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If $A,B \subseteq X$, Show that $B=(A\cap B)\cup(B\setminus A)$

My advances

$B=(A\cap B)\cup (B\setminus A)$

$ =(A \cap B)\cup (B \cap\overline{A})$

$ =(A\cap B)\cup (B\cap X)$

$ =(A \cap B)\cup (B)$

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In your proof you can add this $$(B\cap A)\cup(B\cap \overline A)=B\cap(A\cup \overline A)=B$$

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Another way:
Clearly if $x\in(A\cap B)\cup(B\setminus A)$, then $x\in B$


If $x\in B$, we have two cases:

  • $x\in A$. In this case $x\in(A\cap B)$

  • $x\notin A$. In this case $x\in(B\setminus A)$

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$$(B\cap A)\cup(B\cap \overline A)=(B \cap B)\cap(A\cup \overline A)=B$$

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For me, these are often done mostly mechanically by translating from the set level to the logic level by expanding the definitions, and then using whatever laws of logic seem most appropriate, based on the structure of the formula.$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\followsfrom}{\Leftarrow} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} $

Since this problem looks like a simplification problem, we start by investigating which $\;x\;$ are elements of the right hand side set, and work towards the left hand side: $$\calc x \in (A\cap B)\cup (B\setminus A) \op=\hint{definitions of $\;\cup,\cap,\setminus\;$} (x \in A \land x \in B) \lor (x \in B \land x \not\in A) \op=\hints{logic: $\;\land\;$ distributes over $\;\lor\;$} \hint{-- since we want to end up with a single $\;B\;$} (x \in A \lor x \not\in A) \land x \in B \op=\hint{logic: excluded middle} \true \land x \in B \op=\hint{logic: simplify} x \in B \endcalc$$ So, by set extensionality, we've proved the original statement.

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