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 NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

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

1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.