Certification Problem
Input (TPDB TRS_Standard/Various_04/24)
The rewrite relation of the following TRS is considered.
|
max(L(x)) |
→ |
x |
(1) |
|
max(N(L(0),L(y))) |
→ |
y |
(2) |
|
max(N(L(s(x)),L(s(y)))) |
→ |
s(max(N(L(x),L(y)))) |
(3) |
|
max(N(L(x),N(y,z))) |
→ |
max(N(L(x),L(max(N(y,z))))) |
(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
| [s(x1)] |
= |
· x1 +
|
| [max(x1)] |
= |
· x1 +
|
| [N(x1, x2)] |
= |
· x1 + · x2 +
|
| [L(x1)] |
= |
· x1 +
|
| [0] |
= |
|
all of the following rules can be deleted.
|
max(N(L(0),L(y))) |
→ |
y |
(2) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [s(x1)] |
= |
· x1 +
|
| [max(x1)] |
= |
· x1 +
|
| [N(x1, x2)] |
= |
· x1 + · x2 +
|
| [L(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
max(N(L(s(x)),L(s(y)))) |
→ |
s(max(N(L(x),L(y)))) |
(3) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [max(x1)] |
= |
· x1 +
|
| [N(x1, x2)] |
= |
· x1 + · x2 +
|
| [L(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [max(x1)] |
= |
· x1 +
|
| [N(x1, x2)] |
= |
· x1 + · x2 +
|
| [L(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
|
max(N(L(x),N(y,z))) |
→ |
max(N(L(x),L(max(N(y,z))))) |
(4) |
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.