Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex1_2_Luc02c_GM)

The rewrite relation of the following TRS is considered.

a__2nd(cons(X,cons(Y,Z))) mark(Y) (1)
a__from(X) cons(mark(X),from(s(X))) (2)
mark(2nd(X)) a__2nd(mark(X)) (3)
mark(from(X)) a__from(mark(X)) (4)
mark(cons(X1,X2)) cons(mark(X1),X2) (5)
mark(s(X)) s(mark(X)) (6)
a__2nd(X) 2nd(X) (7)
a__from(X) from(X) (8)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the Weighted Path Order with the following precedence and status
prec(2nd) = 0 status(2nd) = [1] list-extension(2nd) = Lex
prec(from) = 0 status(from) = [1] list-extension(from) = Lex
prec(s) = 0 status(s) = [1] list-extension(s) = Lex
prec(a__from) = 1 status(a__from) = [1] list-extension(a__from) = Lex
prec(mark) = 3 status(mark) = [1] list-extension(mark) = Lex
prec(a__2nd) = 2 status(a__2nd) = [1] list-extension(a__2nd) = Lex
prec(cons) = 0 status(cons) = [2, 1] list-extension(cons) = Lex
and the following Max-polynomial interpretation
[2nd(x1)] = max(0, 4 + 1 · x1)
[from(x1)] = 1 + 1 · x1
[s(x1)] = max(0, 0 + 1 · x1)
[a__from(x1)] = max(0, 1 + 1 · x1)
[mark(x1)] = max(0, 0 + 1 · x1)
[a__2nd(x1)] = max(0, 4 + 1 · x1)
[cons(x1, x2)] = max(0, 0 + 1 · x1, 0 + 1 · x2)
all of the following rules can be deleted.
a__2nd(cons(X,cons(Y,Z))) mark(Y) (1)
a__from(X) cons(mark(X),from(s(X))) (2)
mark(2nd(X)) a__2nd(mark(X)) (3)
mark(from(X)) a__from(mark(X)) (4)
mark(cons(X1,X2)) cons(mark(X1),X2) (5)
mark(s(X)) s(mark(X)) (6)
a__2nd(X) 2nd(X) (7)
a__from(X) from(X) (8)

1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.