DeMorgans Law Application

Question

I'm Trying to prove that

¬(p ↔ q) is equivalent to ¬p ↔ q.

I have done the work for ¬p ↔ q and simplified it to (p ∨ q) ∧ ¬(p ∧ q) .....My trouble is on the other proposition how do i distribute that not(¬)?

$\lnot(p\land q)$ is equivalent to $\lnot p \lor \lnot q$ – J. W. Tanner Sep 12 '19 at 00:37 — J. W. Tanner, Sep 12 '19 at 00:37

Graham Kemp · Accepted Answer · 2019-09-12T00:42:35.913

2

De Morgan's Law: $\neg(p\wedge q) \equiv \neg p\vee\neg q$ and $\neg(p\vee q)\equiv \neg p\wedge \neg q)$

So $\neg(p\leftrightarrow q)~{\equiv \neg((p\to q)\wedge(q\to p))\\\equiv\neg (p\to q)\vee\neg(q\to p)}$

Now, if you do not know how to negate a conditional, first use the equivalence $p\to q\equiv \neg p\vee q$.

$~\\~\\~$

PS: Calling DeMorgan's Law "distributing the not" will only confuse you.

edited Sep 12 '19 at 00:42

answered Sep 12 '19 at 00:37

Graham Kemp

129,094

Ok got it! So After applying DeMorgans law to ¬(p ↔ q) is this valid (¬p↔ ¬q) – Elchavo18 Sep 12 '19 at 00:42
No... not at all. – Graham Kemp Sep 12 '19 at 00:42
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So where you left off in your answer the negation of the equivalence would be ... P ∧ ¬ q? – Elchavo18 Sep 12 '19 at 00:48
Yes... Well, $(p\wedge\neg q)\vee(q\wedge\neg p)$, to be precise. Then apply distribution and so forth.... – Graham Kemp Sep 12 '19 at 00:53

DeMorgans Law Application

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