Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/ExSec4_2_DLMMU04_GM)

The rewrite relation of the following TRS is considered.

a__natsFrom(N) cons(mark(N),natsFrom(s(N))) (1)
a__fst(pair(XS,YS)) mark(XS) (2)
a__snd(pair(XS,YS)) mark(YS) (3)
a__splitAt(0,XS) pair(nil,mark(XS)) (4)
a__splitAt(s(N),cons(X,XS)) a__u(a__splitAt(mark(N),mark(XS)),N,X,XS) (5)
a__u(pair(YS,ZS),N,X,XS) pair(cons(mark(X),YS),mark(ZS)) (6)
a__head(cons(N,XS)) mark(N) (7)
a__tail(cons(N,XS)) mark(XS) (8)
a__sel(N,XS) a__head(a__afterNth(mark(N),mark(XS))) (9)
a__take(N,XS) a__fst(a__splitAt(mark(N),mark(XS))) (10)
a__afterNth(N,XS) a__snd(a__splitAt(mark(N),mark(XS))) (11)
mark(natsFrom(X)) a__natsFrom(mark(X)) (12)
mark(fst(X)) a__fst(mark(X)) (13)
mark(snd(X)) a__snd(mark(X)) (14)
mark(splitAt(X1,X2)) a__splitAt(mark(X1),mark(X2)) (15)
mark(u(X1,X2,X3,X4)) a__u(mark(X1),X2,X3,X4) (16)
mark(head(X)) a__head(mark(X)) (17)
mark(tail(X)) a__tail(mark(X)) (18)
mark(sel(X1,X2)) a__sel(mark(X1),mark(X2)) (19)
mark(afterNth(X1,X2)) a__afterNth(mark(X1),mark(X2)) (20)
mark(take(X1,X2)) a__take(mark(X1),mark(X2)) (21)
mark(cons(X1,X2)) cons(mark(X1),X2) (22)
mark(s(X)) s(mark(X)) (23)
mark(pair(X1,X2)) pair(mark(X1),mark(X2)) (24)
mark(0) 0 (25)
mark(nil) nil (26)
a__natsFrom(X) natsFrom(X) (27)
a__fst(X) fst(X) (28)
a__snd(X) snd(X) (29)
a__splitAt(X1,X2) splitAt(X1,X2) (30)
a__u(X1,X2,X3,X4) u(X1,X2,X3,X4) (31)
a__head(X) head(X) (32)
a__tail(X) tail(X) (33)
a__sel(X1,X2) sel(X1,X2) (34)
a__afterNth(X1,X2) afterNth(X1,X2) (35)
a__take(X1,X2) take(X1,X2) (36)

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.
a__u#(pair(YS,ZS),N,X,XS) mark#(X) (37)
mark#(u(X1,X2,X3,X4)) mark#(X1) (38)
a__splitAt#(s(N),cons(X,XS)) a__u#(a__splitAt(mark(N),mark(XS)),N,X,XS) (39)
mark#(head(X)) mark#(X) (40)
a__tail#(cons(N,XS)) mark#(XS) (41)
mark#(u(X1,X2,X3,X4)) a__u#(mark(X1),X2,X3,X4) (42)
mark#(take(X1,X2)) mark#(X2) (43)
mark#(sel(X1,X2)) mark#(X1) (44)
mark#(splitAt(X1,X2)) a__splitAt#(mark(X1),mark(X2)) (45)
mark#(splitAt(X1,X2)) mark#(X1) (46)
mark#(sel(X1,X2)) mark#(X2) (47)
a__sel#(N,XS) a__head#(a__afterNth(mark(N),mark(XS))) (48)
mark#(take(X1,X2)) mark#(X1) (49)
mark#(tail(X)) a__tail#(mark(X)) (50)
a__afterNth#(N,XS) mark#(XS) (51)
a__sel#(N,XS) mark#(XS) (52)
a__afterNth#(N,XS) a__splitAt#(mark(N),mark(XS)) (53)
a__head#(cons(N,XS)) mark#(N) (54)
mark#(afterNth(X1,X2)) mark#(X2) (55)
a__natsFrom#(N) mark#(N) (56)
a__afterNth#(N,XS) mark#(N) (57)
a__splitAt#(s(N),cons(X,XS)) mark#(N) (58)
mark#(pair(X1,X2)) mark#(X1) (59)
a__splitAt#(s(N),cons(X,XS)) a__splitAt#(mark(N),mark(XS)) (60)
mark#(tail(X)) mark#(X) (61)
mark#(splitAt(X1,X2)) mark#(X2) (62)
a__take#(N,XS) a__fst#(a__splitAt(mark(N),mark(XS))) (63)
mark#(snd(X)) mark#(X) (64)
a__sel#(N,XS) a__afterNth#(mark(N),mark(XS)) (65)
mark#(take(X1,X2)) a__take#(mark(X1),mark(X2)) (66)
a__afterNth#(N,XS) a__snd#(a__splitAt(mark(N),mark(XS))) (67)
mark#(cons(X1,X2)) mark#(X1) (68)
mark#(fst(X)) mark#(X) (69)
a__snd#(pair(XS,YS)) mark#(YS) (70)
a__take#(N,XS) mark#(XS) (71)
mark#(afterNth(X1,X2)) mark#(X1) (72)
mark#(head(X)) a__head#(mark(X)) (73)
a__take#(N,XS) mark#(N) (74)
mark#(natsFrom(X)) a__natsFrom#(mark(X)) (75)
a__splitAt#(s(N),cons(X,XS)) mark#(XS) (76)
mark#(snd(X)) a__snd#(mark(X)) (77)
a__take#(N,XS) a__splitAt#(mark(N),mark(XS)) (78)
mark#(sel(X1,X2)) a__sel#(mark(X1),mark(X2)) (79)
mark#(afterNth(X1,X2)) a__afterNth#(mark(X1),mark(X2)) (80)
mark#(s(X)) mark#(X) (81)
a__u#(pair(YS,ZS),N,X,XS) mark#(ZS) (82)
a__fst#(pair(XS,YS)) mark#(XS) (83)
a__sel#(N,XS) mark#(N) (84)
mark#(pair(X1,X2)) mark#(X2) (85)
mark#(fst(X)) a__fst#(mark(X)) (86)
a__splitAt#(0,XS) mark#(XS) (87)
mark#(natsFrom(X)) mark#(X) (88)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.