Certification Problem

Input (TPDB TRS_Relative/Relative_05/rtL-wl1nz)

The relative rewrite relation R/S is considered where R is the following TRS

top(ok(new(x))) → top(check(x)) (1)
top(ok(old(x))) → top(check(x)) (2)

and S is the following TRS.

bot → new(bot) (3)
check(new(x)) → new(check(x)) (4)
check(old(x)) → old(check(x)) (5)
check(old(x)) → ok(old(x)) (6)
new(ok(x)) → ok(new(x)) (7)
old(ok(x)) → ok(old(x)) (8)

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
[old(x1)] =
1 0 0
0 0 0
1 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[ok(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[check(x1)] =
1 0 0
0 0 0
1 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[bot] =
0 0 0
1 0 0
0 0 0
[new(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[top(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
top(ok(old(x))) → top(check(x)) (2)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[old(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[ok(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[check(x1)] =
1 0 1
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[bot] =
1 0 0
0 0 0
0 0 0
[new(x1)] =
1 0 0
0 1 1
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[top(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
check(old(x)) → old(check(x)) (5)
check(old(x)) → ok(old(x)) (6)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[old(x1)] = 18 · x1 + 0
[ok(x1)] = 4 · x1 + 13
[check(x1)] = 6 · x1 + 0
[bot] = 0
[new(x1)] = 6 · x1 + 0
[top(x1)] = 10 · x1 + 4
all of the following rules can be deleted.
top(ok(new(x))) → top(check(x)) (1)
new(ok(x)) → ok(new(x)) (7)
old(ok(x)) → ok(old(x)) (8)

1.1.1.1 R is empty

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