Certification Problem

Input (TPDB TRS_Innermost/AG01_innermost/#4.27)

The rewrite relation of the following TRS is considered.

p(0) 0 (1)
p(s(x)) x (2)
le(0,y) true (3)
le(s(x),0) false (4)
le(s(x),s(y)) le(x,y) (5)
minus(x,0) x (6)
minus(x,s(y)) if(le(x,s(y)),0,p(minus(x,p(s(y))))) (7)
if(true,x,y) x (8)
if(false,x,y) y (9)
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.
le#(s(x),s(y)) le#(x,y) (10)
minus#(x,s(y)) if#(le(x,s(y)),0,p(minus(x,p(s(y))))) (11)
minus#(x,s(y)) le#(x,s(y)) (12)
minus#(x,s(y)) p#(minus(x,p(s(y)))) (13)
minus#(x,s(y)) minus#(x,p(s(y))) (14)
minus#(x,s(y)) p#(s(y)) (15)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.