Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex1_2_Luc02c_FR)
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(n__s(X))) |
(2) |
|
cons(X1,X2) |
→ |
n__cons(X1,X2) |
(3) |
|
from(X) |
→ |
n__from(X) |
(4) |
|
s(X) |
→ |
n__s(X) |
(5) |
|
activate(n__cons(X1,X2)) |
→ |
cons(activate(X1),X2) |
(6) |
|
activate(n__from(X)) |
→ |
from(activate(X)) |
(7) |
|
activate(n__s(X)) |
→ |
s(activate(X)) |
(8) |
|
activate(X) |
→ |
X |
(9) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
| prec(s) |
= |
1 |
|
status(s) |
= |
[1] |
|
list-extension(s) |
= |
Lex |
| prec(n__from) |
= |
0 |
|
status(n__from) |
= |
[1] |
|
list-extension(n__from) |
= |
Lex |
| prec(n__s) |
= |
0 |
|
status(n__s) |
= |
[1] |
|
list-extension(n__s) |
= |
Lex |
| prec(from) |
= |
3 |
|
status(from) |
= |
[1] |
|
list-extension(from) |
= |
Lex |
| prec(activate) |
= |
7 |
|
status(activate) |
= |
[1] |
|
list-extension(activate) |
= |
Lex |
| prec(2nd) |
= |
0 |
|
status(2nd) |
= |
[1] |
|
list-extension(2nd) |
= |
Lex |
| prec(cons) |
= |
2 |
|
status(cons) |
= |
[1, 2] |
|
list-extension(cons) |
= |
Lex |
| prec(n__cons) |
= |
0 |
|
status(n__cons) |
= |
[1, 2] |
|
list-extension(n__cons) |
= |
Lex |
and the following
Max-polynomial interpretation
| [s(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [n__from(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [n__s(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [from(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [activate(x1)] |
=
|
0 + 1 · x1
|
| [2nd(x1)] |
=
|
max(0, 6 + 1 · x1) |
| [cons(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 0 + 1 · x2) |
| [n__cons(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 0 + 1 · x2) |
all of the following rules can be deleted.
|
2nd(cons(X,n__cons(Y,Z))) |
→ |
activate(Y) |
(1) |
|
from(X) |
→ |
cons(X,n__from(n__s(X))) |
(2) |
|
cons(X1,X2) |
→ |
n__cons(X1,X2) |
(3) |
|
from(X) |
→ |
n__from(X) |
(4) |
|
s(X) |
→ |
n__s(X) |
(5) |
|
activate(n__cons(X1,X2)) |
→ |
cons(activate(X1),X2) |
(6) |
|
activate(n__from(X)) |
→ |
from(activate(X)) |
(7) |
|
activate(n__s(X)) |
→ |
s(activate(X)) |
(8) |
|
activate(X) |
→ |
X |
(9) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.