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I need to prove/disprove the logical equivalences of the following two statements using basic equivalences:

  1. $p \to (q \to r)$ and $(p \to q) \to r$.

  2. $p \to (q \to r)$ and $q \to (p \to r)$.

I also need to verify each of the following by writing an equivalence proof:

  1. $p \to (q \wedge r) = (p \to q) \wedge (p \to r)$.

  2. $(p \to q) \wedge (p \vee q) = q$.

Also, I have two other questions:

  1. Suppose $a$ and $b$ are integers and $a^2 - 5b$ is even. Prove that $b^2 - 5a$ is even.

  2. Prove that for all integers $n$, $n \ge 1$: $1 + 3 + 5 + \cdots + (2n -1) = n^2$.

Thank you if you can help, I can do everything else on my work except these damn proofs!

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closed as off-topic by Lord_Farin, dfeuer, Stefan4024, Norbert, azimut Nov 2 '13 at 22:19

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You need to post questions separately. You've assigned us 6 problems in one post! –  amWhy Dec 2 '12 at 0:47
1  
@amWhy: Well, I don't think posting those as six questions would be better, as some are closely related. At least three or four would make sense (the first two problems are pretty much the same thing, the fifth is number theory and sixth is about mathematical induction). Either way, what I have more of a problem with is that there's no effort shown, it looks like "do my homework for me" kind of question. –  tomasz Dec 2 '12 at 1:22
    
@Omar Qassem: is this a homework question? If so, please consider adding the [homework] tag; we are happy to help with homework questions here, but it useful to know when we’re doing so. –  Peter LeFanu Lumsdaine Dec 2 '12 at 1:29
    
in relation to some of the comments posted - yes it's coursework, and I've tried for hours on these, so please don't say I haven't tried. I would post my whole coursework here if I didn't try - think. –  Omar Qassem Dec 2 '12 at 13:49
3  
This question appears to be off-topic because it contains multiple questions, and hence is unlikely to be of help to future readers. –  Lord_Farin Nov 2 '13 at 21:17

1 Answer 1

up vote 0 down vote accepted

Prove $p \to (q \wedge r) \equiv (p \to q) \wedge (p \to r)$.

$p \to (q \wedge r) \equiv \lnot p \lor (q \land r) \equiv (\lnot p \lor q)\land (\lnot p \lor r) \equiv (p \to q) \wedge (p \to r) $

Prove $(p \to q) \wedge (p \vee q) \equiv q$.

$(p \to q) \wedge (p \vee q) \equiv ( \lnot p \lor q)\land(p\lor q) \equiv q$.

Read your textbook/note to memorize those identities (I have my final on this in a week so I'm studying too).

Prove that for all integers $n$, $n \ge 1$: $1 + 3 + 5 + \cdots + (2n -1) = n^2$.

Use induction: check for base case n=1 n=2 .. this equation holds. Then $n \ge 1$: $1 + 3 + 5 + \cdots + (2n -1) = n^2$ becomes your induction hypothesis, you want to prove that for $n+1$, this equation is still true.

LHS: $$1 + 3 + 5 + \cdots + (2(n+1) -1) = 1+3+5+\cdots+(2n+2-1) $$ $$ = 1+3+5+\cdots+(2n+1)$$ $$ = 1+3+5+\cdots+(2n-1) +(2n+1) $$ $$ = n^2+2n+1$$

RHS: $$(n+1)^2 = n^2 +2n+1$$

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