# Proving set theory subsets using element argument

How do you even prove a set theory subset statement using element argument? I simply just can't find any relevance to the question with the notes i was studying.

Any guidance would be much appreciated.

(A - B) ∩ (C - B) subset of (A ∩ C) - B

The only definition i have is

x element of A - B is logically equivalent to ( x element of A and x not element of B)

If i were to sub in the definition, it would lead me to nowhere where i can use whatever law there is in set theory.

This discrete mathematics is way different from the typical maths i have been doing since young. Any guidance is appreciated.

Note that $$x\in(A-B)\cap (C-B)$$ is equivalent to $$x\in A-B\quad \land \quad x\in C-B$$ and so to $$(x\in A \land x\notin B)\land (x\in C\land x\notin B).$$ Form this you want to show that $x\in(A\cap C)-B$, or equivalently, that $$(x\in A\land x\in C)\land x\notin B.$$ I guess you can take it from here.

You can do this simply by using strict definitions of subsets:

$A \subset B$ is saying $\forall x\in A$ $x\in B$

So:

Let $x\in (A-B)\cap(C-B)$

Then $x\in (A-B)$ and $x\in(C-B)$

Then $x\in A\cap C$ and $x\notin (A\cap C)\cap B$ otherwise contradiction to the assumption $x\in (A-B)\cap(C-B)$

Then $x\in (A\cap C)-B$

• After "Then $x\in (A-B)$ and $x\in (C-B)$", I think it's clearer to just write: "Then $x\in A$, $x\in C$ and $x\notin B$" or "Then $x\in A$ and $x\notin B$, and $x\in C$ and $x\notin B$."
– smcc
Commented Jul 24, 2016 at 12:58

As part of gaining deeper insights I'm trying to answer questions so that errors in my reasoning will hopefully be pointed out.

To proof using the element argument:

${(A-B)\cup(C-B) \subseteq (A \cup C)-B}$

We start as follows

${x \in (A-B)\cap(C-B)}$

${x \in (A \cap \overline{B})\cap(C\cap\overline{B}) }$ difference law

${(x \in A \land x \in \overline{B}) \land (x \in C \land x \in \overline{B})}$ def union

${(x \in A \land x \in C) \land (x \in \overline{B} \land x \in \overline{B})}$ associative + commutative laws

${(x \in A \land x \in C) \land x \in \overline{B}}$ idempotency law

${x \in (A \cap C) \cap \overline{B}}$ def intersection

${x \in (A \cap C)-B}$ difference law

And that concludes the proof.

This is the long version of what Hagen von Eitzen has given.

[ (A - B) ∩ (C - B) ] ⊆ [ (A ∩ C) - B ]

steps:

1. 𝑥∈(𝐴−𝐵)∩(C−𝐵)

2.

(a) 𝑥∈(𝐴−𝐵) and (b) 𝑥∈(C−𝐵)

3.

From (a) 𝑥∈𝐴 and x∉B

From (b) 𝑥∈C and x∉B

4. Hence, 𝑥∈𝐴 and 𝑥∈C and x∉B are true.

5. By definition of intersection: 𝑥∈(𝐴∩C) and x∉B

6. By definition of set difference: 𝑥∈(𝐴∩C)-B