Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex1_Luc02b_Z)
The rewrite relation of the following TRS is considered.
|
from(X) |
→ |
cons(X,n__from(s(X))) |
(1) |
|
first(0,Z) |
→ |
nil |
(2) |
|
first(s(X),cons(Y,Z)) |
→ |
cons(Y,n__first(X,activate(Z))) |
(3) |
|
sel(0,cons(X,Z)) |
→ |
X |
(4) |
|
sel(s(X),cons(Y,Z)) |
→ |
sel(X,activate(Z)) |
(5) |
|
from(X) |
→ |
n__from(X) |
(6) |
|
first(X1,X2) |
→ |
n__first(X1,X2) |
(7) |
|
activate(n__from(X)) |
→ |
from(X) |
(8) |
|
activate(n__first(X1,X2)) |
→ |
first(X1,X2) |
(9) |
|
activate(X) |
→ |
X |
(10) |
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(sel) |
= |
0 |
|
status(sel) |
= |
[1, 2] |
|
list-extension(sel) |
= |
Lex |
| prec(n__first) |
= |
0 |
|
status(n__first) |
= |
[2, 1] |
|
list-extension(n__first) |
= |
Lex |
| prec(activate) |
= |
6 |
|
status(activate) |
= |
[1] |
|
list-extension(activate) |
= |
Lex |
| prec(nil) |
= |
0 |
|
status(nil) |
= |
[] |
|
list-extension(nil) |
= |
Lex |
| prec(first) |
= |
1 |
|
status(first) |
= |
[2, 1] |
|
list-extension(first) |
= |
Lex |
| prec(0) |
= |
0 |
|
status(0) |
= |
[] |
|
list-extension(0) |
= |
Lex |
| prec(cons) |
= |
0 |
|
status(cons) |
= |
[2, 1] |
|
list-extension(cons) |
= |
Lex |
| prec(n__from) |
= |
0 |
|
status(n__from) |
= |
[1] |
|
list-extension(n__from) |
= |
Lex |
| prec(s) |
= |
0 |
|
status(s) |
= |
[1] |
|
list-extension(s) |
= |
Lex |
| prec(from) |
= |
1 |
|
status(from) |
= |
[1] |
|
list-extension(from) |
= |
Lex |
and the following
Max-polynomial interpretation
| [sel(x1, x2)] |
=
|
max(2, 5 + 1 · x1, 1 + 1 · x2) |
| [n__first(x1, x2)] |
=
|
1 + 1 · x1 + 1 · x2
|
| [activate(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [nil] |
=
|
max(0) |
| [first(x1, x2)] |
=
|
1 + 1 · x1 + 1 · x2
|
| [0] |
=
|
max(0) |
| [cons(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 0 + 1 · x2) |
| [n__from(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [s(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [from(x1)] |
=
|
max(0, 0 + 1 · x1) |
all of the following rules can be deleted.
|
from(X) |
→ |
cons(X,n__from(s(X))) |
(1) |
|
first(0,Z) |
→ |
nil |
(2) |
|
first(s(X),cons(Y,Z)) |
→ |
cons(Y,n__first(X,activate(Z))) |
(3) |
|
sel(0,cons(X,Z)) |
→ |
X |
(4) |
|
sel(s(X),cons(Y,Z)) |
→ |
sel(X,activate(Z)) |
(5) |
|
from(X) |
→ |
n__from(X) |
(6) |
|
first(X1,X2) |
→ |
n__first(X1,X2) |
(7) |
|
activate(n__from(X)) |
→ |
from(X) |
(8) |
|
activate(n__first(X1,X2)) |
→ |
first(X1,X2) |
(9) |
|
activate(X) |
→ |
X |
(10) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.