equivalences help

**barc0de** · January 5th 2009, 10:41 AM

I have to verify the following, using equivalences. Can anyone help me please:

a) [( p Þq) and ( p Þnot q)] º not p

b) [(p and q) Û p ] º [p Þ q]

c) not ( p Ûq) º (p Ûnot q)

d) [(p and q Þ r) and (p and not q) Þ not r)] º [p Þ (q Û r)]

**Plato** · January 6th 2009, 11:25 AM

a)
$\begin{gathered} \left( {p \to q} \right) \wedge \left( {p \to \neg q} \right) \hfill \\ \left( {\neg p \vee q} \right) \wedge \left( {\neg p \vee \neg q} \right) \hfill \\ \neg p \wedge \underbrace {\left( {q \vee \neg q} \right)}_{\text{TRUE}} \hfill \\ \neg p \hfill \\ \end{gathered}$

**barc0de** · January 8th 2009, 02:42 AM

Can anyone double check my working please? I need to know if I'm doing these things correctly.

Thanks

[ ( P Ù Q ) Û ] º ( P Þ Q )
[ ( P Ù Q ) Þ ) ] Ù [ P Þ ( P Ù Q ) ] If and only if
[ Ø(P Ù Q ) Ú P ] Ù [ ØP Ú ( P Ù Q ) ] Implication
[ ( ØP ÚØQ ) Ú P ] Ù [ ØP Ú ( P Ù Q ) ] De Morgans
[ ( ØP Ú P ) ÚØQ ] Ù [ ØP Ú ( P Ù Q ) ] Associativity
[ True ÚØQ ] Ù [ ØP Ú ( P Ù Q ) ] Law of Excluded Middle
[ ØP Ú ( P Ù Q ) ] Removal Of Constant
( ØP Ú P ) Ù ( ØP Ú Q ) Distributivity
ØP Ú Q Law of Excluded Middle
P Þ Q Implication

**Soroban** · January 9th 2009, 12:32 PM

Hello, barc0de!

$(c)\;\;\sim(p \Longleftrightarrow q) \;=\;(p \Longleftrightarrow \:\sim q)$

We have: . $\sim\bigg[(p \to q) \wedge (q \to p)\bigg]$

. . . . . . $=\;\;\sim\bigg[(\sim p \vee q) \wedge (\sim q \vee p)\bigg]$

. . . . . . $= \;\;\sim(\sim p \vee q) \:\vee \sim(\sim q \vee p)$

. . . . . . $=\quad(p\: \wedge \sim q)\:\vee (q \:\wedge \sim p)$

. . . . . . $= \;\bigg[(p\: \wedge \sim q) \vee q\bigg] \wedge \bigg[(p \:\wedge \sim q) \:\vee \sim p\bigg]$

. . . . . . $= \;\bigg[(p \vee q) \wedge (q \:\vee \sim q)\bigg] \wedge \bigg[(p \:\vee \sim p) \wedge (\sim q \:\vee \sim p)\bigg]$

. . . . . . $= \;\bigg[(p \vee q) \wedge t\bigg] \wedge \bigg[t \wedge (\sim q \:\vee \sim p)\bigg]$

. . . . . . $= \;(p \vee q) \wedge (\sim q \:\vee \sim p)$

. . . . . . $= \;(q \vee p) \wedge (\sim p \:\vee \sim q)$

. . . . . . $= \;(\sim q \to p) \wedge (p \to \:\sim q)$

. . . . . . $= \;\;p \Longleftrightarrow \:\sim q$

Your work on part (b) is correct . . . Nice work!

**barc0de** · January 9th 2009, 02:35 PM

Thank you very much Soroban for taking the time to check for me. I will post the other two when I have figured them out.

Again thanks,

barc0de

**barc0de** · January 10th 2009, 09:36 AM

Hi Soroban. I don't understand the last bit of your solution. Please could you tell me how you got from this:

$\;(q \vee p) \wedge (\sim p \:\vee \sim q)$

to this:

$ \;(\sim q \to p) \wedge (p \to \:\sim q) $

Thanks

**Plato** · January 10th 2009, 03:04 PM

You are expected to know basic facts.
$\begin{gathered} a \to b \equiv \neg a \vee b \hfill \\ a \to b \equiv \neg b \to \neg a \hfill \\ \neg c \vee \neg d \equiv c \to \neg d \equiv d \to \neg c \hfill \\ \end{gathered}$

Thread: equivalences help

LinkBack

Thread Tools

Display

equivalences help

Similar Math Help Forum Discussions

Proving equivalences

Help with logical equivalences

logical equivalences

Logical Equivalences

Logical equivalences

Search Tags