Question:medium

Negation of $(p \land q) \rightarrow (\sim p \lor r)$ is

Show Hint

The negation of an "If-Then" statement is never another "If-Then" statement. It is always an "And" statement. Remembering $\sim(A \rightarrow B) \equiv A \land \sim B$ allows you to jump straight to the correct form.
Updated On: Jun 4, 2026
  • $p \lor q \lor (\sim r)$
  • $p \land q \land r$
  • $\sim p \land q \land r$
  • $p \land q \land (\sim r)$
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: See the shape of the statement.
The statement is $(p\land q)\rightarrow(\sim p\lor r)$. This is an "if then" of the form $A\rightarrow B$, where $A=(p\land q)$ and $B=(\sim p\lor r)$.

Step 2: Recall how to negate an "if then".
The negation of $A\rightarrow B$ is $A\land\sim B$. In words, "it is not true that A leads to B" means "A is true but B is false".

Step 3: Apply that rule.
\[ \sim\big[(p\land q)\rightarrow(\sim p\lor r)\big]\equiv (p\land q)\land\sim(\sim p\lor r). \]

Step 4: Negate the "or" part with De Morgan.
De Morgan's law says $\sim(X\lor Y)\equiv\sim X\land\sim Y$. So \[ \sim(\sim p\lor r)\equiv\sim(\sim p)\land\sim r\equiv p\land\sim r. \]

Step 5: Put it back together.
\[ (p\land q)\land(p\land\sim r). \]

Step 6: Simplify the repeated $p$.
Since everything is joined by "and", and $p\land p\equiv p$, we drop the extra $p$: \[ p\land q\land\sim r. \]

Step 7: Match the option.
This is $p\land q\land(\sim r)$, which is option (4).
\[ \boxed{p\land q\land(\sim r)} \]
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