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JEE Mains Maths · Mathematical Reasoning

Negation and Simplification

Negate an implication, a biconditional or an and/or combination, and reduce a long statement to a short equivalent one.

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Why this matters

Nineteen PYQs, all multiple choice. Seven negate an implication or a biconditional, five negate an and/or combination with De Morgan's laws, and seven reduce a statement to a shorter equivalent. Three ideas cover the page.

Concept 1 of 3: Negating an implication

An implication is false only when the antecedent is true and the consequent false. So its negation is exactly that case: ∼(A→B)≡A∧∼B\sim(A\rightarrow B)\equiv A\wedge\sim B. It is an 'and', not another implication. A biconditional is false when its two sides differ, so its negation is A↔∼BA\leftrightarrow\sim B.

Definition

  • ∼(A→B)≡A∧∼B\sim(A\rightarrow B)\equiv A\wedge\sim B.
  • ∼(A↔B)≡A↔∼B≡(A∧∼B)∨(∼A∧B)\sim(A\leftrightarrow B)\equiv A\leftrightarrow\sim B\equiv(A\wedge\sim B)\vee(\sim A\wedge B).
  • Simplify the inside first: if the antecedent is a tautology, T→A≡AT\rightarrow A\equiv A.

Negation of an implication

∼(A→B)≡A∧∼B\sim(A\rightarrow B)\equiv A\wedge\sim B

Worked example

Negate (p∧q)→∼r(p\wedge q)\rightarrow\sim r.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q64Moderate

Example 1 · Mathematical Reasoning · Negation and Simplification

The negation of the statement ((A∧(B∨C))⇒(A∨B))⇒A((A \land (B \vee C)) \Rightarrow (A \vee B)) \Rightarrow A is

Not the converse, not the inverse

∼(p→q)\sim(p\rightarrow q) is neither ∼p→∼q\sim p\rightarrow\sim q nor q→pq\rightarrow p. Each of those is true in three rows; p∧∼qp\wedge\sim q is true in only one.
Drill 6 more on negating an implication

Concept 2 of 3: De Morgan's laws

To negate an and/or combination, push the ∼\sim inward: each ∧\wedge becomes ∨\vee, each ∨\vee becomes ∧\wedge, and each letter flips. Then tidy with the distributive law, which often pulls out a common letter and matches an option.

Definition

  • ∼(A∧B)≡∼A∨∼B\sim(A\wedge B)\equiv\sim A\vee\sim B and ∼(A∨B)≡∼A∧∼B\sim(A\vee B)\equiv\sim A\wedge\sim B.
  • ∼(∼A)≡A\sim(\sim A)\equiv A.
  • Distributive law: (A∧B)∨(A∧C)≡A∧(B∨C)(A\wedge B)\vee(A\wedge C)\equiv A\wedge(B\vee C), and the same with ∧,∨\wedge,\vee swapped.

De Morgan's laws

∼(A∧B)≡∼A∨∼B,∼(A∨B)≡∼A∧∼B\sim(A\wedge B)\equiv\sim A\vee\sim B,\qquad\sim(A\vee B)\equiv\sim A\wedge\sim B

Worked example

Negate (p∧q)∨(p∧r)(p\wedge q)\vee(p\wedge r) and simplify.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q74Moderate

Example 2 · Mathematical Reasoning · Negation and Simplification

The negation of the statement (p∨q)∧(q∨(∼r))(p \vee q) \land (q \vee ( \sim r)) is

Flip the connective too

Negating only the letters turns ∼(p∧q)\sim(p\wedge q) into ∼p∧∼q\sim p\wedge\sim q, which is wrong: at p=T, q=Fp=T,\ q=F the true negation holds and this does not. The connective must flip as well.
Drill 4 more on de morgan's laws

Concept 3 of 3: Simplifying to a short equivalent

Long statements in these questions usually reduce to two or three letters. Rewrite every →\rightarrow as ∼A∨B\sim A\vee B, apply De Morgan's laws, then collapse with absorption and the complement laws. When the options are short, a four-row truth table is just as fast: find the rows where the statement is true and match them.

Definition

  • Absorption: p∨(p∧q)≡pp\vee(p\wedge q)\equiv p and p∧(p∨q)≡pp\wedge(p\vee q)\equiv p.
  • Complement: p∧∼p≡Fp\wedge\sim p\equiv F and p∨∼p≡Tp\vee\sim p\equiv T; then X∧F≡FX\wedge F\equiv F and X∨F≡XX\vee F\equiv X.
  • (p∧q)∨(∼p∧q)≡q(p\wedge q)\vee(\sim p\wedge q)\equiv q.

Absorption

p∨(p∧q)≡p,p∧(p∨q)≡pp\vee(p\wedge q)\equiv p,\qquad p\wedge(p\vee q)\equiv p

Worked example

Simplify ∼(p→q)∨(p∧q)\sim(p\rightarrow q)\vee(p\wedge q).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q164Moderate

Example 3 · Mathematical Reasoning · Negation and Simplification

The statement ∼[p∨(∼(p∧q))]\sim \lbrack p \vee ( \sim (p \land q))\rbrack is equivalent to

Check one row before choosing

Slips are easy with five or six connectives. Before choosing, evaluate the original and your answer at one row, say all letters true. If they disagree, the working has an error.
Drill 6 more on simplifying to a short equivalent

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Negating an implication

    Negation of an implication

    ∼(A→B)≡A∧∼B\sim(A\rightarrow B)\equiv A\wedge\sim B
  • De Morgan's laws

    De Morgan's laws

    ∼(A∧B)≡∼A∨∼B,∼(A∨B)≡∼A∧∼B\sim(A\wedge B)\equiv\sim A\vee\sim B,\qquad\sim(A\vee B)\equiv\sim A\wedge\sim B
  • Simplifying to a short equivalent

    Absorption

    p∨(p∧q)≡p,p∧(p∨q)≡pp\vee(p\wedge q)\equiv p,\qquad p\wedge(p\vee q)\equiv p

Watch out for (3)

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