Certification Problem

Input (TPDB TRS_Innermost/AG01_innermost/#4.37a)

The rewrite relation of the following TRS is considered.

f(c(s(x),y)) f(c(x,s(y))) (1)
g(c(x,s(y))) g(c(s(x),y)) (2)
g(s(f(x))) g(f(x)) (3)
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
[f(x1)] = 2 · x1
[c(x1, x2)] = 2 · x1 + 2 · x2
[s(x1)] = 1 + 1 · x1
[g(x1)] = 2 · x1
all of the following rules can be deleted.
g(s(f(x))) g(f(x)) (3)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
f#(c(s(x),y)) f#(c(x,s(y))) (4)
g#(c(x,s(y))) g#(c(s(x),y)) (5)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.