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the problem is verify this statement: $$\lnot ( \lnot p \land q)^\land (p \lor q)=p$$

That turns into $p \lor ( \lnot q \land q)$ - which that last part of the statement is a contradiction and is always false

which means now I have $p \lor F =p$

Does that mean that the original statement is true and valid or does it mean it's false? Could someone please explain.

amWhy
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LittleAuditorium
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  • It means that the equivalence does indeed hold, since you have shown that you can convert $\neg(\neg p \land q) \land (p \lor q)$ into $p$ through the application of elementary equivalence principles. – Bram28 Jan 30 '18 at 19:13

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So, the task is to determine whether $$\lnot( \lnot p \land q)\land (p \lor q) \equiv p \tag{Q}$$

We'll start on the left-hand side (LHS) and use equivalencies to prove that $(Q)$ is true:

$$\begin{align} \lnot( \lnot p \land q)\land (p \lor q) &\equiv (p \lor \lnot q)\land (p \lor q)\tag{De Morgan's} \\ \\ &\equiv p \lor (\underbrace{\lnot q \land q}_{\text{False}})\tag{distributive property} \\ \\ &\equiv p \lor \text {False}\\ \\ &\equiv p \end{align}$$

This means that $$\lnot( \lnot p \land q)\land (p \lor q) \equiv p\tag{Q}$$ is true, since we've shown that the left hand side of $(Q)$ is equivalent to the right hand side, hence the equivalence is true.

amWhy
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  • does ⊥ mean a contradiction or a tautology? contradiction right? So p ∨ contradiction always just equals p? – LittleAuditorium Jan 30 '18 at 19:22
  • It means that $\lnot q \land q$ is always false. (Yes, $p \lor \text {contradiction} \equiv p$, always. Remember in a disjunction, which is what we have, $p \lor \bot \equiv p$. However, if we had had a conjunction at the end, of the form $p \land \bot $, then we would end with $p\land \bot \equiv \bot.$ – amWhy Jan 30 '18 at 19:26
  • That explains a lot thank you. If you don't mind and have time could you type out what you just typed out in the comment above but with tautology's? – LittleAuditorium Jan 30 '18 at 19:39
  • $p \lor \top \equiv \top$, where $\top$ is a tautology. For example, $p \lor (q \lor \lnot q) \equiv p \lor \top \equiv \top.$ Suppose we have $p \land (q \lor \lnot q)$. Then we have $p \land \top \equiv p$. – amWhy Jan 30 '18 at 19:47
  • So you're saying that if we end up p ∨ ⊤=p ... then the statement does not equal p? but if we end up with p ^ T=p ... then the statement does equal p? – LittleAuditorium Jan 31 '18 at 18:56
  • That's correct: For disjunction ($\lor$), we have $p \lor F = p$, $p \lor T = T$. For conjunction, ($\land$), we have $p \land F = F$, $p\land T = p$ – amWhy Jan 31 '18 at 19:02
  • Does that help, @LittleAuditorium ? – amWhy Jan 31 '18 at 19:40
  • Yeah that gives me a much better understanding from before asking this question and your posts. Thank you for helping me out it helped a ton @amWhy – LittleAuditorium Jan 31 '18 at 20:13