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$\begingroup$

Anyone mind helping me break down this statement in discrete mathematics?

Having trouble understanding if I'm doing this correctly or not.

The original statement was:

~(~p -> q) -> (p v r)

And I went on:

Work i've done so far

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    Correct, but the 4th and 5th lines are superfluous : you can go from $\lnot (p \lor q) \to (p \lor r)$ to $\lnot \lnot (p \lor q) \lor (p \lor r)$ and then...2017-01-30

2 Answers 2

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You've done fine. A tip, though, following your third step: third statement:

$$\lnot (p\lor q) \rightarrow (p\lor r)\tag{3}$$

$$ \equiv \lnot\lnot (p\lor q) \lor (p \lor r)\tag 4$$

$$\equiv (p\lor q)\lor (p\lor r)\tag 5$$ $$\equiv p\lor q \lor p \lor r \tag{(6) associativity of $\lor$}$$

$$\equiv \underbrace{p \lor p}_{\large \equiv p} \lor q \lor r\tag{(7) commutativity of $\lor$}$$

$$\equiv p\lor q\lor r\tag 8 $$

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Well that's correct. Now all you need as a final step is: $$ p \lor q \lor r $$ Since you have $$\lor$$ everywhere, you can eliminate brackets.