Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex1_GL02a_iGM)

The rewrite relation of the following TRS is considered.

active(eq(0,0)) mark(true) (1)
active(eq(s(X),s(Y))) mark(eq(X,Y)) (2)
active(eq(X,Y)) mark(false) (3)
active(inf(X)) mark(cons(X,inf(s(X)))) (4)
active(take(0,X)) mark(nil) (5)
active(take(s(X),cons(Y,L))) mark(cons(Y,take(X,L))) (6)
active(length(nil)) mark(0) (7)
active(length(cons(X,L))) mark(s(length(L))) (8)
mark(eq(X1,X2)) active(eq(X1,X2)) (9)
mark(0) active(0) (10)
mark(true) active(true) (11)
mark(s(X)) active(s(X)) (12)
mark(false) active(false) (13)
mark(inf(X)) active(inf(mark(X))) (14)
mark(cons(X1,X2)) active(cons(X1,X2)) (15)
mark(take(X1,X2)) active(take(mark(X1),mark(X2))) (16)
mark(nil) active(nil) (17)
mark(length(X)) active(length(mark(X))) (18)
eq(mark(X1),X2) eq(X1,X2) (19)
eq(X1,mark(X2)) eq(X1,X2) (20)
eq(active(X1),X2) eq(X1,X2) (21)
eq(X1,active(X2)) eq(X1,X2) (22)
s(mark(X)) s(X) (23)
s(active(X)) s(X) (24)
inf(mark(X)) inf(X) (25)
inf(active(X)) inf(X) (26)
cons(mark(X1),X2) cons(X1,X2) (27)
cons(X1,mark(X2)) cons(X1,X2) (28)
cons(active(X1),X2) cons(X1,X2) (29)
cons(X1,active(X2)) cons(X1,X2) (30)
take(mark(X1),X2) take(X1,X2) (31)
take(X1,mark(X2)) take(X1,X2) (32)
take(active(X1),X2) take(X1,X2) (33)
take(X1,active(X2)) take(X1,X2) (34)
length(mark(X)) length(X) (35)
length(active(X)) length(X) (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.
active#(eq(s(X),s(Y))) eq#(X,Y) (37)
active#(length(cons(X,L))) length#(L) (38)
mark#(nil) active#(nil) (39)
mark#(take(X1,X2)) active#(take(mark(X1),mark(X2))) (40)
active#(inf(X)) cons#(X,inf(s(X))) (41)
active#(length(cons(X,L))) s#(length(L)) (42)
active#(length(cons(X,L))) mark#(s(length(L))) (43)
mark#(take(X1,X2)) mark#(X1) (44)
active#(inf(X)) inf#(s(X)) (45)
active#(inf(X)) s#(X) (46)
mark#(take(X1,X2)) mark#(X2) (47)
active#(take(s(X),cons(Y,L))) cons#(Y,take(X,L)) (48)
active#(eq(0,0)) mark#(true) (49)
mark#(length(X)) mark#(X) (50)
mark#(s(X)) active#(s(X)) (51)
mark#(eq(X1,X2)) active#(eq(X1,X2)) (52)
s#(active(X)) s#(X) (53)
active#(take(0,X)) mark#(nil) (54)
take#(active(X1),X2) take#(X1,X2) (55)
active#(inf(X)) mark#(cons(X,inf(s(X)))) (56)
s#(mark(X)) s#(X) (57)
eq#(mark(X1),X2) eq#(X1,X2) (58)
take#(mark(X1),X2) take#(X1,X2) (59)
take#(X1,mark(X2)) take#(X1,X2) (60)
mark#(length(X)) active#(length(mark(X))) (61)
cons#(X1,mark(X2)) cons#(X1,X2) (62)
eq#(X1,mark(X2)) eq#(X1,X2) (63)
active#(take(s(X),cons(Y,L))) take#(X,L) (64)
active#(eq(X,Y)) mark#(false) (65)
mark#(true) active#(true) (66)
length#(active(X)) length#(X) (67)
active#(take(s(X),cons(Y,L))) mark#(cons(Y,take(X,L))) (68)
mark#(cons(X1,X2)) active#(cons(X1,X2)) (69)
cons#(mark(X1),X2) cons#(X1,X2) (70)
mark#(0) active#(0) (71)
eq#(active(X1),X2) eq#(X1,X2) (72)
take#(X1,active(X2)) take#(X1,X2) (73)
mark#(inf(X)) mark#(X) (74)
inf#(active(X)) inf#(X) (75)
inf#(mark(X)) inf#(X) (76)
eq#(X1,active(X2)) eq#(X1,X2) (77)
mark#(take(X1,X2)) take#(mark(X1),mark(X2)) (78)
active#(length(nil)) mark#(0) (79)
mark#(inf(X)) inf#(mark(X)) (80)
cons#(active(X1),X2) cons#(X1,X2) (81)
active#(eq(s(X),s(Y))) mark#(eq(X,Y)) (82)
mark#(false) active#(false) (83)
mark#(inf(X)) active#(inf(mark(X))) (84)
length#(mark(X)) length#(X) (85)
mark#(length(X)) length#(mark(X)) (86)
cons#(X1,active(X2)) cons#(X1,X2) (87)

1.1 Dependency Graph Processor

The dependency pairs are split into 7 components.