Certification Problem

Input (TPDB TRS_Standard/Beerendonk_07/2)

The rewrite relation of the following TRS is considered.

cond(true,x,y) cond(gr(x,y),p(x),s(y)) (1)
gr(0,x) false (2)
gr(s(x),0) true (3)
gr(s(x),s(y)) gr(x,y) (4)
p(0) 0 (5)
p(s(x)) x (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.
cond#(true,x,y) cond#(gr(x,y),p(x),s(y)) (7)
cond#(true,x,y) gr#(x,y) (8)
cond#(true,x,y) p#(x) (9)
gr#(s(x),s(y)) gr#(x,y) (10)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.