Certification Problem

Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex9_Luc04_GM)

The rewrite relation of the following TRS is considered.

a__f(a,b,X) a__f(mark(X),X,mark(X)) (1)
a__c a (2)
a__c b (3)
mark(f(X1,X2,X3)) a__f(mark(X1),X2,mark(X3)) (4)
mark(c) a__c (5)
mark(a) a (6)
mark(b) b (7)
a__f(X1,X2,X3) f(X1,X2,X3) (8)
a__c c (9)
The evaluation strategy is innermost.

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a__f#(a,b,X) a__f#(mark(X),X,mark(X)) (10)
a__f#(a,b,X) mark#(X) (11)
mark#(f(X1,X2,X3)) a__f#(mark(X1),X2,mark(X3)) (12)
mark#(f(X1,X2,X3)) mark#(X1) (13)
mark#(f(X1,X2,X3)) mark#(X3) (14)
mark#(c) a__c# (15)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.