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(I) F = (p ↔ ¬r)

is supposedly (or supposably acc. Joey) euqivalent to:

(II) (p ∧ ¬r) ∨ (¬p ∧ ¬¬r)

However, I can't see it. I always end up at/ circle around (III) (¬p ∨ ¬r) ∧ (r ∨ p)

But I cannot take it from (III) to (II) (or from (I) to (II), obviously.)

It seems to me, I am unfamiliar with one "trick". I have all the rules in the script in front of me but seem to be applying them wrong/inadequately.

Thanks in advance!

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    PS: you cannot take $\rm (III)$ to $\rm (II)$ because they are *not* equivalent. You should not have gotten to $\rm (III)$ in the first place. – 2017-02-09

3 Answers 3

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You seem to understand that $$a \leftrightarrow b \equiv (a \land b) \lor (\lnot a \land \lnot b).$$ (Note: It is also true that $a\leftrightarrow b = (a\rightarrow b) \land (b \rightarrow a))$

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The truth-table above is identical to the truth table for $a\leftrightarrow b$.

But I'll use the first definition I list above.

Now, put $a :=p$, and $b:= \lnot r$. Then we get $$p \leftrightarrow (\lnot r) \equiv (p \land (\lnot r)) \lor (\lnot p \land \lnot(\lnot r))\equiv (p \land \lnot r) \lor (\lnot p \land r)\tag{1}$$

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    Hey, thanks a lot! Actually, the first thing is the part I did/do not understand. Now that you put it, it makes a lot of sense intuitively, but that is the rule I did not have handy. – 2017-02-09
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$(1)\quad(p\Leftrightarrow\neg\,r)$

This can be transformed using the equivalence tautology $(p\Leftrightarrow q)\iff(p\Rightarrow q)\land(q\Rightarrow p)$.

$(2)\quad((p\Rightarrow\neg\,r)\land(\neg\,r\Rightarrow p))$

This can be transformed further using the implication tautology $(p\Rightarrow q)\iff(\neg\,p\lor q)$.

$(3)\quad((\neg\,p\lor \neg\,r)\land(\neg\neg\,r\lor p))$

You can distribute this further as...

$(4)\quad(((\neg\,p\lor\neg\, r)\land\neg\neg\,r)\lor((\neg\,p\lor\neg\, r)\land p)))$

$(5)\quad(((\neg\,p\land\neg\neg\,r)\lor(\neg\, r\land\neg\neg\,r))\lor((\neg\,p\land p)\lor(\neg\, r\land p)))$

Anything of the form $(\neg\,p\land p)$ is a contradiction, therefore always false.

$(6)\quad(((\neg\,p\land\neg\neg\,r)\lor\bot)\lor(\bot\lor(\neg\, r\land p)))$

You can always disjunct with $\bot$ without changing the value, so you can leave it out.

$(7)\quad((\neg\,p\land\neg\neg\,r)\lor(\neg\, r\land p))$

Now you can replace the two and get:

$(8)\quad((\neg\, r\land p)\lor(\neg\,p\land\neg\neg\,r))$

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Tip: Leave double negatives alone when they occur since, for some reason, you want them to remain in the final answer.

A biconditional holds when both arguments have the same truth value.   That is to say that either they are both true, or they are both false.

$$(p~{\gets}\hspace{-1.5ex}{\to}~ \neg r)~~\iff ~~(p\wedge\neg r)\vee(\neg p\wedge\neg\neg r)$$

That is all.

However, you may have defined a biconditional as being that conditional of the arguments hold in both direction.   Then you need to use the definition of conditional, that the antecedent is false or the consequent is true.   Then apply distribution and eliminate the contradictions by disjunctive identity rule.   Everything else is just commutation.

$$\begin{align} (p~{\gets}\hspace{-1.5ex}{\to}~ \neg r)~~&\iff ~~(p\to \neg r)\wedge(p\gets\neg r) \\ &\iff~~ (\neg p\vee \neg r)\wedge(p\vee\neg\neg r) \\ &\iff~~(\neg p\wedge p)\vee(\neg p\wedge \neg\neg r)\vee(\neg r\wedge p)\vee(\neg r\wedge\neg\neg r) \\ &\iff~~\bot\vee(\neg p\wedge \neg\neg r)\vee(\neg r\wedge p)\vee\bot\\&\iff ~~ (p\wedge\neg r)\vee(\neg p\wedge\neg\neg r) \end{align}$$

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    Thank you loads! That was exactly what I needed to grasp it. – 2017-02-09