Certification Problem

Input (TPDB TRS_Standard/GTSSK07/cade11)

The rewrite relation of the following TRS is considered.

f(true,x,y) f(gt(x,y),s(x),s(s(y))) (1)
gt(0,v) false (2)
gt(s(u),0) true (3)
gt(s(u),s(v)) gt(u,v) (4)

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.
f#(true,x,y) f#(gt(x,y),s(x),s(s(y))) (5)
f#(true,x,y) gt#(x,y) (6)
gt#(s(u),s(v)) gt#(u,v) (7)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.