Certification Problem

Input (TPDB TRS_Standard/AG01/#3.1)

The rewrite relation of the following TRS is considered.

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

quot^#(s(x),s(y))	→	quot^#(minus(x,y),s(y))	(5)
quot^#(s(x),s(y))	→	minus^#(x,y)	(6)
minus^#(s(x),s(y))	→	minus^#(x,y)	(7)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
quot^#(s(x),s(y)) → quot^#(minus(x,y),s(y)) (5)

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

[s(x₁)] = x₁ + 2

[minus(x₁, x₂)] = x₁ + 1

[0] = 1

[quot(x₁, x₂)] = 0

[minus^#(x₁, x₂)] = 0

[quot^#(x₁, x₂)] = x₁ + 0

together with the usable rules

minus(x,0) → x (1)

minus(s(x),s(y)) → minus(x,y) (2)

(w.r.t. the implicit argument filter of the reduction pair), the pair
quot^#(s(x),s(y)) → quot^#(minus(x,y),s(y)) (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
minus^#(s(x),s(y)) → minus^#(x,y) (7)

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

[s(x₁)] = x₁ + 2

[minus(x₁, x₂)] = x₁ + 1

[0] = 1

[quot(x₁, x₂)] = 0

[minus^#(x₁, x₂)] = x₁ + x₂ + 0

[quot^#(x₁, x₂)] = x₁ + 0

together with the usable rules

minus(x,0) → x (1)

minus(s(x),s(y)) → minus(x,y) (2)

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

The dependency pairs are split into 0 components.