Certification Problem
Input (TPDB TRS_Standard/SK90/4.22)
The rewrite relation of the following TRS is considered.
|
not(and(x,y)) |
→ |
or(not(x),not(y)) |
(1) |
|
not(or(x,y)) |
→ |
and(not(x),not(y)) |
(2) |
|
and(x,or(y,z)) |
→ |
or(and(x,y),and(x,z)) |
(3) |
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(or) |
= |
0 |
|
status(or) |
= |
[1, 2] |
|
list-extension(or) |
= |
Lex |
| prec(not) |
= |
3 |
|
status(not) |
= |
[1] |
|
list-extension(not) |
= |
Lex |
| prec(and) |
= |
1 |
|
status(and) |
= |
[1, 2] |
|
list-extension(and) |
= |
Lex |
and the following
Max-polynomial interpretation
| [or(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 0 + 1 · x2) |
| [not(x1)] |
=
|
max(0, 0 + 1 · x1) |
| [and(x1, x2)] |
=
|
max(0, 0 + 1 · x1, 0 + 1 · x2) |
all of the following rules can be deleted.
|
not(and(x,y)) |
→ |
or(not(x),not(y)) |
(1) |
|
not(or(x,y)) |
→ |
and(not(x),not(y)) |
(2) |
|
and(x,or(y,z)) |
→ |
or(and(x,y),and(x,z)) |
(3) |
1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.