Can you help me with the inference: if ¬( P & ¬Q ) and Q, then P

I'm taking my classes of symbolic logic, so my question is a bit naïve, but: If this expression is correct: ¬( P & ¬Q), P then Q. Why not the following is not: ¬( P & ¬Q), Q then P. Thank you.

• Maybe "valid" and not correct... If so, use truth table. Oct 20, 2023 at 11:31
• If one of the below answers helped you please consider accepting it as well as upvoting. This will avoid extra answers being posted. Oct 23, 2023 at 19:19

If this expression is correct: ¬( P & ¬Q), P then Q. Why not the following is not: ¬( P & ¬Q), Q then P?

The expression ¬( P & ¬Q) explicitly states that it can not be the case that P is the case whilst Q is not the case. So one can derive from this that if P is the case, then Q must be the case, otherwise, you would be inadvertently stating that P could be the case whilst Q is not the case, resulting in a violation of the expression given.

With that understood, the expression "Q then P" simply states that Q cannot be the case unless P is the case, but this stipulation does not prevent P from being the case and Q not being the case, allowing a violation of the inital expression ¬( P & ¬Q).

To elucidate, let us provide a real world example. Let us assign the phrase [eating chicken] to the variable P and let us assign the phrase [eating meat] to the variable Q. Given ¬( P & ¬Q), we can translate the statement into:

"It is not the case that[¬] (one can be eating chicken [P] and not be eating meat [¬Q])."

From there one could easily extrapolate the following:

"If one is eating chicken, then they are eating meat."

Though one would not be able to extrapolate the following:

"If one is eating meat, then they are eating chicken"

I hope that helps, please let me know if you have any questions regarding my answer.

-Thank you

• Thank you so much! Oct 20, 2023 at 16:50
• +1 For exemplification.
– J D
Oct 20, 2023 at 16:53

R1. Usually, one is introduced to material conditional as `P → Q` in high school geometry because if-then language is so common, and once exposed to truth tables through logic or programming classes, it can be shown that `¬P ∨ Q` is a a rule of inference called material inference by truth tables. (The article does it for you.)

R2. Now, in high school you were probably also taught that there are four forms of material conditional, the conditional, the converse, the inverse, and the contrapositive, and that the conditional and contrapositive are logically equivalent. So it happens to be that `P → Q ≡ ¬Q → ¬P`.

R3. You hopefully also recollect De Morgan's Laws and know that `¬(P ∧ Q) ≡ ¬P ∨ ¬Q`.

So now you know right away why `¬( P ∧ ¬Q) (P1)` cannot be both `P → Q (P2)` and `Q → P (P3)`. P1 by De Morgan's Law (R3) and then by the material inference (R1) becomes the material conditional (P2), but the material conditional is not equal to the converse (P3), only to the contrapositive by R3.

If that English is tough to follow, then the best way is write out the truth tables for P1, P2, and P3.

• Thank you, now I've got it. Oct 20, 2023 at 16:50