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 NaTT @ termCOMP 2023)

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)
half^#(s(s(x)))	→	half^#(x)	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
log^#(s(s(x))) → log^#(s(half(x))) (5)

1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation

[s(x₁)] = x₁ + 2

[log^#(x₁)] = x₁ + 0

[half^#(x₁)] = 0

[half(x₁)] = x₁ + 1

[log(x₁)] = 0

[0] = 28958

together with the usable rules

half(0) → 0 (1)

half(s(s(x))) → s(half(x)) (2)

(w.r.t. the implicit argument filter of the reduction pair), the pair
log^#(s(s(x))) → log^#(s(half(x))) (5)
could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.
The 2^nd component contains the pair
half^#(s(s(x))) → half^#(x) (7)

1.1.2 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation

[s(x₁)] = x₁ + 1

[log^#(x₁)] = x₁ + 0

[half^#(x₁)] = x₁ + 0

[half(x₁)] = x₁ + 11293

[log(x₁)] = 0

[0] = 28958

together with the usable rules

half(0) → 0 (1)

half(s(s(x))) → s(half(x)) (2)

(w.r.t. the implicit argument filter of the reduction pair), the pair
half^#(s(s(x))) → half^#(x) (7)
could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.