# Verify that ~(P ↔ Q) ≡ P ↔ ( ~Q) using logical equivalencies (not truth tables)

I have been able to verify this via truth tables but not with logical equivalencies yet. I understand some of the basic principles of logical equivalencies but I cannot seem to get to the end of this problem where one side equals the other.

• Your question isn't very clear: what logical equivalences are you taking as primitive? The laws of boolean algebras perhaps? Dec 17, 2020 at 22:03
• @RobArthan logical equivalencies such as De Morgan's Laws, for example. Others might include distributing negation, etc
– user863697
Dec 17, 2020 at 22:07
• Well if you have enough equivalences to show that a formula is equivalent to a disjunctive normal form or a conjunctive normal form, you are pretty well there. Dec 17, 2020 at 22:12

Notice that $$A\vee(B\wedge C)\equiv (A\vee B)\wedge(A\vee C)$$.

On the left hand side, we have

$$\neg(P\leftrightarrow Q)\equiv \neg[(P\to Q)\wedge(Q\to P)]\equiv(P\wedge\neg Q)\vee(Q\wedge\neg P)\equiv[(P\wedge\neg Q)\vee Q]\wedge[(P\wedge\neg Q)\vee\neg P]\\\equiv[(P\vee Q)\wedge(\neg Q\vee Q)]\wedge [(P\vee\neg P)\wedge(\neg Q\vee\neg P)]\equiv(P\vee Q)\wedge(\neg Q\vee\neg P)$$

Consider the negation of the right hand side and we have

$$\neg[(P\to\neg Q)\wedge(\neg Q\to P)]\equiv(P\wedge Q)\vee(\neg Q\wedge\neg P)$$

Now we negate it back

$$\neg[(P\wedge Q)\vee(\neg Q\wedge\neg P)]\equiv[\neg(P\wedge Q)]\wedge[\neg(\neg Q\wedge\neg P)]\equiv(\neg P\vee\neg Q)\wedge(Q\vee P)$$

Hence it gives the result that we are asked to verify.