Certification Problem

Input (TPDB TRS_Standard/Mixed_TRS/jones2)

The rewrite relation of the following TRS is considered.

f(empty,l) → l (1)
f(cons(x,k),l) → g(k,l,cons(x,k)) (2)
g(a,b,c) → f(a,cons(b,c)) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[f(x1, x2)] = 24 · x1 + 4 · x2 + 24
[g(x1, x2, x3)] = 26 · x1 + 4 · x2 + 16 · x3 + 24
[empty] = 2
[cons(x1, x2)] = 1 · x1 + 4 · x2 + 0
all of the following rules can be deleted.
f(empty,l) → l (1)

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 1 1
1 0 1
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
0 0 0
0 0 0
0 0 0
[g(x1, x2, x3)] =
1 1 1
1 0 1
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
· x2 +
1 0 1
0 0 0
0 0 0
· x3 +
0 0 0
0 0 0
0 0 0
[cons(x1, x2)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 1
1 1 1
0 1 1
· x2 +
0 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
f(cons(x,k),l) → g(k,l,cons(x,k)) (2)

1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(g) = 1 weight(g) = 4
prec(cons) = 3 weight(cons) = 2
prec(f) = 0 weight(f) = 2
all of the following rules can be deleted.
g(a,b,c) → f(a,cons(b,c)) (3)

1.1.1.1 R is empty

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