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What I am given:

[(p∧r)∧(p→ q)]→q

What I did:

⇔ [(p∧r)∧( ¬p V q)] → q: Implication

⇔ ¬ [(p∧r)∧( ¬p V q)] V q: Implication

⇔ ¬ (p∧r) V ¬ ( ¬p V q) V q: De Morgan

⇔ (¬p V ¬r) V (¬ ¬p ∧ ¬q) V q: De Morgan

⇔(¬p V ¬r) V (p ∧ ¬q) V q: Double Negation

⇔(¬p V ¬r) V (q V p) ∧ (q V ¬q): Distributive

⇔(¬p V ¬r) V (q V p) ∧ T: Tautology

⇔(¬p V ¬r) V (q V p): Identity

⇔(¬p V p) V (q V r): Associative

⇔T V (q V r): Tautology

⇔ T Domination law

I do not think this is right. Can someone provide insight?

Thanks

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  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax. $\endgroup$
    – dantopa
    May 28, 2019 at 18:27
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1 Answer 1

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Your working is completely correct. A slightly more intuitive way to go about it is to try and make the expression false and arrive at an impossibility.

The outer operator is an implication, so if this is false then $q$ must be false and $(p\land r)\land(p\to q)$ must be true, which in turn means that $p,r,p\to q$ must all be true. But then assigning $p,r$ as true means $p\to q$ is false, a contradiction. So the whole expression is a tautology.

As for the bonus question, $p\mid p$ (NAND) is equal to $\neg p$. This can be seen by truth table.

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  • $\begingroup$ That is fantastic to hear! I really appreciate it. What throws me off are the parenthesis. I am so used to algebra and calculus where parenthesis means you can only work inside them before moving on. This is not the case in discrete math, right? $\endgroup$
    – Zevias
    May 28, 2019 at 18:34
  • $\begingroup$ @Zevias Since you are new, I'd like you to upvote and accept my answer. You will see a tick to the left; click it and it should turn green. $\endgroup$ May 28, 2019 at 18:35
  • $\begingroup$ I ticked the green check. I upvote you but it says it won't be publicly displayed since I have less than 15 rep $\endgroup$
    – Zevias
    May 28, 2019 at 18:36
  • $\begingroup$ @Zevias It's OK... ask more good questions. The reputation will come with that. $\endgroup$ May 28, 2019 at 18:37
  • $\begingroup$ SO parenthesis and order of operations is not the same in discrete math? For example, when I first started proving [(p∧r)∧(p→ q)]→q, I thought I had to only work on the left side before utilizing the right handed q. But once I apply implication, I can work with any side in any order? $\endgroup$
    – Zevias
    May 28, 2019 at 19:57

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