Certification Problem

Input (TPDB TRS_Standard/AProVE_07/otto11)

The rewrite relation of the following TRS is considered.

times(x,y) help(x,y,0) (1)
help(x,y,c) if(lt(c,y),x,y,c) (2)
if(true,x,y,c) plus(x,help(x,y,s(c))) (3)
if(false,x,y,c) 0 (4)
lt(0,s(x)) true (5)
lt(s(x),0) false (6)
lt(s(x),s(y)) lt(x,y) (7)
plus(x,0) x (8)
plus(0,x) x (9)
plus(x,s(y)) s(plus(x,y)) (10)
plus(s(x),y) s(plus(x,y)) (11)

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.
times#(x,y) help#(x,y,0) (12)
help#(x,y,c) if#(lt(c,y),x,y,c) (13)
help#(x,y,c) lt#(c,y) (14)
if#(true,x,y,c) plus#(x,help(x,y,s(c))) (15)
if#(true,x,y,c) help#(x,y,s(c)) (16)
lt#(s(x),s(y)) lt#(x,y) (17)
plus#(x,s(y)) plus#(x,y) (18)
plus#(s(x),y) plus#(x,y) (19)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.