The negation of ∼S∨(∼R∧S) is equivalent to

C. S∧R. Step 1: De Morgan: ∼[∼S∨(∼R∧S)]=S∧∼(∼R∧S)=S∧(R∨∼S). Step 2: Distribute: (S∧R)∨(S∧∼S)=(S∧R)∨F=S∧R. Check: S=T,R=T: the original is F∨(F∧T)=F, its negation is T, and S∧R=T. S=T,R=F: the original is F∨(T∧T)=T, its negation is F, and S∧R=F. Trap: (B) S∧∼R forgets to negate ∼R inside the bracket; at S=T,R=F it gives T while the negation is F.

Question 2 of 5 · Mathematical Logic

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The negation of $\sim S \lor (\sim R \land S)$ is equivalent to

Answer (Detailed Solution Below)View answer ▾

Answer: Option C — $S \land R$

Detailed Solution

De Morgan: $\sim [\sim S \lor (\sim R \land S)] = S \land \sim (\sim R \land S) = S \land (R \lor \sim S)$ .

Distribute: $(S \land R) \lor (S \land \sim S) = (S \land R) \lor F = S \land R$ .

✓ Check

S = T, R = T

: the original is

F \lor (F \land T) = F

, its negation is

T

, and

S \land R = T

S = T, R = F

: the original is

F \lor (T \land T) = T

, its negation is

F

, and

S \land R = F

⚠ Common trap

(B)

S \land \sim R

forgets to negate

\sim R

inside the bracket; at

S = T, R = F

it gives

T

while the negation is

F

Quick tip

$\sim [\sim S \lor (\sim R \land S)] = S \land (R \lor \sim S) = S \land R$

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Mathematical Logic — Preparation

This question tests your understanding of Mathematical Logic, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

Mathematical logic deals with statements that are either true or false, and with how they combine. This topic covers statements and logical connectives (negation, conjunction "and", disjunction "or", conditional "if...then" and biconditional "if and only if"), truth tables, tautologies, contradictions and contingencies, and logical equivalence. It also covers the converse, inverse and contrapositive of a conditional statement, quantifiers ("for all" and "there exists"), the negation of compound and quantified statements, De Morgan's laws, and switching circuits. Remember that a conditional statement p -> q is false only when p is true and q is false, and that it is logically equivalent to its contrapositive ~q -> ~p. Mathematical logic is a high-scoring chapter in MHT CET because its questions follow fixed rules and can be checked with truth tables. Practise the mathematical logic questions below, each with a truth table or step-by-step solution.

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The negation of $\sim S \lor (\sim R \land S)$ is equivalent to

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