Certification Problem
Input (TPDB TRS_Relative/Relative_05/rt1-5)
The relative rewrite relation R/S is considered where R is the following TRS
s(a(x)) |
→ |
s(b(x)) |
(1) |
b(b(x)) |
→ |
a(x) |
(2) |
and S is the following TRS.
f(s(x),y) |
→ |
f(x,s(y)) |
(3) |
s(a(x)) |
→ |
a(s(x)) |
(4) |
s(b(x)) |
→ |
b(s(x)) |
(5) |
a(s(x)) |
→ |
s(a(x)) |
(6) |
b(s(x)) |
→ |
s(b(x)) |
(7) |
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
[s(x1)] |
= |
· x1 +
|
[f(x1, x2)] |
= |
· x1 + · x2 +
|
[a(x1)] |
= |
· x1 +
|
[b(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
[s(x1)] |
= |
· x1 +
|
[f(x1, x2)] |
= |
· x1 + · x2 +
|
[a(x1)] |
= |
· x1 +
|
[b(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
f(s(x),y) |
→ |
f(x,s(y)) |
(3) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
[s(x1)] |
= |
· x1 +
|
[a(x1)] |
= |
· x1 +
|
[b(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.