Certification Problem

Input (TPDB TRS_Outermost/Strategy_outermost_added_08/Ex2_8_1ConstSubstFix)

The rewrite relation of the following TRS is considered.

app(app(const,x),y) → x (1)
app(app(app(subst,f),g),x) → app(app(f,x),app(g,x)) (2)
app(app(fix,f),x) → app(app(f,app(fix,f)),x) (3)
The evaluation strategy is outermost

Property / Task

Prove or disprove termination.

Answer / Result

No.

Proof (by AProVE @ termCOMP 2023)

1 Loop

The following loop proves nontermination.

t0 = app(app(app(fix,subst),x'),x)
→ app(app(app(subst,app(fix,subst)),x'),x)
→ app(app(app(fix,subst),x),app(x',x))
= t2
where t2 = t0σ and σ = {x'/x, x/app(x',x)}