Simplifying a compound proposition with De Morgan laws

Key takeaway: This problem tests symbolic logic simplification, especially De Morgan’s laws, distribution, and complements. The goal is to rewrite the proposition into a cleaner equivalent form without changing its truth value.

Identify the structure

We start with the compound statement

$\sim (\sim q \vee p) \wedge (\sim q \vee \sim p)$.

The expression has two conjuncts joined by $\wedge$, so the main task is to simplify each part and then combine them into a single equivalent statement.

The first part, $\sim(\sim q \vee p)$, is a negation of a disjunction. That is the perfect place to use De Morgan’s law.

Apply logical equivalences

Using De Morgan’s law,

$\sim(\sim q \vee p) \equiv q \wedge \sim p$.

So the whole statement becomes

$(q \wedge \sim p) \wedge (\sim q \vee \sim p)$.

Now use associativity of conjunction to group the terms:

$q \wedge \sim p \wedge (\sim q \vee \sim p)$.

At this point, notice that $\sim p$ is already present as a factor. The second parenthesis contains $\sim p$ as one of its options, so the conjunction can be simplified by distribution:

$\sim p \wedge (\sim q \vee \sim p) \equiv \sim p$.

That leaves

$q \wedge \sim p$.

Final simplified form

So the fully simplified equivalent statement is

$\boxed{q \wedge \sim p}$.

This is the cleanest form because it contains no unnecessary nested negations or redundant terms.

Common logic fact to remember

A statement of the form $A \wedge (A \vee B)$ simplifies to $A$. Here, $A$ is $\sim p$, so the entire second half collapses once the first negation is removed.

That kind of pattern appears often in symbolic logic problems, especially when the goal is to reduce a proposition to a shorter equivalent statement.

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Pitfalls the pros know 👇 A common mistake is to negate only one part of $\sim(\sim q \vee p)$ and forget that De Morgan’s law changes both pieces and the connective. Another frequent error is stopping after the first simplification and leaving the answer as $(q \wedge \sim p) \wedge (\sim q \vee \sim p)$, even though the second bracket still reduces. In logic, redundancy matters: if a factor already forces $\sim p$, anything inside a disjunction that includes $\sim p$ becomes unnecessary. Students also sometimes treat $\wedge$ and $\vee$ as if they can be simplified by visual pattern alone, but the truth-table equivalence must remain exact.

What if the problem changes? If the problem were changed to $\sim(\sim q \vee p) \vee (\sim q \vee \sim p)$, the result would be very different because the outer connective becomes a disjunction instead of a conjunction. After De Morgan, the first part still becomes $q \wedge \sim p$, and then the whole expression becomes $(q \wedge \sim p) \vee (\sim q \vee \sim p)$. In that variant, the term $\sim q \vee \sim p$ already covers many cases, so the expression simplifies much less aggressively than the original. A small change in the outer connective can completely change the final logical form.

Tags: De Morgan's law, logical equivalence, compound proposition