Certification Problem

Input (TPDB TRS_Standard/AProVE_07/otto09)

The rewrite relation of the following TRS is considered.

lt(x,0) false (1)
lt(0,s(y)) true (2)
lt(s(x),s(y)) lt(x,y) (3)
plus(x,0) x (4)
plus(x,s(y)) s(plus(x,y)) (5)
quot(x,s(y)) help(x,s(y),0) (6)
help(x,s(y),c) if(lt(c,x),x,s(y),c) (7)
if(true,x,s(y),c) s(help(x,s(y),plus(c,s(y)))) (8)
if(false,x,s(y),c) 0 (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.
lt#(s(x),s(y)) lt#(x,y) (10)
plus#(x,s(y)) plus#(x,y) (11)
quot#(x,s(y)) help#(x,s(y),0) (12)
help#(x,s(y),c) if#(lt(c,x),x,s(y),c) (13)
help#(x,s(y),c) lt#(c,x) (14)
if#(true,x,s(y),c) help#(x,s(y),plus(c,s(y))) (15)
if#(true,x,s(y),c) plus#(c,s(y)) (16)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.