Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex1_2_Luc02c_Z)
The rewrite relation of the following TRS is considered.
|
2nd(cons(X,n__cons(Y,Z))) |
→ |
activate(Y) |
(1) |
|
from(X) |
→ |
cons(X,n__from(s(X))) |
(2) |
|
cons(X1,X2) |
→ |
n__cons(X1,X2) |
(3) |
|
from(X) |
→ |
n__from(X) |
(4) |
|
activate(n__cons(X1,X2)) |
→ |
cons(X1,X2) |
(5) |
|
activate(n__from(X)) |
→ |
from(X) |
(6) |
|
activate(X) |
→ |
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 the naturals
| [from(x1)] |
= |
8 · x1 + 2 |
| [cons(x1, x2)] |
= |
4 · x1 + 1 · x2 + 0 |
| [activate(x1)] |
= |
19 · x1 + 27 |
| [s(x1)] |
= |
1 · x1 + 0 |
| [n__cons(x1, x2)] |
= |
4 · x1 + 1 · x2 + 0 |
| [n__from(x1)] |
= |
4 · x1 + 2 |
| [2nd(x1)] |
= |
18 · x1 + 27 |
all of the following rules can be deleted.
|
activate(n__cons(X1,X2)) |
→ |
cons(X1,X2) |
(5) |
|
activate(n__from(X)) |
→ |
from(X) |
(6) |
|
activate(X) |
→ |
X |
(7) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [from(x1)] |
= |
16 · x1 + 11 |
| [cons(x1, x2)] |
= |
8 · x1 + 2 · x2 + 9 |
| [activate(x1)] |
= |
2 · x1 + 0 |
| [s(x1)] |
= |
1 · x1 + 0 |
| [n__cons(x1, x2)] |
= |
4 · x1 + 1 · x2 + 0 |
| [n__from(x1)] |
= |
4 · x1 + 1 |
| [2nd(x1)] |
= |
1 · x1 + 8 |
all of the following rules can be deleted.
|
2nd(cons(X,n__cons(Y,Z))) |
→ |
activate(Y) |
(1) |
|
cons(X1,X2) |
→ |
n__cons(X1,X2) |
(3) |
|
from(X) |
→ |
n__from(X) |
(4) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [from(x1)] |
= |
31 · x1 + 24 |
| [cons(x1, x2)] |
= |
1 · x1 + 1 · x2 + 1 |
| [s(x1)] |
= |
5 · x1 + 0 |
| [n__from(x1)] |
= |
6 · x1 + 16 |
all of the following rules can be deleted.
|
from(X) |
→ |
cons(X,n__from(s(X))) |
(2) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.