Certification Problem

Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/ExConc_Zan97_GM)

The rewrite relation of the following TRS is considered.

a__f(X) g(h(f(X))) (1)
mark(f(X)) a__f(mark(X)) (2)
mark(g(X)) g(X) (3)
mark(h(X)) h(mark(X)) (4)
a__f(X) f(X) (5)
The evaluation strategy is innermost.

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a__f(X) f(h(g(X))) (6)
f(mark(X)) mark(a__f(X)) (7)
g(mark(X)) g(X) (8)
h(mark(X)) mark(h(X)) (9)
a__f(X) f(X) (5)

1.1 Rule Removal

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

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a__f#(X) f#(h(g(X))) (10)
a__f#(X) h#(g(X)) (11)
f#(mark(X)) a__f#(X) (12)
h#(mark(X)) h#(X) (13)
a__f#(X) f#(X) (14)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.