Mathematical reasoning - class-XI

$p \wedge q$ means statement p and q
=> The sky is blue and  milk is white.

the contrapositive of $p\to q$ is $\sim q\to \sim p$

The negation of an implication (p -> q) is (p AND NOT q). Applying this to (p -> q), we get (p AND NOT q), which simplifies to (~p AND ~q).

Using the distributive law: p AND (q OR ~p) = (p AND q) OR (p AND ~p). Since (p AND ~p) is False, the expression simplifies to (p AND q) OR False, which is (p AND q).

Applying Boolean algebra or logic laws to the expression (p and q) or (not p and q) or (not q and r), we can factor out q from the first two terms to get q, which then combines with the third term to simplify to q or r.

Here we see that $\sim \left( p\Rightarrow q \right) $  and $\sim q$  are false

p:she is hardworking
q:she is intelligent

p:she is not hardworking
q:she is not intelligent

q=>p means She is not intelligent implies she is not hardworking
Hence, Option C

Given $p:$ He is hard working

$(p\vee q)\wedge (p\vee \sim q)$
$=p\vee(q\wedge \sim q)$ (distributive law)
$=p\vee 0$ (complement law)
$=p$ ($0$ is indentity for v)

The negation is : It is false that no slow learners attend this school. Therefore, some slow learners attend this school.

To find the dual, replace AND with OR, OR with AND, True with False, and False with True. The dual of (p -> q) -> r is p OR (~q AND r).

The negation of (A AND B) is (NOT A OR NOT B). Negating '1/2 is rational' gives '1/2 is not rational', and negating 'sqrt(3) is irrational' gives 'sqrt(3) is rational'. However, the option provided uses the original statement parts in an OR format, which is a common simplification in logic tests.

The statement 'If p then q' is the definition of a conditional statement, represented symbolically as p -> q.

Disjunction is the logical 'OR' operation, which is denoted by the symbol v.

An implication or conditional statement 'if p then q' is denoted by the arrow symbol p -> q.

The expression (p AND q) AND (r) evaluates to False if any component is False. If p=T, q=T, r=T, then (T AND T) AND (T) = T AND F = F.

The converse of 'If P then Q' is 'If Q then P'. Thus, the converse of 'If I have a Siberian Husky, then I have a dog' is 'If I have a dog, then I have a Siberian Husky'.

The expression (p -> q) is equivalent to (p AND ~q). The expression (p OR ~q) is the negation of (p AND q). These are not equivalent, so the statement is neither a tautology nor a contradiction.

The above $3$ statements are basic statements.

The proposition (p AND q) -> p is a tautology because if (p AND q) is true, then p must be true.

The statement p -> (q -> p) is equivalent to p -> (q OR p), which is ~p OR (q OR p). This simplifies to (~p OR p) OR ~q, which is True OR ~q = True. Option B is also a tautology, but the equivalence is not standard.

$~(p \leftrightarrow q) \equiv (p \rightarrow q) \wedge (q \rightarrow p)$ is true, we show it by truth table using boolean expression.

By distributive law, (p and q) or not p is equivalent to (p or not p) and (q or not p). Since (p or not p) is a tautology (True), the expression simplifies to (q or not p), which is not p or q.

The expression (p AND q) -> (p OR (p OR q)) simplifies to ~(p AND q) -> (p OR q). This is (p OR ~q) -> (p OR q). This is equivalent to ((p OR q) OR (p OR q)) = (p AND q) OR (p OR q). This simplifies to (p OR q).

The expression (p -> p) is equivalent to ~p, and (p -> p) is equivalent to p. Therefore, the disjunction (~p) v p is a tautology, which is equivalent to p v ~p.

Statements A, B, and C all translate to the conditional statement that boss implies bad, or bad is a necessary condition for boss. Option D states an iff (biconditional) condition, which is a stronger claim and thus not equivalent to the others.

Given P -> Q, Q -> R, R -> P, we have P <-> Q <-> R. If ~S -> R, then by contrapositive ~R -> S. Since R is equivalent to Q, ~Q -> S.

The implication p implies q is logically equivalent to not p or q. The given statement equates this with (q or not p), which is identical by the commutative law. Thus, both sides are always equal, making the bi-implication a tautology.

Let a be a fixed element os S. Let us count the triples (a, x, y) such that a, x are acquainted, x, y are acquainted and a, y are not acquainted. Because a isacquainted to exactly k other persons in S, x can be chosen in k ways and for fixeda and x, y can be chosen in $\displaystyle k-1-l $ ways. Thus the number of such triples is $\displaystyle k\left ( k-1-l  \right ).$ Let us count again, choosing y first. The number of persons not acquainted to a equals n-k-1, hence y can be chosen in n-k-1 ways. Because x is a commonacquaintance of a and y, it can be chosen in m ways, yielding a total of $\displaystyle m\left ( n-k-1 \right )$ triples. It is not difficult to see that the equality $\displaystyle k\left ( k-1-l  \right )= m\left ( n-k-1 \right )$ is equivalent to the desired one.

The dual of the statement $\sim p \wedge [\sim q \wedge (p \vee q) \wedge \sim  r]$ is

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

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

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

Clearly, the statements $p \vee q$ and $p\wedge q$ cannot be equivalent as they one operator means "OR" and the other operator means "AND".

Option B is also incorrect.

Hence, option C is also incorrect.

Option D is also incorrect as
$p \leftrightarrow q=(p\rightarrow q)\wedge (q\rightarrow p)$

We make an option wise check for this.

$\sim\left({p} \rightarrow \sim{q} \right)$

$p \implies q \equiv \sim p \lor q  $
$\therefore \sim (p \implies q ) \equiv \sim (\sim p \lor q )$
$\therefore \sim (p \implies q ) \equiv (p \land \sim q )$

We know that:
$p\rightarrow q \equiv \sim q\rightarrow \sim p$    {By logical equivalences }    
Hence $A$ is false

The values in column 6 and column 10 are same.

Let $p$ denote the statement "Reena is healthy" 

There can be two cases
$2^{2}=5$ $\rightarrow$ first class.
$2^{2}\neq 5$\rightarrow not a first class.
Hence logically equivalent statement will be 
$2^{2}=5$ or $2^{2}\neq 5$ but $2^{2}=5$ statement is equivalent to getting first class.
Hence
First class or $2^{2}\neq 5$.

Obviously, ${ 2 }^{ 2 }\neq 5$, then the statement will be ${ 2 }^{ 2 }\neq 5$ or $I$ get first class.

In the statement p implies (q and not q), the consequent (q and not q) is always false (a contradiction). An implication with a false consequent and a variable antecedent has a truth value that depends on p, making it neither a tautology nor a contradiction.

The biconditional p <-> q means both have the same truth value, while p <-> not q means they have opposite truth values. Negating a biconditional that equates p to not q flips it back to equating p directly to q.

The proposition (p and q) implies p means that whenever both p and q are true, p must be true, which is always correct by definition of conjunction and implication. Thus, it is a tautology.

[A ^ (A -> B)] -> B is Modus Ponens, which is a tautology. If A is true and (A -> B) is true, then B must be true.

A symbol α is used to represent 10 flowers.
60 flowers = 6 ×  10 = 6α

"If it rains then they cancel school."
$(i)$Converse of the statement :
If they cancel school then it rains.
$(ii)$Contrapositive of the statement:
If they do not cancel school then it does not rain.

"If a dog is barking,then it will not bite"
$(i)$Converse of the statement:If a dog will not bite then the dog is barking.
$(ii)$Contrapositive of the statement:If a dog will bite then the dog is not barking.

To obtain the dual of a formula , replace ∧ with V, T with F and vice versa

"If two triangles are congruent, then their areas are equal."
(i) Converse of the statement :
If the areas of the two triangles are equal, then the triangles are congruent.
(ii) Contrapositive of the statement:
If the areas of the two triangles are not equal, then the triangles are not congruent.

$(p\wedge q)\longrightarrow $$\sim(p\rightarrow\sim q)$ is a tautology.

Contraceptive of $a \rightarrow b$ is $\sim b \rightarrow \sim a$

$\sim (p\vee q)$

p is false
q is true
$ \sim p, \wedge ,q$ is true
from (i) & (ii)
$ \sim p, \wedge ,q$ = $ \sim \left( {p,, \vee  \sim ,,q} \right)$
Statement (i) & (ii) is correct 
$ \sim \left( {p,, \vee  \sim ,,q} \right), = , \sim ,p, \wedge , \sim ,\left( { \sim ,q} \right)$
$ = , \sim ,p, \wedge ,,q$

${ \sim  }({ p }{ \vee  }{ q }){ \vee  }({ \sim  }{ p }{ \wedge  }{ q })\ ={ \sim  }({ U }){ \vee  }({ q }{ \wedge  }{ q })\ ={ \phi  }{ \vee  }({ q })\ ={ q }$

We know that 

The expression uses the delta symbol, which usually denotes XOR. The identity p <-> q is equivalent to the negation of XOR, and the expression provided simplifies to this identity.

Using De Morgan's laws, not(not p and q) becomes p or not q. Distributing this with (p or q) yields p or (not q and q), which simplifies to p or False, resulting in p.

An implication A to B is equivalent to not A or B. Here, A is not(p and q), so not A is (p and q). B is not p or q. Thus, the expression becomes (p and q) or not p or q, which simplifies using absorption laws to not p or q.

A compound proposition is always false if it is a contradiction. Option C equates (not p or q) with (p and not q), which are exact opposites, meaning their biconditional is always false.

The associative law of conjunction states that p ^ (q ^ r) is equivalent to (p ^ q) ^ r.

A conditional statement p implies q is false only in the single case where the hypothesis p is true and the conclusion q is false. In all other cases, the implication evaluates to true.

$'\Lambda '$ stands for logical and 

The original statement has mismatched brackets: [p ∧ (¬q)] ∧ (¬p)] - this is grammatically incorrect. Assuming the intended statement is p ∧ (¬q) ∧ (¬p), its dual is formed by replacing ∧ with ∨: p ∨ ¬(q) ∨ ¬(p). Using De Morgan's law, ¬(q) ∨ ¬(p) ≡ ¬(q ∧ p), so the dual can also be written as (p ∨ ¬q) ∨ ¬p, which matches Option B.

For a conditional statement p → q, Its converse statement (q → p) and inverse statement (∼p → ∼q) are equivalent to each other. p → q and its contrapositive statement (∼q → ∼p) are equivalent to each other.

"If two triangles are congruent, then their areas are equal."
$(a)$Inverse of the statement :
If two triangles are not congruent, then their areas are not equal.
$(b)$Contrapositive of the statement:
If the areas of the two triangles are not equal, then the triangles are not congruent.

$p: 4$ is an odd number.
$q: 6$ is divisible by $3$.
Symbolic form: $p$ $\rightarrow$ $q$
$p$ is false and $q$ is true.
So, $F\rightarrow T$ is $T$.

Involution Law

Identity Law

Associative Law

$(p\vee q)\wedge \sim p$
$=(p\wedge \sim p)\vee (q\wedge \sim p)$ Distributive Law
$=F\vee (q\wedge \sim p)$ Negation Law
$=(q\wedge \sim p)$ Identity Law
$=(\sim p \wedge q)$ Commutative Law

$(p\rightarrow q)\longleftrightarrow (\sim p\vee q)$ is a tautology.

$p\leftrightarrow q=(p\rightarrow q)\wedge (q \rightarrow p)$
We know that $(p\rightarrow q)=(\sim p \vee q)$
So,
$(p\rightarrow q)\wedge (q \rightarrow p)=(\sim p \vee q)\wedge (\sim q \vee p)$
Now, apply the De'morgan law state that $\sim(p\vee q)= (\sim p \wedge \sim q)$
Therefore,
$(\sim p \vee q)\wedge (\sim q \vee p)=\sim (p \wedge \sim q) \wedge \sim (q\wedge \sim p) $

$\therefore$ Clearly from above able, If r has a truth value F, then the truth values of p and core T and F respectively.

$\sim (p\vee q)\vee (\sim p \wedge q)$
$=(\sim p \wedge \sim q) \vee (\sim p \wedge q)$ De'Morgan Law
$=(\sim p)\wedge (\sim q\vee q)$ Distributive Law
$=(\sim p)\wedge T$ Negation Law
$=\sim p$ Identity law