Certification Problem

Input (TPDB TRS_Standard/Secret_05_TRS/aprove4)

The rewrite relation of the following TRS is considered.

p(0) s(s(0)) (1)
p(s(x)) x (2)
p(p(s(x))) p(x) (3)
le(p(s(x)),x) le(x,x) (4)
le(0,y) true (5)
le(s(x),0) false (6)
le(s(x),s(y)) le(x,y) (7)
minus(x,y) if(le(x,y),x,y) (8)
if(true,x,y) 0 (9)
if(false,x,y) s(minus(p(x),y)) (10)

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.
p#(p(s(x))) p#(x) (11)
le#(p(s(x)),x) le#(x,x) (12)
le#(s(x),s(y)) le#(x,y) (13)
minus#(x,y) if#(le(x,y),x,y) (14)
minus#(x,y) le#(x,y) (15)
if#(false,x,y) minus#(p(x),y) (16)
if#(false,x,y) p#(x) (17)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.