Certification Problem

Input (TPDB TRS_Standard/GTSSK07/cade03)

The rewrite relation of the following TRS is considered.

minus(x,y) cond(ge(x,s(y)),x,y) (1)
cond(false,x,y) 0 (2)
cond(true,x,y) s(minus(x,s(y))) (3)
ge(u,0) true (4)
ge(0,s(v)) false (5)
ge(s(u),s(v)) ge(u,v) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
minus#(x,y) cond#(ge(x,s(y)),x,y) (7)
minus#(x,y) ge#(x,s(y)) (8)
cond#(true,x,y) minus#(x,s(y)) (9)
ge#(s(u),s(v)) ge#(u,v) (10)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.