MHB Proof: $(A - B)\cup B = A$ iff $B\subseteq A$

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The discussion centers on the proof that $(A - B)\cup B = A$ if and only if $B\subseteq A$. Participants explore the steps of the proof, highlighting that if $B$ is a subset of $A$, the expression simplifies correctly. However, one user expresses confusion, stating they consistently arrive at $B$ instead of $A$. Another contributor clarifies that the expression $(A\cup B)\cap (B^c\cup B)$ simplifies to $(A\cup B)$, which is the universal set. This indicates that the proof holds true under the condition that $B$ is indeed a subset of $A.
Dustinsfl
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$(A - B)\cup B = A$ iff $B\subseteq A$.Suppose $B\subseteq A$.
$$
\begin{array}{lcl}
(A - B)\cup B & = & (A\cap B^c)\cup B\\
& = & (A\cup B)\cap (B^c\cup B)\\
& = & A\cap B\\
& = & B
\end{array}
$$

I keep getting = B not A.
 
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dwsmith said:
$(A - B)\cup B = A$ iff $B\subseteq A$.Suppose $B\subseteq A$.
$$
\begin{array}{lcl}
(A - B)\cup B & = & (A\cap B^c)\cup B\\
& = & (A\cup B)\cap (B^c\cup B)\\
& = & A\cap B\\
& = & B
\end{array}
$$

I keep getting = B not A.

Hi dwsmith, :)

Note that, \[(A\cup B)\cap (B^c\cup B)=(A\cup B)\cap V=(A\cup B)\] where \(V\) is the universal set.

Kind Regards,
Sudharaka.
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...

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