Certification Problem

Input (TPDB TRS_Standard/AG01/#3.8b)

The rewrite relation of the following TRS is considered.

le(0,y) true (1)
le(s(x),0) false (2)
le(s(x),s(y)) le(x,y) (3)
minus(0,y) 0 (4)
minus(s(x),y) if_minus(le(s(x),y),s(x),y) (5)
if_minus(true,s(x),y) 0 (6)
if_minus(false,s(x),y) s(minus(x,y)) (7)
quot(0,s(y)) 0 (8)
quot(s(x),s(y)) s(quot(minus(x,y),s(y))) (9)
log(s(0)) 0 (10)
log(s(s(x))) s(log(s(quot(x,s(s(0)))))) (11)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
quot#(s(x),s(y)) minus#(x,y) (12)
minus#(s(x),y) if_minus#(le(s(x),y),s(x),y) (13)
minus#(s(x),y) le#(s(x),y) (14)
if_minus#(false,s(x),y) minus#(x,y) (15)
log#(s(s(x))) log#(s(quot(x,s(s(0))))) (16)
quot#(s(x),s(y)) quot#(minus(x,y),s(y)) (17)
le#(s(x),s(y)) le#(x,y) (18)
log#(s(s(x))) quot#(x,s(s(0))) (19)

1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.