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The supporting lemma: $$ ((p \land q) \Rightarrow r) \Leftrightarrow(q \Rightarrow(\neg p \lor r)) $$

\begin{align} &(p \land q) \Rightarrow r && \text{Start} \\ &p\rightarrow(q\rightarrow r) && \text{given (outside source)} \\ &\neg p \lor (q\rightarrow r) && \text{implication} \\ &\neg p \lor (\neg q \lor r) && \text{implication} \\ &(\neg p \lor \neg q) \lor r && \text{associative} \\ &(\neg q \lor \neg p) \lor r && \text{commutative} \\ &\neg q \lor (\neg p \lor r) && \text{associative} \\ &\neg q \lor (\neg p \lor r) && \text{associative} \\ &q \rightarrow(\neg p \lor r) && \text{implication} \end{align}

Is to be used to prove the distributive property, where the distributive property is written as:

$$ a \land (b \lor c) \Rightarrow (a \land b) \lor (a \land c) $$

$$ a \lor (b \land c) \Rightarrow (a \lor b) \land (a \lor c) $$

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    I have proof for the supporting lemma (I'll edit that in there, my bad). What I'm having difficulty with is understanding how to use that to prove distribution. – 2017-01-24
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    Your demonstration shows that (p $\rightarrow$ (q $\rightarrow$ r)) is equivalent to (q $\rightarrow$ ($\lnot$p $\lor$ r)). You were only given ((p $\land$ q) $\rightarrow$ r). – 2017-01-24
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    The first step (p -> (q -> r) was a solution from a different problem that's not relevant to this one other than helping prove the equivalence to the given lemma. That's why I commented that it was an outside source. – 2017-01-24
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    So you are saying: using some "equivalence" I'm able to prove the above Lemma. In the same way, I have to prove the distributive properties. Correct ? – 2017-01-24
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    For the lemma yes. Some equivalence was used to prove the lemma, however, I'm not sure if proving the distributive property is a matter of using an equivalence or if it's just using the similar steps from the lemma proof. – 2017-01-24
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    @Jason So what do you have to work with other than Commutation, Association, and Implication? Because as you show, the given Lemma can be proven from these three ... and I see now way to prove Distribution from these three principles alone! Also, the Distributive propoerty is typically stated as an equivalence, rather than an implication. Are you sure you don;t have to show $a \land (b \lor c) \Leftrightarrow (a \land b) \lor (a \land c)$? – 2017-01-24
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    We're allowed to use all the equivalency laws, and all the inference rules. And yes I'm 100% sure that the distributive property is stated as an implication here. My professor wants us to prove it going both ways and wrote out the statement like this on the board. – 2017-01-24

1 Answers 1

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I feel super stupid. The start of this proof is to make the correlation between $q\implies(\lnot p\lor r)$ and $a \land (b\lor c) \implies (a\land b)\lor(a\land c)$ where q is $b \lor c$, $\lnot$p is a (where a takes the negative form), and r is $(a\land b)\lor(a\land c)$

This would give us the following equation:

$(b\lor c) \implies (\lnot a \lor(a\land b)\lor(a\land c))$

$a\rightarrow(a\land b)\lor(a\land c)$ Implication

$a$ Implication

$b\lor c$ Implication

The second part of the of the problem can be solved in a similar fashion.