# Math Help - Logic

1. ## Logic

Express the propositions $p\implies q$ and $q\wedge p$ using only $\downarrow$ and $\sim.$

$p\downarrow q,$ where $p$ and $q$ are propositions, it's defined by the following truth table:

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Let $p_1,p_2,p_3,p_4,p_5,p_6$ be propositions such that $\left[ {(p_1 \veebar p_2 ) \implies (p_3 \implies p_4 )} \right]$ is false. Determine the value of truth of:

1. $(p_5\implies p_6)\vee(p_1\vee p_2).$
2. $[(p_5\implies p_2)\vee\sim p_1]\implies(p_4\vee p_3).$
3. $\sim[(p_6\vee p_5)\wedge(p_1\wedge p_2)\iff(p_4\implies p_3)].$

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Let $p,q$ and $r$ be propositions such that $((\sim p\vee q)\implies r)$ is false. Give the value of truth of the following propositions (justify your answer):

1. $\sim q\implies\sim p.$
2. $r\implies(p\iff\sim(q\vee r)).$

I'm learning these stuff, so if you can help me, it'd be great. Thanks

2. Originally Posted by Krizalid
Express the propositions $p\implies q$ and $q\wedge p$ using only $\downarrow$ and $\sim.$

$p\downarrow q,$ where $p$ and $q$ are propositions, it's defined by the following truth table:

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Let $p_1,p_2,p_3,p_4,p_5,p_6$ be propositions such that $\left[ {(p_1 \veebar p_2 ) \implies (p_3 \implies p_4 )} \right]$ is false. Determine the value of truth of:

1. $(p_5\implies p_6)\vee(p_1\vee p_2).$
2. $[(p_5\implies p_2)\vee\sim p_1]\implies(p_4\vee p_3).$
3. $\sim[(p_6\vee p_5)\wedge(p_1\wedge p_2)\iff(p_4\implies p_3)].$

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Let $p,q$ and $r$ be propositions such that $((\sim p\vee q)\implies r)$ is false. Give the value of truth of the following propositions (justify your answer):

1. $\sim q\implies\sim p.$
2. $r\implies(p\iff\sim(q\vee r)).$

I'm learning these stuff, so if you can help me, it'd be great. Thanks
what does veebar mean? Is that "exclusive or" ?

3. Exactly.

4. Hello,

I'm not sure i understood well these notations o.O

What if you make a table of truth for A => B, which is ¬P v Q ?

A=1, B=1, (A => B) = 1
A=0, (A => B) = 1
A=1, B=0, (A => B) = 0

For the first one, find the values of p1 to p4 such as the proposition is false. Then you can determine the table of truth of i. ii. and iii. according to the values of p1 to p4...

5. $p\downarrow q$ is $(\sim p)\wedge (\sim q)$; hence $q\wedge p$ is $(\sim q)\downarrow(\sim p)$.

$p\Rightarrow q$ is $\sim(p\wedge(\sim q))$ so that would be $\sim((\sim p)\downarrow q)$.

6. Originally Posted by Krizalid
Express the propositions $p\implies q$ and $q\wedge p$ using only $\downarrow$ and $\sim.$

$p\downarrow q,$ where $p$ and $q$ are propositions, it's defined by the following truth table:

Looking at your truth table, $p \downarrow q$ = $\sim p \wedge \sim q$

So we can get it to have the same truth value as p -> q in a truth table if we make it look like this: $\sim (\sim \sim p \wedge \sim q)$

Which translates to $\sim (\sim p \downarrow q)$

The reason this works is because
$\sim (\sim \sim p \wedge\sim q) = \sim( p \wedge \sim q)$ by double negation

$\sim( p \wedge \sim q) = \sim p \vee q$ by De Morgan's Laws

$\sim p \vee q = p \to q$ by implication

edit: just fixed it, I had transposed my wedges and vees

7. Originally Posted by Krizalid
Express the propositions $p\implies q$ and $q\wedge p$ using only $\downarrow$ and $\sim.$
$p\downarrow q,$ where $p$ and $q$ are propositions, it's defined by the following truth table:
This is also know as Sheffer’s stroke named for H.M.Sheffer but actually C.S.Peirce discovered it. Here are some equivalences to help you.

${\neg P} \equiv {P \downarrow P}$
${P \wedge Q} \equiv {\left( {P \downarrow P} \right) \downarrow \left( {Q \downarrow Q} \right)}$
${P \vee Q} \equiv {\left( {P \downarrow Q} \right) \downarrow \left( {P \downarrow Q} \right)}$
${P \to Q} \equiv {\left[ {\left( {P \downarrow P} \right) \downarrow Q} \right] \downarrow \left[ {\left( {P \downarrow P} \right) \downarrow Q} \right]}$

8. Hello, Krizalid!

Express the propositions $p\to q$ and $q\wedge p$ using only $\downarrow$ and $\sim$

where $p\downarrow q,$ is defined by the following truth table:

. . . $\begin{array}{cc|c} p & q & p \downarrow q \\ \hline
T & T & F \\ T & F & F \\ F & T & F \\ F & F &T \end{array}$

I don't know of any direct method; I used some good guessing . . .

For the first one, we know the pattern: . $\begin{array}{cc|c} p & q & p\to q \\ \hline T & T & {\color{blue}T} \\ T & F & {\color{blue}F} \\ F & T & {\color{blue}T} \\ F & F & {\color{blue}T} \end{array}$

Now we must come up with $TFTT$ using only $\downarrow$ and $\sim$

With a little testing I found that $\sim p \downarrow q$ has $FTFF$

. . $\begin{array}{cc|ccc} p & q & \sim p & \downarrow & q \\ \hline
T & T & F & {\color{blue}F} & T \\ T & F & F & {\color{blue}T} & F \\ F & T & T & {\color{blue}F} & T \\ F & F & T & {\color{blue}F} & F \end{array}\quad\hdots\quad\text{Now negate it }\quad\hdots$
. . $\begin{array}{c}\sim(\sim p \downarrow q) \\ \hline {\color{blue}T} \\ {\color{blue}F} \\ {\color{blue}T} \\ {\color{blue}T} \end{array}$

Therefore: . $\sim(\sim p \downarrow q)\,\text{ is equivalent to }\,p \to q$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For the second one: . $\begin{array}{cc|c} p & q & q \wedge p \\ \hline T & T & {\color{blue}T} \\ T & F & {\color{blue}F}\\ F & T & {\color{blue}F} \\ F & F & {\color{blue}F} \end{array}$
I found that . $\sim p \downarrow \sim q$ .produces the same pattern: . $\begin{array}{ccc}\sim p & \downarrow & \sim q \\ \hline F & {\color{blue}T} & F \\ F & {\color{blue}F} & T \\ T & {\color{blue}F} & F \\ T & {\color{blue}F} & T \end{array}$

Therefore: . $\sim p \downarrow \sim q\,\text{ is equivalent to }\,q \wedge p$

9. Originally Posted by Plato
This is also know as Sheffer’s stroke named for H.M.Sheffer but actually C.S.Peirce discovered it. Here are some equivalences to help you.

${\neg P} \equiv {P \downarrow P}$
${P \wedge Q} \equiv {\left( {P \downarrow P} \right) \downarrow \left( {Q \downarrow Q} \right)}$
${P \vee Q} \equiv {\left( {P \downarrow Q} \right) \downarrow \left( {P \downarrow Q} \right)}$
${P \to Q} \equiv {\left[ {\left( {P \downarrow P} \right) \downarrow Q} \right] \downarrow \left[ {\left( {P \downarrow P} \right) \downarrow Q} \right]}$
And $P\leftrightarrow Q \equiv \left[\left(P\downarrow P\right)\downarrow Q\right]\downarrow\left[P\downarrow\left(Q\downarrow Q\right)\right]$.

The truth table is exactly the same as the logic of a NOR gate in electronics.

10. Hello, Krizalid!

I have explanations for the third problem,
. . but I don't know if they are acceptable for your course.

I also used the alternate defintion of implication:
. . $p \implies q\;\text{ is equivalnent to: }\,\sim p \vee q$

Let $p,q, r$ be propositions such that: . $(\sim p\vee q)\implies r$ .is false.
Give the value of truth of the following propositions. Justify your answers.

$(i)\;\;\sim q \implies\: \sim p$

$(ii)\;\;r \implies [p \iff \sim (q \vee r)]$
An implication has this truth table: . $\begin{array}{cc|c} p & q & p \to q \\ \hline T & T & T \\ T & F & F \\ F & T & T \\ F & F & T\end{array}$

Note:

. . An implication is false if the first part is true and the second part is false.

. . An implication is true if the first part is false.

$\text{We are told that: }\;\underbrace{(\sim p \vee q)}_T \implies \underbrace{r}_F\:\text{ is false.}$

Hence: . $\sim p \vee q\:\text{ is true, and }\:r\text{ is false.}$

(i) If $(\sim p \vee q)$ is true, then $(q\, \vee \sim p)$ is true.

. . But $(q\,\vee \sim p)$ is equivalent to: $\sim q \implies \,\sim p$

. . Therefore: . $\sim q \implies \,\sim p$ is true.

(ii) We are given: . $r \implies [p \iff \sim(q \vee r)]$

. . And we are told that $r$ is false.

. . $\text{So we have: }\:\underbrace{r}_F \implies \underbrace{[p \iff \sim(q \vee r)]}_{\text{doesn't matter}}$

. . Since the first part is false, the implication is true.