Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex3_3_25_Bor03_GM)

The rewrite relation of the following TRS is considered.

a__app(nil,YS) mark(YS) (1)
a__app(cons(X,XS),YS) cons(mark(X),app(XS,YS)) (2)
a__from(X) cons(mark(X),from(s(X))) (3)
a__zWadr(nil,YS) nil (4)
a__zWadr(XS,nil) nil (5)
a__zWadr(cons(X,XS),cons(Y,YS)) cons(a__app(mark(Y),cons(mark(X),nil)),zWadr(XS,YS)) (6)
a__prefix(L) cons(nil,zWadr(L,prefix(L))) (7)
mark(app(X1,X2)) a__app(mark(X1),mark(X2)) (8)
mark(from(X)) a__from(mark(X)) (9)
mark(zWadr(X1,X2)) a__zWadr(mark(X1),mark(X2)) (10)
mark(prefix(X)) a__prefix(mark(X)) (11)
mark(nil) nil (12)
mark(cons(X1,X2)) cons(mark(X1),X2) (13)
mark(s(X)) s(mark(X)) (14)
a__app(X1,X2) app(X1,X2) (15)
a__from(X) from(X) (16)
a__zWadr(X1,X2) zWadr(X1,X2) (17)
a__prefix(X) prefix(X) (18)

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.
a__app#(nil,YS) mark#(YS) (19)
a__app#(cons(X,XS),YS) mark#(X) (20)
a__from#(X) mark#(X) (21)
a__zWadr#(cons(X,XS),cons(Y,YS)) mark#(X) (22)
a__zWadr#(cons(X,XS),cons(Y,YS)) mark#(Y) (23)
a__zWadr#(cons(X,XS),cons(Y,YS)) a__app#(mark(Y),cons(mark(X),nil)) (24)
mark#(app(X1,X2)) mark#(X2) (25)
mark#(app(X1,X2)) mark#(X1) (26)
mark#(app(X1,X2)) a__app#(mark(X1),mark(X2)) (27)
mark#(from(X)) mark#(X) (28)
mark#(from(X)) a__from#(mark(X)) (29)
mark#(zWadr(X1,X2)) mark#(X2) (30)
mark#(zWadr(X1,X2)) mark#(X1) (31)
mark#(zWadr(X1,X2)) a__zWadr#(mark(X1),mark(X2)) (32)
mark#(prefix(X)) mark#(X) (33)
mark#(prefix(X)) a__prefix#(mark(X)) (34)
mark#(cons(X1,X2)) mark#(X1) (35)
mark#(s(X)) mark#(X) (36)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.