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 AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [2nd(x1)] |
= |
1 + 2 · x1
|
| [cons(x1, x2)] |
= |
1 + 1 · x1 + 1 · x2
|
| [n__cons(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [activate(x1)] |
= |
2 + 2 · x1
|
| [from(x1)] |
= |
1 + 2 · x1
|
| [n__from(x1)] |
= |
1 · x1
|
| [s(x1)] |
= |
1 · x1
|
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) |
|
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
| prec(from) |
= |
1 |
|
stat(from) |
= |
mul
|
| prec(cons) |
= |
0 |
|
stat(cons) |
= |
mul
|
| π(from) |
= |
[1] |
| π(cons) |
= |
[1,2] |
| π(n__from) |
= |
1 |
| π(s) |
= |
1 |
all of the following rules can be deleted.
|
from(X) |
→ |
cons(X,n__from(s(X))) |
(2) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.