Showing the sets are equivalent using set builder notation.

Proof: Just observe the following sequence of equalities.
A×(B−C)={(x,y):(x∈A)∧(y∈(B−C))}(def. of ×)={(x,y):(x∈A)∧(y∈B)∧(y∉C)}(def. of -)={(x,y):(x∈A)∧(x∈A)∧(y∈B)∧(y∉C)}(P=P∧P)={(x,y):((x∈A)∧(y∈B))∧((x∈A)∧(y∉C))}(rearrange)={(x,y):(x∈A)∧(y∈B)}∩{(x,y):(x∈A)∧(y∉C)}(def. of ∩)={(x,y):(x∈A)∧(y∈B)}−{(x,y):(x∈A)∧(y∉C)}(Basic properties of ∩)=(A×B)−(A×C)(def. of ×)\begin{align*}
A \times (B – C)&=\{(x,y):(x\in A)\land(y\in(B-C))\} \textbf{(def. of $\times$)} \\
&=\{(x,y):(x\in A)\land (y\in B)\land(y\notin C)\} \textbf{(def. of -)} \\
&=\{(x,y):(x\in A)\land(x\in A)\land (y\in B)\land(y\notin C)\} \textbf{(P=$P\land P$)} \\
&=\{(x,y):((x\in A)\land (y\in B))\land((x\in A)\land(y\notin C))\}\textbf{(rearrange)} \\
&=\{(x,y):(x\in A)\land (y\in B)\}\cap\{(x,y):(x\in A)\land(y\notin C)\}\textbf{(def. of $\cap$)} \\
&=\{(x,y):(x\in A)\land (y\in B)\}-\{(x,y):(x\in A)\land(y\notin C)\}\textbf{(Basic properties of $\cap$)} \\
&=(A \times B) – (A \times C)\textbf{(def. of $\times$)}
\end{align*}

The last two lines is wrong but I don’t know how to arrive at that part.

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\{(x,y):(x\in A)\land (y\in B)\}\cap\{(x,y):(x\in A)\land(y\notin C)\} =\{(x,y):(x\in A)\land (y\in B)\}-\{(x,y):(x\in A)\land(y\in C)\}\{(x,y):(x\in A)\land (y\in B)\}\cap\{(x,y):(x\in A)\land(y\notin C)\} =\{(x,y):(x\in A)\land (y\in B)\}-\{(x,y):(x\in A)\land(y\in C)\}
– mfl
2 days ago

I don’t follow the – – from the intersection
– HighSchool15
2 days ago

Look only at y.y. So, y\in B\cap \bar{C} \iff y\in B, y\notin C \iff y\in B\setminus C.y\in B\cap \bar{C} \iff y\in B, y\notin C \iff y\in B\setminus C.
– mfl
2 days ago

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