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Compound Proposition Simplification

User40405

New member
Oct 7, 2018
2
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: Capture.PNG
 

steenis

Well-known member
MHB Math Helper
Jul 30, 2016
250
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

New member
Oct 7, 2018
2
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.

- - - Updated - - -

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

Olinguito

Well-known member
Apr 22, 2018
251
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

Well-known member
MHB Math Helper
Jul 30, 2016
250
I. study the theory

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