集合定律求证：请求证指定集合等式并验证推导及指导后续

Hey there! Let's break down this set equality proof properly, starting by checking your current step and then walking through the correct derivation step-by-step.

First: Validate your current progress #

Your derived expression $C- [(A^{c}) ∩ (A^{c} ∪ B^{c})]{c}$ isn't correct. Here's why:
The original left-hand side (LHS) is:
$$C - \left[ A^c ∪ (A ∪ B)^c \right]^c$$
When applying De Morgan's Law to the outer complement $\left[ X ∪ Y \right]^c$ (where $X = A^c$ and $Y = (A∪B)^c$), we should get $X^c ∩ Y^c$, not $X ∩ (A^c ∪ B^c)$. You mixed up which parts of the expression need their complements taken.

Let's restart the proof with clear, law-labeled steps:

Step-by-Step Proof #

We need to prove:
$$C - \left[ A^c ∪ (A ∪ B)^c \right]^c = C ∩ A^c$$

1. Rewrite set difference as intersection with complement
   Recall that for any sets $X, Y$, $X - Y = X ∩ Y^c$. Apply this to the LHS:
   $$\text{LHS} = C ∩ \left( \left[ A^c ∪ (A ∪ B)^c \right]^c \right)^c$$

2. Apply Double Complement Law
   The double complement of a set is the set itself: $(Z^c)c = Z$. This simplifies the expression to:
   $$\text{LHS} = C ∩ \left[ A^c ∪ (A ∪ B)^c \right]$$

3. Apply De Morgan's Law to $(A ∪ B)^c$
   De Morgan's Law states $(X ∪ Y)^c = X^c ∩ Y^c$. Here, $X=A$, $Y=B$, so:
   $$(A ∪ B)^c = A^c ∩ B^c$$
   Substitute back into the expression:
   $$\text{LHS} = C ∩ \left[ A^c ∪ (A^c ∩ B^c) \right]$$

4. Apply Absorption Law
   The Absorption Law tells us $X ∪ (X ∩ Y) = X$ (any element in both $X$ and $Y$ is already in $X$). Let $X = A^c$, $Y = B^c$:
   $$A^c ∪ (A^c ∩ B^c) = A^c$$
   Substitute this in:
   $$\text{LHS} = C ∩ A^c$$

This matches exactly the right-hand side (RHS) of the original equality, so we've proven the statement holds!

Quick recap of the misstep #

You incorrectly applied De Morgan's Law to the outer complement: you kept $A^c$ unchanged instead of taking its complement, and misexpanded $(A∪B)^c$ within the intersection. The key simplification was resolving the double complement from the set difference first, which makes the rest of the proof straightforward.

内容的提问来源于stack exchange，提问作者JpDGe