Certification Problem

Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex2_Luc02a_GM)

The rewrite relation of the following TRS is considered.

a__terms(N) cons(recip(a__sqr(mark(N))),terms(s(N))) (1)
a__sqr(0) 0 (2)
a__sqr(s(X)) s(a__add(a__sqr(mark(X)),a__dbl(mark(X)))) (3)
a__dbl(0) 0 (4)
a__dbl(s(X)) s(s(a__dbl(mark(X)))) (5)
a__add(0,X) mark(X) (6)
a__add(s(X),Y) s(a__add(mark(X),mark(Y))) (7)
a__first(0,X) nil (8)
a__first(s(X),cons(Y,Z)) cons(mark(Y),first(X,Z)) (9)
mark(terms(X)) a__terms(mark(X)) (10)
mark(sqr(X)) a__sqr(mark(X)) (11)
mark(add(X1,X2)) a__add(mark(X1),mark(X2)) (12)
mark(dbl(X)) a__dbl(mark(X)) (13)
mark(first(X1,X2)) a__first(mark(X1),mark(X2)) (14)
mark(cons(X1,X2)) cons(mark(X1),X2) (15)
mark(recip(X)) recip(mark(X)) (16)
mark(s(X)) s(mark(X)) (17)
mark(0) 0 (18)
mark(nil) nil (19)
a__terms(X) terms(X) (20)
a__sqr(X) sqr(X) (21)
a__add(X1,X2) add(X1,X2) (22)
a__dbl(X) dbl(X) (23)
a__first(X1,X2) first(X1,X2) (24)
The evaluation strategy is innermost.

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a__terms#(N) a__sqr#(mark(N)) (25)
a__terms#(N) mark#(N) (26)
a__sqr#(s(X)) a__add#(a__sqr(mark(X)),a__dbl(mark(X))) (27)
a__sqr#(s(X)) a__sqr#(mark(X)) (28)
a__sqr#(s(X)) mark#(X) (29)
a__sqr#(s(X)) a__dbl#(mark(X)) (30)
a__dbl#(s(X)) a__dbl#(mark(X)) (31)
a__dbl#(s(X)) mark#(X) (32)
a__add#(0,X) mark#(X) (33)
a__add#(s(X),Y) a__add#(mark(X),mark(Y)) (34)
a__add#(s(X),Y) mark#(X) (35)
a__add#(s(X),Y) mark#(Y) (36)
a__first#(s(X),cons(Y,Z)) mark#(Y) (37)
mark#(terms(X)) a__terms#(mark(X)) (38)
mark#(terms(X)) mark#(X) (39)
mark#(sqr(X)) a__sqr#(mark(X)) (40)
mark#(sqr(X)) mark#(X) (41)
mark#(add(X1,X2)) a__add#(mark(X1),mark(X2)) (42)
mark#(add(X1,X2)) mark#(X1) (43)
mark#(add(X1,X2)) mark#(X2) (44)
mark#(dbl(X)) a__dbl#(mark(X)) (45)
mark#(dbl(X)) mark#(X) (46)
mark#(first(X1,X2)) a__first#(mark(X1),mark(X2)) (47)
mark#(first(X1,X2)) mark#(X1) (48)
mark#(first(X1,X2)) mark#(X2) (49)
mark#(cons(X1,X2)) mark#(X1) (50)
mark#(recip(X)) mark#(X) (51)
mark#(s(X)) mark#(X) (52)

1.1 Reduction Pair Processor

Using the
prec(a__terms#) = 4 stat(a__terms#) = lex
prec(a__sqr#) = 4 stat(a__sqr#) = lex
prec(mark#) = 0 stat(mark#) = mul
prec(s) = 0 stat(s) = mul
prec(a__add#) = 1 stat(a__add#) = lex
prec(a__sqr) = 4 stat(a__sqr) = lex
prec(a__dbl) = 2 stat(a__dbl) = lex
prec(0) = 2 stat(0) = mul
prec(a__first#) = 0 stat(a__first#) = mul
prec(terms) = 4 stat(terms) = lex
prec(sqr) = 4 stat(sqr) = lex
prec(add) = 3 stat(add) = mul
prec(dbl) = 2 stat(dbl) = lex
prec(first) = 2 stat(first) = mul
prec(a__terms) = 4 stat(a__terms) = lex
prec(a__add) = 3 stat(a__add) = mul
prec(a__first) = 2 stat(a__first) = mul
prec(nil) = 2 stat(nil) = mul

π(a__terms#) = [1]
π(a__sqr#) = [1]
π(mark) = 1
π(mark#) = [1]
π(s) = [1]
π(a__add#) = [2,1]
π(a__sqr) = [1]
π(a__dbl) = [1]
π(a__dbl#) = 1
π(0) = []
π(a__first#) = [2]
π(cons) = 1
π(terms) = [1]
π(sqr) = [1]
π(add) = [1,2]
π(dbl) = [1]
π(first) = [1,2]
π(recip) = 1
π(a__terms) = [1]
π(a__add) = [1,2]
π(a__first) = [1,2]
π(nil) = []

the pairs
a__terms#(N) mark#(N) (26)
a__sqr#(s(X)) a__add#(a__sqr(mark(X)),a__dbl(mark(X))) (27)
a__sqr#(s(X)) a__sqr#(mark(X)) (28)
a__sqr#(s(X)) mark#(X) (29)
a__sqr#(s(X)) a__dbl#(mark(X)) (30)
a__dbl#(s(X)) a__dbl#(mark(X)) (31)
a__add#(0,X) mark#(X) (33)
a__add#(s(X),Y) a__add#(mark(X),mark(Y)) (34)
a__add#(s(X),Y) mark#(X) (35)
a__add#(s(X),Y) mark#(Y) (36)
mark#(terms(X)) a__terms#(mark(X)) (38)
mark#(terms(X)) mark#(X) (39)
mark#(sqr(X)) a__sqr#(mark(X)) (40)
mark#(sqr(X)) mark#(X) (41)
mark#(add(X1,X2)) a__add#(mark(X1),mark(X2)) (42)
mark#(add(X1,X2)) mark#(X1) (43)
mark#(add(X1,X2)) mark#(X2) (44)
mark#(dbl(X)) a__dbl#(mark(X)) (45)
mark#(dbl(X)) mark#(X) (46)
mark#(first(X1,X2)) a__first#(mark(X1),mark(X2)) (47)
mark#(first(X1,X2)) mark#(X1) (48)
mark#(first(X1,X2)) mark#(X2) (49)
mark#(s(X)) mark#(X) (52)
could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.