Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/divExp)

The rewrite relation of the following TRS is considered.

minus(x,x) 0 (1)
minus(s(x),s(y)) minus(x,y) (2)
minus(0,x) 0 (3)
minus(x,0) x (4)
div(s(x),s(y)) s(div(minus(x,y),s(y))) (5)
div(0,s(y)) 0 (6)
f(x,0,b) x (7)
f(x,s(y),b) div(f(x,minus(s(y),s(0)),b),b) (8)

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#(s(x),s(y)) minus#(x,y) (9)
div#(s(x),s(y)) div#(minus(x,y),s(y)) (10)
div#(s(x),s(y)) minus#(x,y) (11)
f#(x,s(y),b) div#(f(x,minus(s(y),s(0)),b),b) (12)
f#(x,s(y),b) f#(x,minus(s(y),s(0)),b) (13)
f#(x,s(y),b) minus#(s(y),s(0)) (14)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.