Discrete Mathematics again

[mamma_mia66]

Construct a formal proof of the theorem:

If (p-> q), (neg [r] -> s), and (neg [q] V neg , then (p-> r).

[refer to table of logical equivalences (p62) and the table of logical implication (p62)]The tables are in the textbook Kenneth Ross, 5th edition, Discrete Math's.

First I have to construct the proof line: I am gessing that the hypotheses are like this, but am not sure for their order:
(neg [r] -> s) /\ (p -> q) /\ (neg [q] V neg ) => (p -> r)

Then I need to make the "formal proof of it" using the tables.

Please someone help me with any sugestion how to keep going with this problem.
Thank you:

 

[lippyka]

) Proof: 1. (neg [r] -> s) /\ (p -> q) /\ (neg [q] V neg ) [Given]2. (neg [r] -> s) [Conjunction Elimination from 1]3. (p -> q) [Conjunction Elimination from 1]4. (neg [q] V neg ) [Conjunction Elimination from 1]5. (neg [q] -> neg ) [Implication Equivalence from Table of Logical Equivalences] 6. (neg [q] -> neg ) /\ (p -> q) [Conjunction Introduction from 3, 5] 7. (neg [q] -> (neg /\ (p -> q))) [Implication Equivalence from Table of Logical Equivalences] 8. (neg /\ (p -> q)) -> (p -> r) [Implication Introduction from 2, 7] 9. (p -> r) [Implication Elimination from 8] Therefore, (p -> r) by Modus Ponens.