Certification Problem

Input (TPDB TRS_Standard/Various_04/24)

The rewrite relation of the following TRS is considered.

max(L(x))	→	x	(1)
max(N(L(0),L(y)))	→	y	(2)
max(N(L(s(x)),L(s(y))))	→	s(max(N(L(x),L(y))))	(3)
max(N(L(x),N(y,z)))	→	max(N(L(x),L(max(N(y,z)))))	(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.

max^#(N(L(x),N(y,z)))	→	max^#(N(L(x),L(max(N(y,z)))))	(5)
max^#(N(L(x),N(y,z)))	→	max^#(N(y,z))	(6)
max^#(N(L(s(x)),L(s(y))))	→	max^#(N(L(x),L(y)))	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
max^#(N(L(x),N(y,z))) → max^#(N(y,z)) (6)

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

[s(x₁)] = 0

[0] = 0

[max(x₁)] = 0

[max^#(x₁)] = x₁ + 0

[L(x₁)] = 1

[N(x₁, x₂)] = x₁ + x₂ + 23676

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
max^#(N(L(x),N(y,z))) → max^#(N(y,z)) (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
max^#(N(L(s(x)),L(s(y)))) → max^#(N(L(x),L(y))) (7)

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

[s(x₁)] = x₁ + 1

[0] = 0

[max(x₁)] = 0

[max^#(x₁)] = x₁ + 0

[L(x₁)] = x₁ + 33954

[N(x₁, x₂)] = x₁ + x₂ + 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
max^#(N(L(s(x)),L(s(y)))) → max^#(N(L(x),L(y))) (7)
could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.