# Proving the Distributive Property with a Lemma

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:

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

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

• 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. Commented Jan 24, 2017 at 5:25
• 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). Commented Jan 24, 2017 at 7:23
• 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. Commented Jan 24, 2017 at 7:38
• 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 ? Commented Jan 24, 2017 at 10:02
• 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. Commented Jan 24, 2017 at 17:57

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.