Certification Problem

Input (TPDB TRS_Standard/AotoYamada_05/023)

The rewrite relation of the following TRS is considered.

app(id,x)	→	x	(1)
app(plus,0)	→	id	(2)
app(app(plus,app(s,x)),y)	→	app(s,app(app(plus,x),y))	(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.

app^#(app(plus,app(s,x)),y)	→	app^#(app(plus,x),y)	(4)
app^#(app(plus,app(s,x)),y)	→	app^#(plus,x)	(5)
app^#(app(plus,app(s,x)),y)	→	app^#(s,app(app(plus,x),y))	(6)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair
app^#(app(plus,app(s,x)),y) → app^#(app(plus,x),y) (4)

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

[s] = 1

[0] = 19306

[app^#(x₁, x₂)] = x₁ + 0

[plus] = 8041

[id] = 1

[app(x₁, x₂)] = x₁ + x₂ + 12122

together with the usable rules

app(id,x) → x (1)

app(app(plus,app(s,x)),y) → app(s,app(app(plus,x),y)) (3)

app(plus,0) → id (2)

(w.r.t. the implicit argument filter of the reduction pair), the pair
app^#(app(plus,app(s,x)),y) → app^#(app(plus,x),y) (4)
could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.