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.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

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

1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[top(x1)] =
0
0
+
2 2
0 2
· x1
[ok(x1)] =
0
2
+
2 0
0 1
· x1
[new(x1)] =
0
0
+
1 0
0 2
· x1
[check(x1)] =
0
0
+
2 0
0 2
· x1
[old(x1)] =
0
2
+
1 0
2 1
· x1
[bot] =
0
0
all of the following rules can be deleted.
top(ok(new(x))) → top(check(x)) (1)

1.1.1 R is empty

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