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$$¬(p ↔ q) ⇔ (p ↔ ¬q)$$

I started with the left side.

$LS: ¬(p ↔ q)$

$\Leftrightarrow ¬ ((p → q）∧(ｑ → ｐ）） $ Biconditional Law

$\Leftrightarrow　¬（（¬ｐｖｑ）∧（¬ｑｖｐ）） $ Conditional Law

$\Leftrightarrow（¬（¬ｐｖｑ））ｖ（¬（¬ｑｖｐ））） $ DeMorgan's

$\Leftrightarrow　（ｐ∧¬ｑ）ｖ（ｑ∧¬ｐ）） $ DeMorgan's

I am having trouble figuring out where to go from here.

Were there also any mistakes in any of my steps?

Thanks.

Ethan
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Isaiah
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2 Answers2

2

$$\neg(p\leftrightarrow q)$$ $$\neg (p \to q \wedge q\to p)$$ $$\neg(p\to q)\vee\neg (q\to p)$$ $$(p\wedge\neg q)\vee(q\wedge\neg p)$$ $$(p\vee q)\wedge(\neg q\wedge \neg p)$$ $$(\neg q\to p)\wedge(p\to\neg q)$$ $$p\leftrightarrow \neg q$$

ajotatxe
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2

So far: all good.

Next: use Distribution, Negation, then Identity.

Graham Kemp
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  • How would you distribute:（ｐ∧¬ｑ）ｖ（ｑ∧¬ｐ）.

    I know that p V (q ∧ r) <=> (p V q) ∧ (p V r)

     – Isaiah Nov 07 '19 at 06:29
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    So treat the first conjunction as one term.$${\color{blue}{(p\wedge \neg q)}~\vee~(\neg p~\wedge ~q)\=(\color{blue}{(p\wedge\neg q)}\vee\neg p)\wedge(\color{blue}{(p\wedge\neg q)}\vee q)\ =\ldots}$$Now distribute those two disjunction. [Or you might use Absorption if you like.] – Graham Kemp Nov 07 '19 at 06:41