Certification Problem

Input (TPDB TRS_Standard/AProVE_07/otto07)

The rewrite relation of the following TRS is considered.

lt(0,s(x)) true (1)
lt(x,0) false (2)
lt(s(x),s(y)) lt(x,y) (3)
minus(x,y) help(lt(y,x),x,y) (4)
help(true,x,y) s(minus(x,s(y))) (5)
help(false,x,y) 0 (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.
lt#(s(x),s(y)) lt#(x,y) (7)
minus#(x,y) help#(lt(y,x),x,y) (8)
minus#(x,y) lt#(y,x) (9)
help#(true,x,y) minus#(x,s(y)) (10)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.