Prove that $(A \cup B)\backslash (C \backslash A)=A \cup (B \backslash C)$

Question

For all sets $A, B$ and $C$, I need to prove that it holds

$$(A \cup B)\backslash (C \backslash A)=A \cup (B \backslash C)$$

I tried to prove on this way:

$$C\backslash A= X$$

$$(A \cup B) \backslash X = \\ =AX \cup BX = \\ =AX \cup B\backslash C\backslash A=\\ = A\cup (B\backslash C)$$

Is this correct?

Could you please help me to prove this correctly if I'm wrong?

Answer 1

not quite. You should try doing it by contentions...so if $x\in(A\cup B)-(C-A)$ then ($x\in A$ or $x\in B$) and $x\not\in(C-A)$. if $x\in A$ then you are done because then $x\in A\cup (B-C)$. So assume $x\not\in A$. Then $x\in B$, then again also $x\not\in C-A$ which means $x\not\in C$ or $x\in A$, but $x\not\in A$, so it must mean $x\not\in C$. so, $x\in B-C$ and therefore $x\in A\cup(B-C)$. This proves $(A\cup B)-(C-A)\subset A\cup (B-C)$.

Try proving $A\cup (B-C)\subset (A\cup B)-(C-A)$ by the same means as above.

Answer 2

If you recall that $X\backslash Y$ means exactly the same as $X\cap Y^c$, where the exponent $c$ indicates complement, then

\begin{align} (A \cup B)\backslash (C \backslash A)&=(A\cup B)\cap (C\cap A^c)^c \\ &=(A\cup B)\cap (A\cup C^c) \\ &=A\cup (B\cap C^c) \\ &=A \cup (B \backslash C). \end{align}

This assumes you can use the distributive property.

Answer 3

Alternative approach:

Use a truth table.

Let LHS denote $~(A \cup B) ~\setminus ~(C ~\setminus ~A).$
Let RHS denote $~A \cup ~(B ~\setminus ~C).$

\begin{array}{| c | c | c | c | c | c | c | c |} \hline A & B & C & (A \cup B) & (C\setminus A) & \color{red}{\text{LHS}} & (B\setminus C) & \color{red}{\text{RHS}} \\ \hline T & T & T & T & F & \color{red}{T} & F & \color{red}{T} \\ \hline T & T & F & T & F & \color{red}{T} & T & \color{red}{T} \\ \hline T & F & T & T & F & \color{red}{T} & F & \color{red}{T} \\ \hline T & F & F & T & F & \color{red}{T} & F & \color{red}{T} \\ \hline F & T & T & T & T & \color{red}{F} & F & \color{red}{F} \\ \hline F & T & F & T & F & \color{red}{T} & T & \color{red}{T} \\ \hline F & F & T & F & T & \color{red}{F} & F & \color{red}{F} \\ \hline F & F & F & F & F & \color{red}{F} & F & \color{red}{F} \\ \hline \end{array}

Since the two red columns match, row by row, the LHS is true if and only if the RHS is true. So, the assertion is proven.