Certification Problem

Input (TPDB TRS_Standard/SK90/4.51)

The rewrite relation of the following TRS is considered.

f(a) g(h(a)) (1)
h(g(x)) g(h(f(x))) (2)
k(x,h(x),a) h(x) (3)
k(f(x),y,x) f(x) (4)

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] = 0
[g(x1)] = 2 · x1
[h(x1)] = 1 · x1
[k(x1, x2, x3)] = 2 + 2 · x1 + 2 · x2 + 1 · x3
all of the following rules can be deleted.
k(x,h(x),a) h(x) (3)
k(f(x),y,x) f(x) (4)

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.
f#(a) h#(a) (5)
h#(g(x)) h#(f(x)) (6)
h#(g(x)) f#(x) (7)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.