Certification Problem

Input (TPDB TRS_Standard/SK90/2.17)

The rewrite relation of the following TRS is considered.

sum(0)	→	0	(1)
sum(s(x))	→	+(sum(x),s(x))	(2)
sum1(0)	→	0	(3)
sum1(s(x))	→	s(+(sum1(x),+(x,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.

sum1^#(s(x))	→	sum1^#(x)	(5)
sum^#(s(x))	→	sum^#(x)	(6)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
sum^#(s(x)) → sum^#(x) (6)

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

[s(x₁)] = x₁ + 1

[sum(x₁)] = 0

[0] = 0

[sum1^#(x₁)] = 0

[+(x₁, x₂)] = 0

[sum^#(x₁)] = x₁ + 0

[sum1(x₁)] = 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
sum^#(s(x)) → sum^#(x) (6)
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
sum1^#(s(x)) → sum1^#(x) (5)

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

[s(x₁)] = x₁ + 1

[sum(x₁)] = 0

[0] = 0

[sum1^#(x₁)] = x₁ + 0

[+(x₁, x₂)] = 0

[sum^#(x₁)] = 0

[sum1(x₁)] = 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
sum1^#(s(x)) → sum1^#(x) (5)
could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.