Mathematical Reasoning

Let $Δ, \nabla \in {\land, \lor}$ be such that $p \nabla q \Rightarrow ((p Δ q) \nabla r)$ is a tautology. Then $(p \nabla q) Δ r$ is logically equivalent to:

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Contributor-Level 9

Case – I $Δ \equiv \nabla \equiv \land$

$(p \land q) \to ((p \land q) \land r)$

it can be false if r is false,

so not a tautology

Case – II If $Δ \equiv \nabla \equiv \lor$

$(p \lor q) ? \to ((p \lor q) \lor r) \equiv$ tautology

then $(p \lor q) \lor r \equiv (p Δ r) \lor q$

Case – III If $Δ \equiv v, \nabla \equiv \land$ &nbs

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Case – I $Δ \equiv \nabla \equiv \land$

$(p \land q) \to ((p \land q) \land r)$

it can be false if r is false,

so not a tautology

Case – II If $Δ \equiv \nabla \equiv \lor$

$(p \lor q) ? \to ((p \lor q) \lor r) \equiv$ tautology

then $(p \lor q) \lor r \equiv (p Δ r) \lor q$

Case – III If $Δ \equiv v, \nabla \equiv \land$

then $(p \land q) \to {(p \lor q) \land r}$

Not a tautology

Case – IV If $Δ \equiv \land, \nabla \equiv \lor$

$(p \land q) \to {(p \land q) \lor r}$

Not a tautology

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New answer posted

4 weeks ago

Negation of the Boolean statement $(p \lor q) \Rightarrow ((\sim r) \lor p)$ is equivalent to

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Contributor-Level 10

pva $(\sim r v p)$

$\sim (p \lor q) \lor (\sim r v p)$

its negation as asked in question

$(p \lor q) \land (\sim p \land r)$

= $(p \land \sim p \land r) \lor (q \land r \land \sim p)$

= $(p \land r \land \sim p) [a s p \land \sim p i s f a l s e]$

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New answer posted

4 weeks ago

The remainder when (2021)²⁰²³ is divided by 7 is:

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$2021 \equiv - 2$ mod (7)

$\Rightarrow {(2021)}^{2023} \equiv {(- 2)}^{2023} m o d (7)$ …… (i)

Now, ${(- 2)}^{3} \equiv - 1 m o d (7)$

$\Rightarrow {(- 2)}^{2023} \equiv (- 2) m o d (7) \equiv 5 m o d (7)$ ……. (ii)

(i) & (ii)

$\Rightarrow {(2021)}^{2023} \equiv 5 m o d (7)$

$∴$ Remainder = 5

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New answer posted

4 weeks ago

Let p, q, r be three logical statements. Consider the compound statements

$S_{1} : ((\sim p) \lor q) \lor ((\sim p) \lor r) a n d$

$S_{2} : p \to (q \lor r)$

Then, which of the following is NOT true?

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Contributor-Level 9

kindly consider the following Image

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New answer posted

4 weeks ago

The statement $(\sim (p \Leftrightarrow \sim q)) \land q i s :$

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Contributor-Level 9

$\sim (p \Leftrightarrow \sim q) \land q = (p \Leftrightarrow \sim q) \land q i s :$

$∴ (\sim (P \Leftrightarrow \sim Q)) \land$ q is equivalent to $(p \Rightarrow q) \land$

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New answer posted

a month ago

The negation of the statement “If a quadrilateral is a square then it is a rhombus”.

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Contributor-Level 10

$\sim (p \to q) = p \land \sim q$

A quadrilateral is a square and it is not a rhombus

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New answer posted

a month ago

Let p, q, r be three statements such that the truth value of ~(~p ∨ ~q) → (~q ∨ r) is F. Then the truth values of p, q, r are respectively:

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Contributor-Level 9

$\sim (\sim p \lor \sim q) \to (\sim q \lor r) \Rightarrow p \land q \to \sim q \lor r$

$T F T \Rightarrow F \to T \to T$

$F T T \Rightarrow F \to T \to T$

$T T T \Rightarrow T \to T \to T$

$T T F \Rightarrow T \to F \to F$

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New answer posted

a month ago

Let p : Sindhu plays to win q : Sindhu gets Bharath Ratna

Then the contrapositive of $" \sim (\sim q \land p) "$ is

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alok kumar singh

Contributor-Level 10

$\sim (\sim q \land p) = (\sim (\sim q)) \lor (\sim p)$

$= \sim p \lor q$

$= p \to q$

$∴ c o n t r a p o s i t i v e i s \sim q \to \sim p$

Hence, if $\sim$ q then $\sim$ p

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New answer posted

a month ago

The statement p → (q → p) is equivalent is:

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Contributor-Level 10

Kindly consider the following figure

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New answer posted

a month ago

The negation of the compound proposition p∨(~p∨q) is

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