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I have been doing some discrete math on my own and I came upon this example problem:

prove without using a truth table the following statement:

$$ ((\neg p \lor q) \land (\neg q \lor r)) \land (p \land \neg r) $$

I know that the way to solve this is by distributing the $(p \land \neg r)$ across the dis-junctions. So for example, $p \land (\neg p \lor q) = (p \land \neg p) \lor (p \land q) = False \lor (p \land q) = (p \land q)$. But I am not sure how to distribute it properly across all the terms.

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    Isn't this false if $p$ is false? – 2017-01-25
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    It's supposed to be a contradiction and no, the right is $p \land \neg r$. – 2017-01-25
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    Actually, the truth value of this formula is false for all values of $p,q,r$. – 2017-01-25
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    Proving a *contradiction* ??? – 2017-01-25
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    I think that you have to show (without t-t) that it is a *contradiction*, which amounts at proving that it is equivalent to $p \land \lnot p$... – 2017-01-25
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    @MauroALLEGRANZA I guess he means to prove that it is a contradiction. – 2017-01-25
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    Note that $\neg p\vee q$ and $\neg q\vee r$ are equivalent to $p\to q$ and $q\to r$, respectively. Your conjunction contains these two, along with $p$ and $\neg r$. From $p$ and $p\to q$ you can conclude $q$; from $q$ and $q\to r$ you can conclude $r$. – 2017-01-25

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Since it ends in $\dotsb\land(p\land\lnot r)$, this could only be true if $p\land\lnot r$ is true — that is, if $p$ is true and $r$ is false.

Substituting that into the first term, we get that for it to be true, we must have: \begin{align}(\lnot{\rm T}\lor q)&\land(\lnot q\lor{\rm F})\\ ({\rm F}\lor q)&\land(\lnot q\lor{\rm F})\\ q&\land\lnot q \end{align} which is clearly never true. Thus, the statement as a whole is never true, i.e., it is a contradiction.

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    Could you explain the intermediate steps (how did you get the first line?) that's what I am not clear on how to do – 2017-01-25
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    You mean, how I got to $(\lnot{\rm T}\lor q)\land(\lnot q\lor{\rm F})$? I was substituting in True for $p$ and False for $r$, since I determined that the statement is true only if $p$ and $r$ have those values. – 2017-01-25
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The trick is to group them differently:

$(¬p∨q)∧[(¬q∨r))∧(p∧¬r)] \equiv (¬p∨q) ∧ [((p∧¬r) ∧ ¬q) ] ∨ ((p∧¬r) ∧ r)]$ by Distributivity.

But $(p∧¬r)∧r \equiv p∧(¬r∧r) \equiv (p∧F) \equiv F$, and we can cancel this disjunct to get:

$(¬p∨q) ∧ (p∧¬r ∧ ¬q)$.

Again, by Distributivity :

$(¬p∨q) ∧ (p∧¬r ∧ ¬q) \equiv [ ¬p ∧ (p∧¬r ∧ ¬q) ] ∨ [ q ∧ (p∧¬r ∧ ¬q) ]$

and again :

$[(¬p ∧ p) ∧ (¬r ∧ ¬q)] ∨ [(q ∧ ¬q) ∧ (p ∧ ¬r)] \equiv [ F ∧ (¬r ∧ ¬q) ] ∨ [ F ∧ (p∧¬r) ] \equiv (F ∨ F) \equiv F$

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I know this is an old topic, but there's actually (at least in my opinion) a much simpler way to prove falsity for this. Your formula is equivalent to (p implies q) and (q implies r) and (not(p implies r)), where I used ((p implies q) iff (not(p) or q). By hypothetical syllogism, if you have (p implies q) and (q implies r), then you can derive (p implies r). But, then you have (p implies r) and (not(p implies r)), which is a contradiction.