Mathematical Reasoning

The given statement is:

$p \wedge (q \wedge \sim(p \wedge q))$

We want to find the negation of this statement. To do that, we negate the whole expression:

$\sim (p \wedge (q \wedge \sim(p \wedge q)))$

Now, we will use De Morgan's laws to simplify the expression. De Morgan's laws state:

1. $\sim (A \wedge B) \equiv \sim A \vee \sim B$
2. $\sim (A \vee B) \equiv \sim A \wedge \sim B$

Applying the first De Morgan's law to the outer conjunction of the negated expression:

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

Now we need to address the inner conjunction $(q \wedge \sim (p \wedge q))$. To simplify this part of the expression, we again apply the first De Morgan's law:

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

Distribute the disjunction over the inner conjunction:

$ \sim \mathrm{p} \vee((\sim \mathrm{q} \vee \mathrm{p}) \wedge(\sim \mathrm{q} \vee \mathrm{q})) $

Apply the law of excluded middle (A ∨ ¬A = T) on the last term:

$ \sim \mathrm{p} \vee(\sim \mathrm{q} \vee \mathrm{p}) $

Apply De Morgan's law on the inner disjunction:

$ \sim(\mathrm{p} \wedge \mathrm{q}) \vee \mathrm{p} $

Now, comparing this with the provided options, it matches with Option B :

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

So, the negation of the given statement is :

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

The given expression is

$P \to \sim Q \wedge R$

$ \equiv ( \sim P) \vee ( \sim Q \wedge R)$

$ \equiv ( \sim P \vee \sim Q) \wedge ( \sim P \vee R)$

${S1}:\left( {(p \vee q) \Rightarrow r} \right) \Leftrightarrow (p \Rightarrow r)$

${S2}:\left( {(p \vee q) \Rightarrow r} \right) \Leftrightarrow \left( {(p \Rightarrow r) \vee (q \Rightarrow r)} \right)$

In $S1:$ If $p=F, q=T, r=F$ then $S_1$ is false

In $S2:$ If $p=T, q=F, r=F$ then $S_2$ is false

$\therefore$ Neither S1 nor S2 is a tautology

Making truth table (Let $(p \wedge \sim q) \Rightarrow(p \Rightarrow \sim q)=E$ )

$\therefore E$ is a tautology

Making truth table $(E \equiv \sim(p \wedge(p \Rightarrow \sim q))$

$\&$

$\therefore \sim(p \wedge(p \Rightarrow \sim q))$ is equivalent to $\sim p \vee q$

Other Method :

$ \sim \left( {p \vee \left( {p \to \sim q} \right.} \right)$

$ = \sim p \vee \sim \left( {p \to \sim q} \right)$

$ = \sim p \vee \sim \left( { \sim p \vee \sim q} \right)$

$= \sim p \vee \left( {p \wedge q} \right)$

$ = \left( { \sim p \vee p} \right) \wedge \left( { \sim p \vee q} \right)$

$ = t \wedge \left( { \sim p \vee q} \right)$

$ = \sim p \vee q$

$(p\, \wedge \,q) \Rightarrow (p\, \wedge \,r)$ is equivalent to $(p\, \wedge \,q) \Rightarrow r$

p : Ramesh listens to music

q : Ramesh is out of his village

r : It is Sunday

s : It is Saturday

p $\to$ q conveys the same p only if q

Statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday"

$p \Rightarrow (( \sim \,q)\, \wedge \,(r\, \vee \,s))$

$ * ,\, \odot \in \{ \wedge ,\, \vee \} $

Now for $(p * q) \odot (p \odot \sim q)$ is tautology

(A) $( \vee , \wedge ):(p \vee q) \wedge (p \wedge \sim q)$ not a tautology

(B) $( \vee , \vee ):(p \vee q) \vee (p \vee \sim q)$

$ = P \vee T$ is tautology

(C) $( \wedge , \wedge ):(p \wedge q) \wedge (p \wedge \sim q)$

$ = (p \wedge p) \wedge (q \wedge \sim q) = p \wedge F$ not a tautology (Fallasy)

(D) $( \wedge , \vee ):(p \wedge q) \vee (p \vee \sim q)$ not a tautology

The truth table

Clearly p $\wedge$ q $\Leftrightarrow$ p $\wedge$ $\sim$ q $\equiv$ $\sim$ p

$\therefore$ r = q

$p \Leftrightarrow (q \Rightarrow p)$

$ \sim (p \Leftrightarrow (q \Leftrightarrow p))$

$ \equiv p \Leftrightarrow \, \sim (q \Rightarrow p)$

$ \equiv p \Leftrightarrow (q \wedge \sim p)$

$ \equiv (p \Rightarrow (q \wedge \sim p)) \wedge ((q \wedge \sim p) \Rightarrow p))$

$ \equiv ( \sim p \vee (q \wedge \sim p)) \wedge (( \sim q \vee p) \vee p))$

$ \equiv (( \sim p \vee q) \wedge \sim p) \wedge ( \sim q \vee p)$

$ \equiv \sim p \wedge ( \sim q \vee p)$

$ \equiv ( \sim p \wedge \sim q) \vee ( \sim p \wedge p)$

$ \equiv ( \sim p \wedge \sim q) \vee c$

$ \equiv ( \sim p \wedge \sim q)$

$\sim$ (p $ \Leftrightarrow $ $\sim$ q) $\wedge$ q

= (p $ \Leftrightarrow $ q) $\wedge$ q

$\therefore$ ($\sim$ (p $ \Leftrightarrow $ $\sim$ q)) $\wedge$ q is equivalent to (p $ \Rightarrow $ q) $\wedge$ p.

P : Ramu is intelligent

Q : Ramu is rich

R : Ramu is not honest

Given statement, "Ramu is intelligent and honest if and only if Ramu is not rich"

$ = (P \wedge \sim R) \Leftrightarrow \, \sim Q$

So, negation of the statement is

$ \sim [(P \wedge \sim R) \Leftrightarrow \, \sim Q]$

$ = \sim [\{ \sim (P \wedge \sim R) \vee \, \sim Q\} \wedge \{ Q \vee (P \wedge \sim R)\} ]$

$ = ((P \wedge \sim R) \wedge Q) \vee ( \sim Q \wedge ( \sim P \vee R))$

$\therefore$ Given statement is a tautology.

Given,

(p $\vee$ q) $\Rightarrow$ (($\sim$ r) $\vee$ p)

Negation is

$\sim$ ((p $\vee$ q) $\Rightarrow$ ($\sim$ r) $\vee$ p))

= (p $\vee$ q) $\wedge$ $\sim$ (($\sim$ r) $\vee$ p)

= (p $\vee$ q) $\wedge$ (r $\wedge$ $\sim$ p)

[(p $\wedge$ $\sim$ p) $\vee$ (q $\wedge$ $\sim$ p)] $\wedge$ r

= q $\wedge$ $\sim$ p $\wedge$ r

${S_1}:( \sim p \vee q) \vee ( \sim p \vee r)$

$ \cong ( \sim p \vee q \vee r)$

${S_2}: \sim p \vee (q \vee r)$

Both are same

So, option (C) is incorrect.

$\because$ (($\sim$ q) $\wedge$ p) $\vee$ (p $\vee$ ($\sim$ p))

= ($\sim$ q $\wedge$ p) $\vee$ t (t is tautology)

$\equiv$ t

$\therefore$ option (C) is correct.

Making truth table

$\therefore$ ($\sim$ (p $\wedge$ q)) $\vee$ q $\equiv$ p $\to$ (p $\vee$ q)

Clearly r must be equal to $\sim$ p

$\because$ $\sim$ p $\vee$ $\sim$ p = $\sim$ p

and (p $\wedge$ q) $\vee$ $\sim$ p = p

$\therefore$ $\sim$ p $\Rightarrow$ p = tautology.

Let $S:(( \sim q) \wedge p) \Rightarrow (( \sim p) \vee q)$

$ \Rightarrow S:\, \sim (( \sim q) \wedge p) \vee (( \sim p) \vee q)$

$ \Rightarrow S:(q \vee ( \sim p)) \vee (( \sim p) \vee q)$

$ \Rightarrow S:( \sim p) \vee q$

$ \Rightarrow S:p \Rightarrow q$

So, negation of S will be $ \sim (p \Rightarrow q)$

$\because$ Given statement is

(A $\wedge$ C) $\to$ B

Then its negation is

$\sim$ {(A $\wedge$ C) $\to$ B}

or $\sim$ {$\sim$ (A $\wedge$ C) $\vee$ B}

$\therefore$ (A $\wedge$ C) $\wedge$ ($\sim$ B)

or ($\sim$ B) $\wedge$ (A $\wedge$ C)

Negative of (p $\vee$ r) $\Rightarrow$ (q $\vee$ r)

$ \equiv $ $\sim$ ((p $\vee$ r) $\Rightarrow$ (q $\vee$ r)) $\equiv$ (p $\vee$ r) $ \wedge $ ($\sim$ (q $\vee$ r))

$\equiv$ (p $\vee$ r) $\wedge$ ($\sim$ q $\wedge$ $\sim$ r) $\equiv$ (p $\vee$ r) $\wedge$ $\sim$ r) $\wedge$ $\sim$ q

$\equiv$ ((p $\wedge$ $\sim$ r)) $\vee$ (r $\wedge$ $\sim$ r) $\wedge$ $\sim$ q $\equiv$ (p $\wedge$ $\sim$ r) $\wedge$ f) $\wedge$ $\sim$ q

$\equiv$ (p $\wedge$ $\sim$ r) $\wedge$ ($\sim$ q) $\equiv$ p $\wedge$ $\sim$ q $\wedge$ $\sim$ r