Proving $[x \in A_i \land x \in (\forall 1\le k\le i-1: {A_k}^c)] \land [x \in A_j \land x \in (\forall 1\le k\le j-1: {A_k}^c)]$ =$\emptyset$ Prove that if $B_1 = A_1$, $B_2 = A_2 - A_1$, $B_3 = A_3 - (A_1 \cup A_2)$, then ${B_i} \cap {B_j} = \emptyset$ ($i\neq j$). 
My attempt:
$x \in (B_i \cap B_j)
=x \in [ {A_i} - (A_1 \bigcup A_2 \bigcup \cdots \bigcup A_{i-1})] \bigcap x \in [ A_j - (A_1 \bigcup A_2 \cup \cdots \bigcup A_{j-1}) ]$
$= x\in [[A_i- (\overset {i-1}\bigcup \limits_{k=1} A_k)] \bigcap [A_j -(\overset {j-1}\bigcup \limits_{k=1} A_k)]]$
$\equiv [x\in A_i \land x \notin (\overset {i-1}\bigcup \limits_{k=1} A_k)] \land [x \in A_j \land x \notin (\overset {j-1}\bigcup \limits_{k=1} A_k)] \space\space By \space definition\space of\space intersection,\space complement.$ 
$\equiv [x \in A_i \land x \in (\overset {i-1} \bigcup \limits_{k=1} A_k)^c] \land [x \in A_j \land x \in (\overset{j-1} \bigcup \limits_{k=1} A_k)^c]$
$\equiv [x \in A_i \land x \in (\overset {i-1} \bigcap \limits_{k=1} {A_k}^c)] \land [x \in A_j \land x \in (\overset{j-1} \bigcap \limits_{k=1} {A_k}^c)]$ by generalized de morgan's theorem.
$\equiv [x \in A_i \land x \in (\forall 1\le  k\le i-1: {A_k}^c)] \land [x \in A_j \land x \in (\forall 1\le  k\le j-1: {A_k}^c)]$ by the definition of the intersection of of sets in an arbitrary family of sets.
I don't know how to develop this further from here. It would lead to the empty set if  either $\forall 1\le  k\le i$ or $\forall 1\le  k\le j$ was in the step, but note that it's $\forall 1\le  k\le i-1$ and $\forall 1\le  k\le j-1$ in the step. So I don't think $[x \in A_i \land x \in (\forall 1\le  k\le i-1: {A_k}^c)] \land [x \in A_j \land x \in (\forall 1\le  k\le j-1: {A_k}^c)]$ leads to the empty set. 
Can you show $[x \in A_i \land x \in (\forall 1\le  k\le i-1: {A_k}^c)] \land [x \in A_j \land x \in (\forall 1\le  k\le j-1: {A_k}^c)]$ leads to the empty set? 
[EDIT] Now the proof is completed.
Since $p \neq q$, if i>j from $[x \in A_i \land x \in (\overset {i-1} \bigcap \limits_{k=1} {A_k}^c)] \land [x \in A_j \land x \in (\overset{j-1} \bigcap \limits_{k=1} {A_k}^c)]$ 
$\equiv [x \in A_i \land x \in ($~$A_1 \land$ ~$A_2\land \cdots \land$ ~$A_{j-1}$ $\land$ ~$A_j$ $\land \cdots \land$ ~$A_{i-1}$) $\land x \in A_j] \land x \in $ (~$A_1$ $\land$ ~$A_2$ $\land \cdots \land$ ~$A_{j-1}$)] 
$\equiv [x \in A_i \land x \in ($~$A_1 \land$ ~$A_2\land \cdots \land$ ~$A_{j-1}$ $\land \cdots \land$ ~$A_{i-1}$) $\land$ ~$A_j$ $\land x \in A_j] \land x \in $ (~$A_1$ $\land$ ~$A_2$ $\land \cdots \land$ ~$A_{j-1}$)] 
$\equiv [x \in A_i \land x \in ($~$A_1 \land$ ~$A_2\land \cdots \land$ ~$A_{j-1}$ $\land \cdots \land$ ~$A_{i-1}$) $\land$ c] $\land x \in $ (~$A_1$ $\land$ ~$A_2$ $\land \cdots \land$ ~$A_{j-1}$)] 
$\equiv c$
Similarly for j>i, from $[x \in A_i \land x \in (\overset {i-1} \bigcap \limits_{k=1} {A_k}^c)] \land [x \in A_j \land x \in (\overset{j-1} \bigcap \limits_{k=1} {A_k}^c)] \equiv c$ 
FYI
$p \land$ ~p $\Leftrightarrow c$ 
"Let t, c and p be a tautology, a contradiction, and an arbitrary statement, respectively. Then 
(a) $p\land t \Leftrightarrow p$, $p\lor t \Leftrightarrow t$   (b) $p \lor c \Leftrightarrow p$, $p\land c \Leftrightarrow c$ (c) $c \Rightarrow p$, $p \Rightarrow t$"
 A: $$B_k=A_k\backslash \bigcup_{i=1}^{k-1}A_i\subset A_k.$$
Then, if $i\neq j$, the fact that $B_i\cap B_j=\emptyset$ follow. Indeed, if $i<j$, then $$B_j=A_j\backslash \underbrace{\bigcup_{i=1}^{j-1}A_i}_{\supset B_i}.$$
A: Suppose $x \in B_i \cap B_j$, where $i \neq j$. Suppose that $i < j$ for concreteness. Then $x \in A_j$ but in none of the $A_k$ for $k < j$, by definition of $B_j = A_j \setminus \cup_{k=1}^{j-1} A_k$. So $x$ is not in $A_i$ in particular, but then $x$ cannot be in $B_i$ either! contradiction.
A: You're quite a way already, let me see if I can help you continue in the same direction.$
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$
The approach you've taken is to calculate which $\;x\;$ are elements of $\;B_i \cap B_j\;$, by expanding the definitions of $\;B_\_\;$, $\;\cap\;$ and $\;-\;$.  So what if we do the same thing for $\;\bigcup\;$, and see where that leads us?
In a slightly different notation that I am more used to, we calculate for each $\;x, i, j\;$:
$$\calc
    x \in B_i \cap B_j
\op=\hint{definition of $\;B_\_\;$, twice}
    x \in (A_i - \langle \cup k : 1 \le k \lt i : A_k \rangle)
    \;\cap\;
    (A_j - \langle \cup k : 1 \le k \lt j : A_k \rangle)
\op=\hint{definition of $\;\cap\;$; definition of $\;-\;$, twice}
    x \in A_i \land \lnot (x \in \langle \cup k : 1 \le k \lt i : A_k \rangle)
    \;\land\;
    \\ &
    x \in A_j \land \lnot (x \in \langle \cup k : 1 \le k \lt j : A_k \rangle)
\op=\hint{definition of $\;\cup\;$, twice}
    x \in A_i \land \lnot \langle \exists k : 1 \le k \lt i : x \in A_k \rangle
    \;\land\;
    \\ &
    x \in A_j \land \lnot \langle \exists k : 1 \le k \lt j : x \in A_k \rangle
\op=\hint{logic: DeMorgan -- since we'd like to combine the quantifications}
    x \in A_i \land \langle \forall k : 1 \le k \lt i : x \notin A_k \rangle
    \;\land\;
    \\ &
    x \in A_j \land \langle \forall k : 1 \le k \lt j : x \notin A_k \rangle
\op=\hint{logic: combine the quantifications}
    x \in A_i \land x \in A_j \;\land\; \langle \forall k : 1 \le k \lt i \;\lor\; 1 \le k \lt j : x \notin A_k \rangle
\op=\hint{arithmetic}
    x \in A_i \land x \in A_j \;\land\; \langle \forall k : 1 \le k \lt i \max j : x \notin A_k \rangle
    \tag{*}
\endcalc$$
Now it looks like we have done everything that we can do 'mechanically', and it is time to investigate how we can use the assumption $\;i \not= j\;$ to do something with the expression $\;i \max j\;$.  The simplest thing seems to use the fact that $\;i \not= j\;$ is the same as $\;i \min j \;\lt\; i \max j\;$.  So we choose $\;k:= i \min j\;$ and continue our calculation:
$$\calc
    \tag{*}
    x \in A_i \land x \in A_j \;\land\; \langle \forall k : 1 \le k \lt i \max j : x \notin A_k \rangle
\op\then
        \hints{choose $\;k:= i \min j\;$, allowed since $\;i \min j \;\lt\; i \max j\;$,}
        \hint{which is equivalent to our assumption $\;i \not= j\;$}
    x \in A_i \land x \in A_j \;\land\; x \notin A_{i \min j}
\op=\hint{case split: use the fact that $\;i \min j = i\;$ or $\;i \min j = j\;$}
    (i \min j = i \;\land\; x \in A_i \land x \in A_j \;\land\; x \notin A_i)
    \;\lor\;
    \\ &
    (i \min j = j \;\land\; x \in A_i \land x \in A_j \;\land\; x \notin A_j)
\op=\hint{both sides of $\;\lor\;$ are a contradiction}
    \false
\endcalc$$
With this calculation we've proven that if $\;i \not= j\;$, then $\;\langle \forall x :: x \in B_i \cap B_j \;\then\; \false \rangle\;$, i.e., $\;B_i \cap B_j = \emptyset\;$.
A: So $B_{k} = A_{k} - (\cup_{i = 1}^{k - 1} A_{i})$.
Suppose $k \neq j$.  We want to show $B_{k} \neq B_{j}$.
Notice that for sets $A$ and $B$, $A - B = A \cap B^{c}$.
Then $B_{k} = A_{k} - (\cup_{i = 1}^{k - 1} A_{i}) = A_{k} \cap (\cup_{i = 1}^{k - 1} A_{i})^{c} = A_{k} \cap (\cap_{i = 1}^{k - 1} A_{i}^{c})$ (where the last equality is by DeMorgan's laws).
Now, if $k \neq j$, we can assume without loss of generality that $k < j$.  So we have an expression above for $B_{k}$.  Let's get one for $B_{j}$:
$B_{j} = A_{j} \cap (\cap_{i = 1}^{j - 1} A_{i}^{c})$.  Since $k < j$, $A_{k}^{c}$ appears in the intersection that's in $B_{j}$'s definition.  That means $B_{j} \subseteq A_{k}^{c}$.
But $B_{k} \subseteq A_{k}$ (just look at how we defined $B_{k}$.  This shows that $B_{j} \cap B_{k} = \emptyset$.
