Certification Problem

Input (TPDB TRS_Innermost/AG01_innermost/#4.30a)

The rewrite relation of the following TRS is considered.

minus(x,0) x (1)
minus(s(x),s(y)) minus(x,y) (2)
le(0,y) true (3)
le(s(x),0) false (4)
le(s(x),s(y)) le(x,y) (5)
quot(0,s(y)) 0 (6)
quot(s(x),s(y)) s(quot(minus(s(x),s(y)),s(y))) (7)
The evaluation strategy is innermost.

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
minus#(s(x),s(y)) minus#(x,y) (8)
le#(s(x),s(y)) le#(x,y) (9)
quot#(s(x),s(y)) quot#(minus(s(x),s(y)),s(y)) (10)
quot#(s(x),s(y)) minus#(s(x),s(y)) (11)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.