Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex4_7_56_Bor03_Z)

The rewrite relation of the following TRS is considered.

from(X) cons(X,n__from(s(X))) (1)
after(0,XS) XS (2)
after(s(N),cons(X,XS)) after(N,activate(XS)) (3)
from(X) n__from(X) (4)
activate(n__from(X)) from(X) (5)
activate(X) X (6)

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(activate) = 5 status(activate) = [1] list-extension(activate) = Lex
prec(after) = 6 status(after) = [1, 2] list-extension(after) = Lex
prec(0) = 0 status(0) = [] list-extension(0) = Lex
prec(cons) = 0 status(cons) = [2, 1] list-extension(cons) = Lex
prec(n__from) = 0 status(n__from) = [1] list-extension(n__from) = Lex
prec(s) = 0 status(s) = [1] list-extension(s) = Lex
prec(from) = 4 status(from) = [1] list-extension(from) = Lex
and the following Max-polynomial interpretation
[activate(x1)] = max(1, 0 + 1 · x1)
[after(x1, x2)] = max(2, 0 + 1 · x1, 0 + 1 · x2)
[0] = max(0)
[cons(x1, x2)] = max(1, 0 + 1 · x1, 0 + 1 · x2)
[n__from(x1)] = max(1, 0 + 1 · x1)
[s(x1)] = max(0, 0 + 1 · x1)
[from(x1)] = max(1, 0 + 1 · x1)
all of the following rules can be deleted.
from(X) cons(X,n__from(s(X))) (1)
after(0,XS) XS (2)
after(s(N),cons(X,XS)) after(N,activate(XS)) (3)
from(X) n__from(X) (4)
activate(n__from(X)) from(X) (5)
activate(X) X (6)

1.1 R is empty

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