Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/ExAppendixB_AEL03_FR)

The rewrite relation of the following TRS is considered.

from(X) cons(X,n__from(n__s(X))) (1)
2ndspos(0,Z) rnil (2)
2ndspos(s(N),cons(X,Z)) 2ndspos(s(N),cons2(X,activate(Z))) (3)
2ndspos(s(N),cons2(X,cons(Y,Z))) rcons(posrecip(Y),2ndsneg(N,activate(Z))) (4)
2ndsneg(0,Z) rnil (5)
2ndsneg(s(N),cons(X,Z)) 2ndsneg(s(N),cons2(X,activate(Z))) (6)
2ndsneg(s(N),cons2(X,cons(Y,Z))) rcons(negrecip(Y),2ndspos(N,activate(Z))) (7)
pi(X) 2ndspos(X,from(0)) (8)
plus(0,Y) Y (9)
plus(s(X),Y) s(plus(X,Y)) (10)
times(0,Y) 0 (11)
times(s(X),Y) plus(Y,times(X,Y)) (12)
square(X) times(X,X) (13)
from(X) n__from(X) (14)
s(X) n__s(X) (15)
activate(n__from(X)) from(activate(X)) (16)
activate(n__s(X)) s(activate(X)) (17)
activate(X) X (18)

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.
2ndspos#(s(N),cons(X,Z)) activate#(Z) (19)
2ndspos#(s(N),cons(X,Z)) 2ndspos#(s(N),cons2(X,activate(Z))) (20)
2ndspos#(s(N),cons2(X,cons(Y,Z))) activate#(Z) (21)
2ndspos#(s(N),cons2(X,cons(Y,Z))) 2ndsneg#(N,activate(Z)) (22)
2ndsneg#(s(N),cons(X,Z)) activate#(Z) (23)
2ndsneg#(s(N),cons(X,Z)) 2ndsneg#(s(N),cons2(X,activate(Z))) (24)
2ndsneg#(s(N),cons2(X,cons(Y,Z))) activate#(Z) (25)
2ndsneg#(s(N),cons2(X,cons(Y,Z))) 2ndspos#(N,activate(Z)) (26)
pi#(X) from#(0) (27)
pi#(X) 2ndspos#(X,from(0)) (28)
plus#(s(X),Y) plus#(X,Y) (29)
plus#(s(X),Y) s#(plus(X,Y)) (30)
times#(s(X),Y) times#(X,Y) (31)
times#(s(X),Y) plus#(Y,times(X,Y)) (32)
square#(X) times#(X,X) (33)
activate#(n__from(X)) activate#(X) (34)
activate#(n__from(X)) from#(activate(X)) (35)
activate#(n__s(X)) activate#(X) (36)
activate#(n__s(X)) s#(activate(X)) (37)

1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.