Certification Problem

Input (TPDB TRS_Standard/AG01/#3.19)

The rewrite relation of the following TRS is considered.

minus(x,0) x (1)
minus(s(x),s(y)) minus(x,y) (2)
double(0) 0 (3)
double(s(x)) s(s(double(x))) (4)
plus(0,y) y (5)
plus(s(x),y) s(plus(x,y)) (6)
plus(s(x),y) plus(x,s(y)) (7)
plus(s(x),y) s(plus(minus(x,y),double(y))) (8)
plus(s(plus(x,y)),z) s(plus(plus(x,y),z)) (9)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
minus#(s(x),s(y)) minus#(x,y) (10)
double#(s(x)) double#(x) (11)
plus#(s(x),y) plus#(x,y) (12)
plus#(s(x),y) plus#(x,s(y)) (13)
plus#(s(x),y) double#(y) (14)
plus#(s(x),y) minus#(x,y) (15)
plus#(s(x),y) plus#(minus(x,y),double(y)) (16)
plus#(s(plus(x,y)),z) plus#(plus(x,y),z) (17)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.