Is NOT distributive in mathematical logic?

Question

Consider the following proposition: If P, then Q AND R.

Would the contrapositive therefore be If NOT(Q) AND NOT(R), then NOT(P)?

See http://matheducators.stackexchange.com/questions/926/whence-the-everything-is-linear-phenomenon-and-what-can-we-do-about-it. — user21820, Aug 12 '16 at 11:53

score 6 · Accepted Answer · edited Aug 11 '16 at 19:18

6

No.

$$\mathrm{Not} (Q \text{ and } R) \quad\text{is the same as}\quad (\mathrm{Not}(Q)) \text{ or } (\mathrm{Not}(R))$$

What is the opposite of having two things? It's lacking at least one of those things.

edited Aug 11 '16 at 19:18

answered Aug 11 '16 at 19:14

Mike Pierce

amWhy · Answer 2 · 2016-08-11T19:32:13.920

Remember that Parentheses are crucial in logic.

Assuming your initial statement is meant to be: If $P,$ then ($Q$ AND $R$)

[which is very different than "If ($P$ then $Q$) AND $R$]:

Then the contrapositive would be: If Not ($Q$ AND $R$), then NOT $(P).$

Equivalently, by DeMorgan's Law, it becomes: If(Not Q or Not R) then NOT $P$.

In symbols, $$P \rightarrow (Q \land R) \equiv \lnot (Q\land R) \rightarrow \lnot P \equiv (\lnot Q \lor \lnot R) \rightarrow \lnot P$$

Use the given link for a more thorough explanation and understanding of DeMorgan's Law's.

2 Answers2