# task 3:x <-> (y1 -> y2)

+1 vote
(x->(y1->y2))&((y1->y2)->x)

(!x|(!y1|y2))& ((!y1|y2)->x)

(!x|!y1|!y3) & ((y1&!y2)|x)

i got solution :(!x|!y1|!y2)&(x|y1)&(x|!y2)   but its  not correct

at which part of my solution is wrong.

## 2 Answers

0 votes

The first part is correct, i.e.,

`   (x->(y1->y2)) & ((y1->y2)->x) <-> (!x|(!y1|y2)) & ((!y1|y2)->x)`

is valid. The second part is wrong (there is also a y3 that should be y2, but even with that typo corrected, it remains wrong). Converting !x|(!y1|y2) to CNF should go like this:

`   !x|(!y1|y2) = !x|!y1|y2`

while you wrote

```   !x|(!y1|y2) = !x|!y1|!y3
```
by (167k points)
+1 vote

On your third line, you have added a negation in front of y2 that is not present on the second line. So if I follow the formula from the second line:

(!x|(!y1|y2))& ((!y1|y2)->x)

(!x | !y1 | y2)&((y1&!y2) | x)

(!x | !y1 | y2) & (x | y1) & (x | !y2)

Other than the extra negation in front of y2, there seems to be no other error.

by (1.5k points)

+2 votes
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+4 votes
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