Certification Problem

Input (TPDB TRS_Standard/AProVE_07/thiemann27)

The rewrite relation of the following TRS is considered.

le(0,y) true (1)
le(s(x),0) false (2)
le(s(x),s(y)) le(x,y) (3)
int(x,y) if(le(x,y),x,y) (4)
if(true,x,y) cons(x,int(s(x),y)) (5)
if(false,x,y) nil (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.
le#(s(x),s(y)) le#(x,y) (7)
int#(x,y) if#(le(x,y),x,y) (8)
int#(x,y) le#(x,y) (9)
if#(true,x,y) int#(s(x),y) (10)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.