | x2 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
f(x, f(y, x)) → f(f(x, x), f(a, y))
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, f(y, x)) → F(f(x, x), f(a, y))
The TRS R consists of the following rules:
f(x, f(y, x)) → f(f(x, x), f(a, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) 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.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F_{f_2,f_2}(x, f_{f_2,f_2}(y, x)) → F_{f_2,f_2}(f_{f_2,f_2}(x, x), f_{a,f_2}(a, y))
F_{f_2,f_2}(x, f_{a,f_2}(y, x)) → F_{f_2,f_2}(f_{f_2,f_2}(x, x), f_{a,a}(a, y))
F_{a,f_2}(x, f_{f_2,a}(y, x)) → F_{f_2,f_2}(f_{a,a}(x, x), f_{a,f_2}(a, y))
F_{a,f_2}(x, f_{a,a}(y, x)) → F_{f_2,f_2}(f_{a,a}(x, x), f_{a,a}(a, y))
The TRS R consists of the following rules:
f_{f_2,f_2}(x, f_{f_2,f_2}(y, x)) → f_{f_2,f_2}(f_{f_2,f_2}(x, x), f_{a,f_2}(a, y))
f_{f_2,f_2}(x, f_{a,f_2}(y, x)) → f_{f_2,f_2}(f_{f_2,f_2}(x, x), f_{a,a}(a, y))
f_{a,f_2}(x, f_{f_2,a}(y, x)) → f_{f_2,f_2}(f_{a,a}(x, x), f_{a,f_2}(a, y))
f_{a,f_2}(x, f_{a,a}(y, x)) → f_{f_2,f_2}(f_{a,a}(x, x), f_{a,a}(a, y))
Q is empty.
We have to consider all (P,Q,R)-chains.
(7) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F_{f_2,f_2}(x, f_{f_2,f_2}(y, x)) → F_{f_2,f_2}(f_{f_2,f_2}(x, x), f_{a,f_2}(a, y))
The TRS R consists of the following rules:
f_{f_2,f_2}(x, f_{f_2,f_2}(y, x)) → f_{f_2,f_2}(f_{f_2,f_2}(x, x), f_{a,f_2}(a, y))
f_{f_2,f_2}(x, f_{a,f_2}(y, x)) → f_{f_2,f_2}(f_{f_2,f_2}(x, x), f_{a,a}(a, y))
f_{a,f_2}(x, f_{f_2,a}(y, x)) → f_{f_2,f_2}(f_{a,a}(x, x), f_{a,f_2}(a, y))
f_{a,f_2}(x, f_{a,a}(y, x)) → f_{f_2,f_2}(f_{a,a}(x, x), f_{a,a}(a, y))
Q is empty.
We have to consider all (P,Q,R)-chains.
(9) 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 rules of the TRS R:
f_{a,f_2}(x, f_{f_2,a}(y, x)) → f_{f_2,f_2}(f_{a,a}(x, x), f_{a,f_2}(a, y))
Used ordering: Polynomial interpretation [POLO]:
POL(F_{f_2,f_2}(x1, x2)) = x1 + x2
POL(a) = 0
POL(f_{a,a}(x1, x2)) = 2·x1 + 2·x2
POL(f_{a,f_2}(x1, x2)) = 2·x1 + x2
POL(f_{f_2,a}(x1, x2)) = 1 + x1 + 2·x2
POL(f_{f_2,f_2}(x1, x2)) = x1 + x2
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F_{f_2,f_2}(x, f_{f_2,f_2}(y, x)) → F_{f_2,f_2}(f_{f_2,f_2}(x, x), f_{a,f_2}(a, y))
The TRS R consists of the following rules:
f_{f_2,f_2}(x, f_{f_2,f_2}(y, x)) → f_{f_2,f_2}(f_{f_2,f_2}(x, x), f_{a,f_2}(a, y))
f_{f_2,f_2}(x, f_{a,f_2}(y, x)) → f_{f_2,f_2}(f_{f_2,f_2}(x, x), f_{a,a}(a, y))
f_{a,f_2}(x, f_{a,a}(y, x)) → f_{f_2,f_2}(f_{a,a}(x, x), f_{a,a}(a, y))
Q is empty.
We have to consider all (P,Q,R)-chains.
(11) SemLabProof (SOUND transformation)
We found the following model for the rules of the TRSs R and P.
Interpretation over the domain with elements from 0 to 1.
a: 0
f_{a,f_2}: 0
F_{f_2,f_2}: 0
f_{a,a}: 1
f_{f_2,f_2}: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F_{f_2,f_2}.0-0(x, f_{f_2,f_2}.0-0(y, x)) → F_{f_2,f_2}.0-0(f_{f_2,f_2}.0-0(x, x), f_{a,f_2}.0-0(a., y))
F_{f_2,f_2}.0-0(x, f_{f_2,f_2}.1-0(y, x)) → F_{f_2,f_2}.0-0(f_{f_2,f_2}.0-0(x, x), f_{a,f_2}.0-1(a., y))
F_{f_2,f_2}.1-0(x, f_{f_2,f_2}.0-1(y, x)) → F_{f_2,f_2}.0-0(f_{f_2,f_2}.1-1(x, x), f_{a,f_2}.0-0(a., y))
F_{f_2,f_2}.1-0(x, f_{f_2,f_2}.1-1(y, x)) → F_{f_2,f_2}.0-0(f_{f_2,f_2}.1-1(x, x), f_{a,f_2}.0-1(a., y))
The TRS R consists of the following rules:
f_{f_2,f_2}.0-0(x, f_{f_2,f_2}.0-0(y, x)) → f_{f_2,f_2}.0-0(f_{f_2,f_2}.0-0(x, x), f_{a,f_2}.0-0(a., y))
f_{f_2,f_2}.0-0(x, f_{f_2,f_2}.1-0(y, x)) → f_{f_2,f_2}.0-0(f_{f_2,f_2}.0-0(x, x), f_{a,f_2}.0-1(a., y))
f_{f_2,f_2}.1-0(x, f_{f_2,f_2}.0-1(y, x)) → f_{f_2,f_2}.0-0(f_{f_2,f_2}.1-1(x, x), f_{a,f_2}.0-0(a., y))
f_{f_2,f_2}.1-0(x, f_{f_2,f_2}.1-1(y, x)) → f_{f_2,f_2}.0-0(f_{f_2,f_2}.1-1(x, x), f_{a,f_2}.0-1(a., y))
f_{f_2,f_2}.0-0(x, f_{a,f_2}.0-0(y, x)) → f_{f_2,f_2}.0-1(f_{f_2,f_2}.0-0(x, x), f_{a,a}.0-0(a., y))
f_{f_2,f_2}.0-0(x, f_{a,f_2}.1-0(y, x)) → f_{f_2,f_2}.0-1(f_{f_2,f_2}.0-0(x, x), f_{a,a}.0-1(a., y))
f_{f_2,f_2}.1-0(x, f_{a,f_2}.0-1(y, x)) → f_{f_2,f_2}.0-1(f_{f_2,f_2}.1-1(x, x), f_{a,a}.0-0(a., y))
f_{f_2,f_2}.1-0(x, f_{a,f_2}.1-1(y, x)) → f_{f_2,f_2}.0-1(f_{f_2,f_2}.1-1(x, x), f_{a,a}.0-1(a., y))
f_{a,f_2}.0-1(x, f_{a,a}.0-0(y, x)) → f_{f_2,f_2}.1-1(f_{a,a}.0-0(x, x), f_{a,a}.0-0(a., y))
f_{a,f_2}.0-1(x, f_{a,a}.1-0(y, x)) → f_{f_2,f_2}.1-1(f_{a,a}.0-0(x, x), f_{a,a}.0-1(a., y))
f_{a,f_2}.1-1(x, f_{a,a}.0-1(y, x)) → f_{f_2,f_2}.1-1(f_{a,a}.1-1(x, x), f_{a,a}.0-0(a., y))
f_{a,f_2}.1-1(x, f_{a,a}.1-1(y, x)) → f_{f_2,f_2}.1-1(f_{a,a}.1-1(x, x), f_{a,a}.0-1(a., y))
Q is empty.
We have to consider all (P,Q,R)-chains.
(13) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 4 less nodes.
(14) TRUE