MHB Compound Proposition Simplification

[User40405]

Hi all

I need to complete this question for an assignment, but I cannot seem to understand how to simplify the compound proposition with logical equivalences. If anyone here understands how to complete this question, please could you show me how, as it would be greatly appreciated. Thank you.

Here is the question: View attachment 8461

 

[steenis]

You know that $a \to b$ is equivalent to $\neg a \vee b$

Therefore $[(p \vee q) \wedge \neg p] \to q$ is equivalent to $\neg [(p \vee q) \wedge \neg p] \vee q$

Now you can simplify the last expression.

 

[User40405]

You know that $a \to b$ is equivalent to $\neg a \vee b$

Therefore $[(p \vee q) \wedge \neg p] \to q$ is equivalent to $\neg [(p \vee q) \wedge \neg p] \vee q$

Now you can simplify the last expression.

Thank you so so much!

I have been checking guides the entire day and yesterday. I can now get to where you got with it, but I cannot simplify the last expression (pvq). I do not know how to change this and get the simplified form.

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Because (pvq) is equivalent to (qvp). But how does that help me?

 

[Olinguito]

Or use the distributive law in the first block:
$$(p\vee q)\wedge\neg p\ \equiv\ (p\wedge\neg p)\vee(q\wedge\neg p)\ \equiv\ q\wedge\neg p.$$

 

[steenis]

I. study the theory

II. use $\neg (a \wedge b)$ is equivalent with $\neg a \vee \neg b$

 

[Celes]

Hey guys I have the same problem given by our professor but we weren't even taught anything that we can use with simplifying the given compound proposition.

I was hoping someone can help and show how to simplify that given, it would be a really great help.

 

[Kansas Boy]

I don't know what "study the theory" or "use" mean here. I do know that you have "p is true OR q is true" AND "p is NOT true". It follows that q must be true.