Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex1_2_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,n__cons(Y,Z))) rcons(posrecip(activate(Y)),2ndsneg(N,activate(Z))) (3)
2ndsneg(0,Z) rnil (4)
2ndsneg(s(N),cons(X,n__cons(Y,Z))) rcons(negrecip(activate(Y)),2ndspos(N,activate(Z))) (5)
pi(X) 2ndspos(X,from(0)) (6)
plus(0,Y) Y (7)
plus(s(X),Y) s(plus(X,Y)) (8)
times(0,Y) 0 (9)
times(s(X),Y) plus(Y,times(X,Y)) (10)
square(X) times(X,X) (11)
from(X) n__from(X) (12)
s(X) n__s(X) (13)
cons(X1,X2) n__cons(X1,X2) (14)
activate(n__from(X)) from(activate(X)) (15)
activate(n__s(X)) s(activate(X)) (16)
activate(n__cons(X1,X2)) cons(activate(X1),X2) (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.
from#(X) cons#(X,n__from(n__s(X))) (19)
2ndspos#(s(N),cons(X,n__cons(Y,Z))) activate#(Z) (20)
2ndspos#(s(N),cons(X,n__cons(Y,Z))) 2ndsneg#(N,activate(Z)) (21)
2ndspos#(s(N),cons(X,n__cons(Y,Z))) activate#(Y) (22)
2ndsneg#(s(N),cons(X,n__cons(Y,Z))) activate#(Z) (23)
2ndsneg#(s(N),cons(X,n__cons(Y,Z))) 2ndspos#(N,activate(Z)) (24)
2ndsneg#(s(N),cons(X,n__cons(Y,Z))) activate#(Y) (25)
pi#(X) from#(0) (26)
pi#(X) 2ndspos#(X,from(0)) (27)
plus#(s(X),Y) plus#(X,Y) (28)
plus#(s(X),Y) s#(plus(X,Y)) (29)
times#(s(X),Y) times#(X,Y) (30)
times#(s(X),Y) plus#(Y,times(X,Y)) (31)
square#(X) times#(X,X) (32)
activate#(n__from(X)) activate#(X) (33)
activate#(n__from(X)) from#(activate(X)) (34)
activate#(n__s(X)) activate#(X) (35)
activate#(n__s(X)) s#(activate(X)) (36)
activate#(n__cons(X1,X2)) activate#(X1) (37)
activate#(n__cons(X1,X2)) cons#(activate(X1),X2) (38)

1.1 Dependency Graph Processor

The dependency pairs are split into 4 components.