Certification Problem

Input (TPDB TRS_Innermost/Mixed_innermost/cade05)

The rewrite relation of the following TRS is considered.

minus(x,x) 0 (1)
minus(x,y) cond(equal(min(x,y),y),x,y) (2)
cond(true,x,y) s(minus(x,s(y))) (3)
min(0,v) 0 (4)
min(u,0) 0 (5)
min(s(u),s(v)) s(min(u,v)) (6)
equal(0,0) true (7)
equal(s(x),0) false (8)
equal(0,s(y)) false (9)
equal(s(x),s(y)) equal(x,y) (10)
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#(x,y) cond#(equal(min(x,y),y),x,y) (11)
minus#(x,y) equal#(min(x,y),y) (12)
minus#(x,y) min#(x,y) (13)
cond#(true,x,y) minus#(x,s(y)) (14)
min#(s(u),s(v)) min#(u,v) (15)
equal#(s(x),s(y)) equal#(x,y) (16)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.