Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex23_Luc06_FR)

The rewrite relation of the following TRS is considered.

f(f(a)) c(n__f(n__g(n__f(n__a)))) (1)
f(X) n__f(X) (2)
g(X) n__g(X) (3)
a n__a (4)
activate(n__f(X)) f(activate(X)) (5)
activate(n__g(X)) g(activate(X)) (6)
activate(n__a) a (7)
activate(X) X (8)

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
[f(x1)] = 1 · x1
[a] = 1
[c(x1)] = 1 · x1
[n__f(x1)] = 1 · x1
[n__g(x1)] = 1 · x1
[n__a] = 0
[g(x1)] = 1 · x1
[activate(x1)] = 1 · x1 + 2
all of the following rules can be deleted.
f(f(a)) c(n__f(n__g(n__f(n__a)))) (1)
a n__a (4)
activate(n__a) a (7)
activate(X) X (8)

1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
activate#(n__f(X)) f#(activate(X)) (9)
activate#(n__f(X)) activate#(X) (10)
activate#(n__g(X)) g#(activate(X)) (11)
activate#(n__g(X)) activate#(X) (12)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.