Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex4_7_37_Bor03_Z)

The rewrite relation of the following TRS is considered.

from(X) cons(X,n__from(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)
zWquot(X1,X2) n__zWquot(X1,X2) (12)
activate(n__from(X)) from(X) (13)
activate(n__zWquot(X1,X2)) zWquot(X1,X2) (14)
activate(X) X (15)

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) (16)
sel#(s(N),cons(X,XS)) sel#(N,activate(XS)) (17)
minus#(s(X),s(Y)) minus#(X,Y) (18)
quot#(s(X),s(Y)) minus#(X,Y) (19)
quot#(s(X),s(Y)) quot#(minus(X,Y),s(Y)) (20)
zWquot#(cons(X,XS),cons(Y,YS)) activate#(YS) (21)
zWquot#(cons(X,XS),cons(Y,YS)) activate#(XS) (22)
zWquot#(cons(X,XS),cons(Y,YS)) quot#(X,Y) (23)
activate#(n__from(X)) from#(X) (24)
activate#(n__zWquot(X1,X2)) zWquot#(X1,X2) (25)

1.1 Dependency Graph Processor

The dependency pairs are split into 3 components.