Certification Problem

Input (TPDB TRS_Standard/HirokawaMiddeldorp_04/t002)

The rewrite relation of the following TRS is considered.

leq(0,y) true (1)
leq(s(x),0) false (2)
leq(s(x),s(y)) leq(x,y) (3)
if(true,x,y) x (4)
if(false,x,y) y (5)
-(x,0) x (6)
-(s(x),s(y)) -(x,y) (7)
mod(0,y) 0 (8)
mod(s(x),0) 0 (9)
mod(s(x),s(y)) if(leq(y,x),mod(-(s(x),s(y)),s(y)),s(x)) (10)

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.
leq#(s(x),s(y)) leq#(x,y) (11)
-#(s(x),s(y)) -#(x,y) (12)
mod#(s(x),s(y)) if#(leq(y,x),mod(-(s(x),s(y)),s(y)),s(x)) (13)
mod#(s(x),s(y)) leq#(y,x) (14)
mod#(s(x),s(y)) mod#(-(s(x),s(y)),s(y)) (15)
mod#(s(x),s(y)) -#(s(x),s(y)) (16)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.