Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex4_7_37_Bor03_FR)

The rewrite relation of the following TRS is considered.

from(X) cons(X,n__from(n__s(X))) (1)
sel(0,cons(X,XS)) X (2)
sel(s(N),cons(X,XS)) sel(N,activate(XS)) (3)
minus(X,0) 0 (4)
minus(s(X),s(Y)) minus(X,Y) (5)
quot(0,s(Y)) 0 (6)
quot(s(X),s(Y)) s(quot(minus(X,Y),s(Y))) (7)
zWquot(XS,nil) nil (8)
zWquot(nil,XS) nil (9)
zWquot(cons(X,XS),cons(Y,YS)) cons(quot(X,Y),n__zWquot(activate(XS),activate(YS))) (10)
from(X) n__from(X) (11)
s(X) n__s(X) (12)
zWquot(X1,X2) n__zWquot(X1,X2) (13)
activate(n__from(X)) from(activate(X)) (14)
activate(n__s(X)) s(activate(X)) (15)
activate(n__zWquot(X1,X2)) zWquot(activate(X1),activate(X2)) (16)
activate(X) X (17)

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.
sel#(s(N),cons(X,XS)) activate#(XS) (18)
sel#(s(N),cons(X,XS)) sel#(N,activate(XS)) (19)
minus#(s(X),s(Y)) minus#(X,Y) (20)
quot#(s(X),s(Y)) minus#(X,Y) (21)
quot#(s(X),s(Y)) quot#(minus(X,Y),s(Y)) (22)
quot#(s(X),s(Y)) s#(quot(minus(X,Y),s(Y))) (23)
zWquot#(cons(X,XS),cons(Y,YS)) activate#(YS) (24)
zWquot#(cons(X,XS),cons(Y,YS)) activate#(XS) (25)
zWquot#(cons(X,XS),cons(Y,YS)) quot#(X,Y) (26)
activate#(n__from(X)) activate#(X) (27)
activate#(n__from(X)) from#(activate(X)) (28)
activate#(n__s(X)) activate#(X) (29)
activate#(n__s(X)) s#(activate(X)) (30)
activate#(n__zWquot(X1,X2)) activate#(X2) (31)
activate#(n__zWquot(X1,X2)) activate#(X1) (32)
activate#(n__zWquot(X1,X2)) zWquot#(activate(X1),activate(X2)) (33)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.