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#### Joystar1977

##### Active member

- Jul 24, 2013

- 119

( p ^ ( ~ ( ~ p V q))) V ( p ^ q) = p

- Thread starter Joystar1977
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- Thread starter
- #1

- Jul 24, 2013

- 119

( p ^ ( ~ ( ~ p V q))) V ( p ^ q) = p

- Jan 30, 2012

- 2,492

- Thread starter
- #3

- Jul 24, 2013

- 119

p arrow q = slash bar q arrow slash bar p

Does this solve p and q being propositions and using logical equivalences to show that

( p ^ (~ (~ p V q))) V (p ^ q) = p

- Jan 30, 2012

- 2,492

I assume by "slash bar" you mean negation. In plain text, you can use the following notations.Can I write this p and q propositions as follows and use this logical equivalence that involves the following:

p arrow q = slash bar q arrow slash bar p

~ : negation (not)

/\ or ^ : conjunction (and)

\/ or v : disjunction (or)

-> : implication (if-then)

Or copy and paste characters from this page.

You probably mean contraposition: p -> q = ~q -> ~p.

The phrase "solve p and q being propositions" makes no sense to me. You need to prove an equality (in this case, better called equivalence) using other equalities.Does this solve p and q being propositions and using logical equivalences to show that

( p ^ (~ (~ p V q))) V (p ^ q) = p

I would not use contraposition here because the left-hand side has no implications. Start by applying De Morgan's law to ~(~p \/ q); then remove double negation.

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- #5

- Jul 24, 2013

- 119

Can I use the following logical equivalences where p and q are propositions where it shows I am using DeMorgan's Law?

1. p double arrow q = (p arrow q) ^ (q arrow p)

2. p double arrow q = slash bar p double arrow slash bar q

3. p double arrow q = (p ^q) V (slash bar p ^ slash bar q)

4. slash bar (p double arrow q) = p double arrow slash bar q

- Jan 30, 2012

- 2,492

These equivalences, while correct, have nothing to do with De Morgan's law.Can I use the following logical equivalences where p and q are propositions where it shows I am using DeMorgan's Law?

Could you say how many completed examples from your textbook or lecture notes that deal with proving equivalences like the one in post #1 have you seen and understood? Look specifically at applications of De Morgan's law. Then do the same for ~(~p \/ q), which is a subformula of (p ^ (~(~p V q))) V (p ^ q).

And please consider the notation suggestion in post #4 or use LaTeX because otherwise your formulas are hard to read.

[tex]\begin{array}{c}\text{Let }p\text{ and }q\text{ be propositions.} \\

\text{Use logical equivalences to show that:} \\ \bigg\{p \wedge \big[\sim(\sim\!p \vee q)\big]\bigg\} \vee (p \wedge q) \:\Leftrightarrow \: p \end{array}[/tex]

[tex]\begin{array}{cccccccc}

1. & \bigg\{p \wedge \big[\sim(\sim\!p \vee q)\big]\bigg\} \vee (p \wedge q) && 1. & \text{Given} \\

2. & \bigg\{p \wedge \big(p\: \wedge \sim\!q\big)\bigg\} \vee (p \wedge q) && 2. & \text{DeMorgan} \\

3. & \bigg\{(p \wedge p)\: \wedge \sim\! q\bigg\} \vee (p \wedge q) && 3. & \text{Assoc.} \\

4. & (p\: \wedge \sim\!q) \vee (p \wedge q) && 4. & s \wedge s \,=\,s \\ \\

5. & p \wedge (\sim\!q \vee q) && 5. & \text{Distr.} \\ \\

6. & p \wedge T && 6. & s \:\vee \sim\!s \,=\,T \\ \\

7. & p && 7. & s \wedge T \,=\,s \end{array}[/tex]

- Thread starter
- #8

- Jul 24, 2013

- 119

( p ^ q) p V q

( p V q) p ^ q

There isn't anything else mentioning De Morgan's Laws. I will try to copy and paste from Microsoft Word so I can get the symbols correct when I write them down.

- - - Updated - - -These equivalences, while correct, have nothing to do with De Morgan's law.

Could you say how many completed examples from your textbook or lecture notes that deal with proving equivalences like the one in post #1 have you seen and understood? Look specifically at applications of De Morgan's law. Then do the same for ~(~p \/ q), which is a subformula of (p ^ (~(~p V q))) V (p ^ q).

And please consider the notation suggestion in post #4 or use LaTeX because otherwise your formulas are hard to read.

Thanks Soroban for showing me this!

Hello, Joystar1977!

[tex]\begin{array}{cccccccc}

1. & \bigg\{p \wedge \big[\sim(\sim\!p \vee q)\big]\bigg\} \vee (p \wedge q) && 1. & \text{Given} \\

2. & \bigg\{p \wedge \big(p\: \wedge \sim\!q\big)\bigg\} \vee (p \wedge q) && 2. & \text{DeMorgan} \\

3. & \bigg\{(p \wedge p)\: \wedge \sim\! q\bigg\} \vee (p \wedge q) && 3. & \text{Assoc.} \\

4. & (p\: \wedge \sim\!q) \vee (p \wedge q) && 4. & s \wedge s \,=\,s \\ \\

5. & p \wedge (\sim\!q \vee q) && 5. & \text{Distr.} \\ \\

6. & p \wedge T && 6. & s \:\vee \sim\!s \,=\,T \\ \\

7. & p && 7. & s \wedge T \,=\,s \end{array}[/tex]

- Aug 30, 2012

- 1,123

I know it's pretty obvious but I just want to make certain: [tex]\text{~} s \wedge s = T[/tex] represents "Tautology." Is this correct?

7. & p && 7. & s \wedge T \,=\,s \end{array}[/tex]

-Dan

Addendum: Whoops! I didn't see all of Joystar1977's last post. I didn't mean to butt in.

- Jan 30, 2012

- 2,492

No, in step 6 Soroban has $s\lor\neg s=T$, which means that the truth value of $s\lor\neg s$ is T (Truth) regardless of the truth value of $s$. By definition, $s\lor\neg s$ is a tautology.I know it's pretty obvious but I just want to make certain: [tex]\text{~} s \wedge s = T[/tex] represents "Tautology." Is this correct?