Certification Problem

Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex25_Luc06_GM)

The rewrite relation of the following TRS is considered.

a__f(f(X)) a__c(f(g(f(X)))) (1)
a__c(X) d(X) (2)
a__h(X) a__c(d(X)) (3)
mark(f(X)) a__f(mark(X)) (4)
mark(c(X)) a__c(X) (5)
mark(h(X)) a__h(mark(X)) (6)
mark(g(X)) g(X) (7)
mark(d(X)) d(X) (8)
a__f(X) f(X) (9)
a__c(X) c(X) (10)
a__h(X) h(X) (11)
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)] = 1 · x1
[f(x1)] = 1 · x1
[a__c(x1)] = 1 · x1
[g(x1)] = 1 · x1
[d(x1)] = 1 · x1
[a__h(x1)] = 1 · x1 + 1
[mark(x1)] = 1 · x1
[c(x1)] = 1 · x1
[h(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
a__h(X) a__c(d(X)) (3)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a__f(x1)] = 1 · x1
[f(x1)] = 1 · x1
[a__c(x1)] = 1 · x1
[g(x1)] = 1 · x1
[d(x1)] = 1 · x1
[mark(x1)] = 1 · x1 + 1
[c(x1)] = 1 · x1
[h(x1)] = 1 · x1
[a__h(x1)] = 1 · x1
all of the following rules can be deleted.
mark(c(X)) a__c(X) (5)
mark(g(X)) g(X) (7)
mark(d(X)) d(X) (8)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a__f#(f(X)) a__c#(f(g(f(X)))) (12)
mark#(f(X)) a__f#(mark(X)) (13)
mark#(f(X)) mark#(X) (14)
mark#(h(X)) a__h#(mark(X)) (15)
mark#(h(X)) mark#(X) (16)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.