Certification Problem

Input (TPDB TRS_Standard/Mixed_TRS/jones6)

The rewrite relation of the following TRS is considered.

f(a,empty) → g(a,empty) (1)
f(a,cons(x,k)) → f(cons(x,a),k) (2)
g(empty,d) → d (3)
g(cons(x,k),d) → g(k,cons(x,d)) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1, x2)] =
1 1 0
1 0 0
0 0 0
· x1 +
1 1 0
1 0 1
0 0 0
· x2 +
0 0 0
1 0 0
1 0 0
[cons(x1, x2)] =
1 1 0
0 0 0
0 0 0
· x1 +
1 0 0
0 1 0
0 0 1
· x2 +
1 0 0
1 0 0
0 0 0
[empty] =
0 0 0
1 0 0
1 0 0
[g(x1, x2)] =
1 1 0
1 0 0
0 0 0
· x1 +
1 0 1
0 1 1
0 0 1
· x2 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
g(empty,d) → d (3)
g(cons(x,k),d) → g(k,cons(x,d)) (4)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 1 1
0 0 0
0 0 0
· x2 +
1 0 0
0 0 0
0 0 0
[cons(x1, x2)] =
1 1 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 1
0 1 0
· x2 +
1 0 0
0 0 0
1 0 0
[empty] =
0 0 0
0 0 0
1 0 0
[g(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 1
0 0 0
0 0 0
· x2 +
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
f(a,cons(x,k)) → f(cons(x,a),k) (2)

1.1.1 Rule Removal

Using the Weighted Path Order with the following precedence and status
prec(g) = 0 status(g) = [2, 1] list-extension(g) = Lex
prec(f) = 0 status(f) = [2, 1] list-extension(f) = Lex
prec(empty) = 0 status(empty) = [] list-extension(empty) = Lex
and the following Max-polynomial interpretation
[g(x1, x2)] = max(2, 2 + 1 · x1, 2 + 1 · x2)
[f(x1, x2)] = max(6, 4 + 1 · x1, 2 + 1 · x2)
[empty] = max(1)
all of the following rules can be deleted.
f(a,empty) → g(a,empty) (1)

1.1.1.1 R is empty

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