Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex2_Luc03b_Z)
The rewrite relation of the following TRS is considered.
|
fst(0,Z) |
→ |
nil |
(1) |
|
fst(s(X),cons(Y,Z)) |
→ |
cons(Y,n__fst(activate(X),activate(Z))) |
(2) |
|
from(X) |
→ |
cons(X,n__from(s(X))) |
(3) |
|
add(0,X) |
→ |
X |
(4) |
|
add(s(X),Y) |
→ |
s(n__add(activate(X),Y)) |
(5) |
|
len(nil) |
→ |
0 |
(6) |
|
len(cons(X,Z)) |
→ |
s(n__len(activate(Z))) |
(7) |
|
fst(X1,X2) |
→ |
n__fst(X1,X2) |
(8) |
|
from(X) |
→ |
n__from(X) |
(9) |
|
add(X1,X2) |
→ |
n__add(X1,X2) |
(10) |
|
len(X) |
→ |
n__len(X) |
(11) |
|
activate(n__fst(X1,X2)) |
→ |
fst(X1,X2) |
(12) |
|
activate(n__from(X)) |
→ |
from(X) |
(13) |
|
activate(n__add(X1,X2)) |
→ |
add(X1,X2) |
(14) |
|
activate(n__len(X)) |
→ |
len(X) |
(15) |
|
activate(X) |
→ |
X |
(16) |
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
| [cons(x1, x2)] |
= |
8 · x1 + 1 · x2 + 1 |
| [n__from(x1)] |
= |
17 · x1 + 8 |
| [fst(x1, x2)] |
= |
2 · x1 + 4 · x2 + 1 |
| [s(x1)] |
= |
1 · x1 + 0 |
| [activate(x1)] |
= |
2 · x1 + 0 |
| [n__add(x1, x2)] |
= |
4 · x1 + 1 · x2 + 0 |
| [0] |
= |
19 |
| [from(x1)] |
= |
26 · x1 + 9 |
| [n__len(x1)] |
= |
2 · x1 + 0 |
| [len(x1)] |
= |
4 · x1 + 0 |
| [add(x1, x2)] |
= |
8 · x1 + 1 · x2 + 0 |
| [n__fst(x1, x2)] |
= |
1 · x1 + 2 · x2 + 1 |
| [nil] |
= |
28 |
all of the following rules can be deleted.
|
fst(0,Z) |
→ |
nil |
(1) |
|
fst(s(X),cons(Y,Z)) |
→ |
cons(Y,n__fst(activate(X),activate(Z))) |
(2) |
|
add(0,X) |
→ |
X |
(4) |
|
len(nil) |
→ |
0 |
(6) |
|
len(cons(X,Z)) |
→ |
s(n__len(activate(Z))) |
(7) |
|
from(X) |
→ |
n__from(X) |
(9) |
|
activate(n__fst(X1,X2)) |
→ |
fst(X1,X2) |
(12) |
|
activate(n__from(X)) |
→ |
from(X) |
(13) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [cons(x1, x2)] |
= |
1 · x1 + 1 · x2 + 4 |
| [n__from(x1)] |
= |
2 · x1 + 0 |
| [fst(x1, x2)] |
= |
8 · x1 + 4 · x2 + 16 |
| [s(x1)] |
= |
1 · x1 + 7 |
| [activate(x1)] |
= |
3 · x1 + 1 |
| [n__add(x1, x2)] |
= |
8 · x1 + 1 · x2 + 0 |
| [from(x1)] |
= |
3 · x1 + 19 |
| [n__len(x1)] |
= |
23 · x1 + 7 |
| [len(x1)] |
= |
24 · x1 + 8 |
| [add(x1, x2)] |
= |
24 · x1 + 1 · x2 + 0 |
| [n__fst(x1, x2)] |
= |
2 · x1 + 2 · x2 + 0 |
all of the following rules can be deleted.
|
from(X) |
→ |
cons(X,n__from(s(X))) |
(3) |
|
add(s(X),Y) |
→ |
s(n__add(activate(X),Y)) |
(5) |
|
fst(X1,X2) |
→ |
n__fst(X1,X2) |
(8) |
|
len(X) |
→ |
n__len(X) |
(11) |
|
activate(n__add(X1,X2)) |
→ |
add(X1,X2) |
(14) |
|
activate(n__len(X)) |
→ |
len(X) |
(15) |
|
activate(X) |
→ |
X |
(16) |
1.1.1 Rule Removal
Using the
Weighted Path Order with the following precedence and status
| prec(n__add) |
= |
0 |
|
status(n__add) |
= |
[2, 1] |
|
list-extension(n__add) |
= |
Lex |
| prec(add) |
= |
0 |
|
status(add) |
= |
[2, 1] |
|
list-extension(add) |
= |
Lex |
and the following
Max-polynomial interpretation
| [n__add(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 4 + 1 · x2) |
| [add(x1, x2)] |
=
|
max(1, 1 + 1 · x1, 5 + 1 · x2) |
all of the following rules can be deleted.
|
add(X1,X2) |
→ |
n__add(X1,X2) |
(10) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.