Certification Problem

Input (TPDB TRS_Standard/GTSSK07/cade05t)

The rewrite relation of the following TRS is considered.

minus(x,y) cond(equal(min(x,y),y),x,y) (1)
cond(true,x,y) s(minus(x,s(y))) (2)
min(0,v) 0 (3)
min(u,0) 0 (4)
min(s(u),s(v)) s(min(u,v)) (5)
equal(0,0) true (6)
equal(s(x),0) false (7)
equal(0,s(y)) false (8)
equal(s(x),s(y)) equal(x,y) (9)

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#(equal(min(x,y),y),x,y) (10)
minus#(x,y) equal#(min(x,y),y) (11)
minus#(x,y) min#(x,y) (12)
cond#(true,x,y) minus#(x,s(y)) (13)
min#(s(u),s(v)) min#(u,v) (14)
equal#(s(x),s(y)) equal#(x,y) (15)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.