Certification Problem

Input (TPDB TRS_Standard/AProVE_06/tower)

The rewrite relation of the following TRS is considered.

plus(0,x) x (1)
plus(s(x),y) s(plus(p(s(x)),y)) (2)
times(0,y) 0 (3)
times(s(x),y) plus(y,times(p(s(x)),y)) (4)
exp(x,0) s(0) (5)
exp(x,s(y)) times(x,exp(x,y)) (6)
p(s(0)) 0 (7)
p(s(s(x))) s(p(s(x))) (8)
tower(x,y) towerIter(x,y,s(0)) (9)
towerIter(0,y,z) z (10)
towerIter(s(x),y,z) towerIter(p(s(x)),y,exp(y,z)) (11)

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.
plus#(s(x),y) p#(s(x)) (12)
plus#(s(x),y) plus#(p(s(x)),y) (13)
times#(s(x),y) p#(s(x)) (14)
times#(s(x),y) times#(p(s(x)),y) (15)
times#(s(x),y) plus#(y,times(p(s(x)),y)) (16)
exp#(x,s(y)) exp#(x,y) (17)
exp#(x,s(y)) times#(x,exp(x,y)) (18)
p#(s(s(x))) p#(s(x)) (19)
tower#(x,y) towerIter#(x,y,s(0)) (20)
towerIter#(s(x),y,z) exp#(y,z) (21)
towerIter#(s(x),y,z) p#(s(x)) (22)
towerIter#(s(x),y,z) towerIter#(p(s(x)),y,exp(y,z)) (23)

1.1 Dependency Graph Processor

The dependency pairs are split into 5 components.