Certification Problem

Input (TPDB TRS_Standard/AG01/#3.40)

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)
plus(0,y) y (5)
plus(s(x),y) s(plus(x,y)) (6)
plus(minus(x,s(0)),minus(y,s(s(z)))) plus(minus(y,s(s(z))),minus(x,s(0))) (7)
plus(plus(x,s(0)),plus(y,s(s(z)))) plus(plus(y,s(s(z))),plus(x,s(0))) (8)

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) (9)
quot#(s(x),s(y)) quot#(minus(x,y),s(y)) (10)
quot#(s(x),s(y)) minus#(x,y) (11)
plus#(s(x),y) plus#(x,y) (12)
plus#(minus(x,s(0)),minus(y,s(s(z)))) plus#(minus(y,s(s(z))),minus(x,s(0))) (13)
plus#(plus(x,s(0)),plus(y,s(s(z)))) plus#(plus(y,s(s(z))),plus(x,s(0))) (14)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.