Negation of set statement

Question

What is the negation of the following statement?

$$B \subseteq A \Rightarrow C \subsetneq A$$

So far, I have:

$$B \subseteq A \land \neg (C \subsetneq A)$$

How do I simplify it further?

Answer 1

[Note: Following comments, this answer assumes $\subsetneq$ means $\not\subseteq$, or not a subset. Apparently $\subsetneq$ is undefined.]

Let $P=B\subseteq A$ and $Q=C\subsetneq A$. Then the negation of $P\Rightarrow Q$ is $P\not\Rightarrow Q$, which is called Material nonimplication, and has the logical equivalence $P\land \lnot Q$.

\begin{array} {c|c|c|c} P&Q&P\Rightarrow Q&P\not\Rightarrow Q\\ \hline T&T&T&F\\ T&F&F&T\\ F&T&T&F\\ F&F&T&F \end{array} In your case, $C \subsetneq A$ has the simple negation to $C \subseteq A$, so we arrive at: $$B \subseteq A\land C \subseteq A\to B \cup C \subseteq A$$

Answer 2

The expression $\neg(C\subsetneqq A)$ says that $C$ is not a proper subset of $A$. There are therefore two possibilities: either $C$ is not a subset of $A$ at all; or $C$ is a subset of $A$ but not a proper one, in which case $C=A$. Thus, the expression is equivalent to

$$C\nsubseteq A\lor C=A\;.$$

Your whole expression is therefore equivalent to

$$B\subseteq A\land\big(C\nsubseteq A\lor C=A\big)\;.\tag{1}$$

You can apply distributivity of $\land$ over $\lor$ to get

$$(B\subseteq A\land C\nsubseteq A)\lor B\subseteq A=C\;.$$

Whether either of these is actually simpler than $B\subseteq A\land\neg(C\subsetneqq A)$ is a matter of taste, opinion, and how you’re intending to use the expression; there isn’t any really nice equivalent expression. Expression $(1)$ is arguably the clearest, since it shows explicitly the two ways in which $C$ can fail to be a proper subset of $A$.