Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.26)

The rewrite relation of the following TRS is considered.

p(0) 0 (1)
p(s(x)) x (2)
le(0,y) true (3)
le(s(x),0) false (4)
le(s(x),s(y)) le(x,y) (5)
minus(x,y) if(le(x,y),x,y) (6)
if(true,x,y) 0 (7)
if(false,x,y) s(minus(p(x),y)) (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.
le#(s(x),s(y)) le#(x,y) (9)
minus#(x,y) if#(le(x,y),x,y) (10)
minus#(x,y) le#(x,y) (11)
if#(false,x,y) minus#(p(x),y) (12)
if#(false,x,y) p#(x) (13)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.