Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.22)

The rewrite relation of the following TRS is considered.

quot(0,s(y),s(z))	→	0	(1)
quot(s(x),s(y),z)	→	quot(x,y,z)	(2)
quot(x,0,s(z))	→	s(quot(x,s(z),s(z)))	(3)

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.

quot^#(x,0,s(z))	→	quot^#(x,s(z),s(z))	(4)
quot^#(s(x),s(y),z)	→	quot^#(x,y,z)	(5)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

quot^#(s(x),s(y),z) → quot^#(x,y,z) (5)

quot^#(x,0,s(z)) → quot^#(x,s(z),s(z)) (4)

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

[s(x₁)] = x₁ + 1

[0] = 1

[quot(x₁, x₂, x₃)] = 0

[quot^#(x₁, x₂, x₃)] = x₁ + 0

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

The dependency pairs are split into 0 components.