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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Mathematics
Chapter
Mathematical Reasoning

BITSAT 2023 · Paper 1 · Mathematics · Mathematical Reasoning

Negation of the Boolean expression p (q p) is

Negation of the Boolean expression p⇔(q⇒p)\mathrm{p} \Leftrightarrow(\mathrm{q} \Rightarrow \mathrm{p}) is

Options

  1. A(∼p)∧q(\sim p) \wedge q
  2. B(p)∧(∼q)(\mathrm{p}) \wedge(\sim \mathrm{q})
  3. C(∼p)∨(∼q)(\sim \mathrm{p}) \vee(\sim \mathrm{q})
  4. D(∼p)∧(∼q)(\sim p) \wedge(\sim q)

Correct answer

D. (∼p)∧(∼q)(\sim p) \wedge(\sim q)

Step-by-step solution

Given expression is p⇔(q⇒p)p \Leftrightarrow(q \Rightarrow p)
∼(p↔(q→p))\sim(\mathrm{p} \leftrightarrow(\mathrm{q} \rightarrow \mathrm{p}))
∼(p↔q)=(p∧∼q)∨(q∧∼p)\sim(\mathrm{p} \leftrightarrow \mathrm{q})=(\mathrm{p} \wedge \sim \mathrm{q}) \vee(\mathrm{q} \wedge \sim \mathrm{p})
∼(p↔(q→p))=(p∧∼(q→p))∨((q→p)∧∼p)\sim(\mathrm{p} \leftrightarrow(\mathrm{q} \rightarrow \mathrm{p}))=(\mathrm{p} \wedge \sim(\mathrm{q} \rightarrow \mathrm{p})) \vee((\mathrm{q} \rightarrow \mathrm{p}) \wedge \sim \mathrm{p})
(p∧∼(q→p))=p∧(q∧∼p)=(p∧∼p)∧q=c(\mathrm{p} \wedge \sim(\mathrm{q} \rightarrow \mathrm{p}))=\mathrm{p} \wedge(\mathrm{q} \wedge \sim \mathrm{p})=(p \wedge \sim p) \wedge q=c
(q→p)∧∼p=(∼q∨p)∧∼p=∼p∧(∼q∨p)(q \rightarrow p) \wedge \sim p=(\sim q \vee p) \wedge \sim p=\sim p \wedge(\sim q \vee p)
=(∼p∧∼q)∨(∼p∧p)=∼p∧∼q=(\sim \mathrm{p} \wedge \sim \mathrm{q}) \vee(\sim \mathrm{p} \wedge \mathrm{p})=\sim \mathrm{p} \wedge \sim \mathrm{q}
∼(p↔(q˙→p))=c∨(∼p∧∼q)=∼p∧∼q\sim(p \leftrightarrow(\dot{q} \rightarrow p))=c \vee(\sim p \wedge \sim q)=\sim p \wedge \sim q
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