Logic

**Krizalid** · April 3rd 2008, 01:08 PM

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:

-------

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:

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

-------

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):

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

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

**angel.white** · April 3rd 2008, 01:14 PM

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:

-------

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:

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

-------

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):

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

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

what does veebar mean? Is that "exclusive or" ?

**Krizalid** · April 3rd 2008, 01:15 PM

Exactly.

**Moo** · April 3rd 2008, 01:17 PM

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...

**JaneBennet** · April 3rd 2008, 01:41 PM

$p\downarrow q$ 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 $\sim((\sim p)\downarrow q)$ .

**angel.white** · April 3rd 2008, 01:41 PM

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

**Plato** · April 3rd 2008, 01:54 PM

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.

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

**Soroban** · April 3rd 2008, 06:44 PM

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:

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

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 $FTFF$

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

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

**JaneBennet** · April 3rd 2008, 06:58 PM

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.

**Soroban** · April 3rd 2008, 09:45 PM

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.

Thread: Logic

LinkBack

Thread Tools

Display

Logic

Similar Math Help Forum Discussions

Can someone check my logic (sentential logic)

logic

logic

logic

Logic

Search Tags