Certification Problem

Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex16_Luc06_GM)

The rewrite relation of the following TRS is considered.

a__f(X,X) a__f(a,b) (1)
a__b a (2)
mark(f(X1,X2)) a__f(mark(X1),X2) (3)
mark(b) a__b (4)
mark(a) a (5)
a__f(X1,X2) f(X1,X2) (6)
a__b b (7)
The evaluation strategy is innermost.

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a__f(x1, x2)] = 2 + 2 · x1 + 2 · x2
[a] = 0
[b] = 0
[a__b] = 0
[mark(x1)] = 2 · x1
[f(x1, x2)] = 1 + 2 · x1 + 1 · x2
all of the following rules can be deleted.
a__f(X1,X2) f(X1,X2) (6)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a__f(x1, x2)] = 1 · x1 + 2 · x2
[a] = 0
[b] = 0
[a__b] = 0
[mark(x1)] = 2 · x1
[f(x1, x2)] = 1 + 2 · x1 + 1 · x2
all of the following rules can be deleted.
mark(f(X1,X2)) a__f(mark(X1),X2) (3)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a__f(x1, x2)] = 1 · x1 + 1 · x2
[a] = 0
[b] = 0
[a__b] = 2
[mark(x1)] = 2 + 1 · x1
all of the following rules can be deleted.
a__b a (2)
mark(a) a (5)
a__b b (7)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a__f(x1, x2)] = 1 · x1 + 1 · x2
[a] = 0
[b] = 0
[mark(x1)] = 1 + 1 · x1
[a__b] = 0
all of the following rules can be deleted.
mark(b) a__b (4)

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a__f#(X,X) a__f#(a,b) (8)

1.1.1.1.1.1 Usable Rules Processor

We restrict the rewrite rules to the following usable rules of the DP problem.

There are no rules.

1.1.1.1.1.1.1 Innermost Lhss Removal Processor

We restrict the innermost strategy to the following left hand sides.

There are no lhss.

1.1.1.1.1.1.1.1 Instantiation Processor

We instantiate the pair to the following set of pairs

There are no rules.

1.1.1.1.1.1.1.1.1 P is empty

There are no pairs anymore.