Certification Problem

Input (TPDB TRS_Standard/AG01/#3.7)

The rewrite relation of the following TRS is considered.

half(0) → 0 (1)
half(s(s(x))) → s(half(x)) (2)
log(s(0)) → 0 (3)
log(s(s(x))) → s(log(s(half(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
[half(x1)] = 1 · x1
[0] = 0
[s(x1)] = 1 · x1 + 1
[log(x1)] = 1 · x1
all of the following rules can be deleted.
half(s(s(x))) → s(half(x)) (2)
log(s(0)) → 0 (3)

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.
log#(s(s(x))) → log#(s(half(x))) (5)
log#(s(s(x))) → half#(x) (6)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.