Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex4_Zan97_GM)

The rewrite relation of the following TRS is considered.

a__f(X) cons(mark(X),f(g(X))) (1)
a__g(0) s(0) (2)
a__g(s(X)) s(s(a__g(mark(X)))) (3)
a__sel(0,cons(X,Y)) mark(X) (4)
a__sel(s(X),cons(Y,Z)) a__sel(mark(X),mark(Z)) (5)
mark(f(X)) a__f(mark(X)) (6)
mark(g(X)) a__g(mark(X)) (7)
mark(sel(X1,X2)) a__sel(mark(X1),mark(X2)) (8)
mark(cons(X1,X2)) cons(mark(X1),X2) (9)
mark(0) 0 (10)
mark(s(X)) s(mark(X)) (11)
a__f(X) f(X) (12)
a__g(X) g(X) (13)
a__sel(X1,X2) sel(X1,X2) (14)

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.
mark#(f(X)) mark#(X) (15)
mark#(g(X)) mark#(X) (16)
mark#(sel(X1,X2)) a__sel#(mark(X1),mark(X2)) (17)
a__sel#(s(X),cons(Y,Z)) mark#(Z) (18)
a__g#(s(X)) mark#(X) (19)
a__f#(X) mark#(X) (20)
mark#(sel(X1,X2)) mark#(X1) (21)
a__sel#(s(X),cons(Y,Z)) a__sel#(mark(X),mark(Z)) (22)
a__g#(s(X)) a__g#(mark(X)) (23)
mark#(g(X)) a__g#(mark(X)) (24)
a__sel#(0,cons(X,Y)) mark#(X) (25)
mark#(cons(X1,X2)) mark#(X1) (26)
mark#(f(X)) a__f#(mark(X)) (27)
a__sel#(s(X),cons(Y,Z)) mark#(X) (28)
mark#(sel(X1,X2)) mark#(X2) (29)
mark#(s(X)) mark#(X) (30)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.