# What's Wrong in this Proof Logic?

Trying to show that the empty set $\emptyset \subseteq A$, for any set $A$.

Let $x \in \emptyset$, then by definition, $x \in \emptyset \iff (x \neq x)$.

$x \in \emptyset \implies (x \neq x) \lor P$ where $P$ is any statement

Let $P$ be $(x \in A)$, then $$x \in \emptyset \implies (x \neq x) \lor (x \in A).$$ But $(x \neq x)$ is false, then we can write $$x \in \emptyset \implies (x \in A),$$ which is equivalent to $\emptyset \subseteq A$.

Is this correct?

• While nothing you have written is wrong, it seems to take an unusual detour. Why not say, "false implies true and false implies false are always true", so $x \in \emptyset \to x \in A$ is always true. – Mees de Vries Nov 10 '16 at 17:51
• You should define $\forall x: x\notin \emptyset$. – Dan Christensen Nov 10 '16 at 17:54
• It really bugs me when people say "by definition" without actually instantiating a definition. – DanielV Nov 10 '16 at 18:04
• "DanielV", just look carefully, the definition is there. ;-) – Joseph Nov 10 '16 at 18:14

It is easier to note that $x \in \emptyset$ is false, hence $x \in \emptyset \Rightarrow x \in A$ is true for an arbitrary $x$. Every "$\Rightarrow$"-conclusion that you make based on a false premise is true trivially. This is easy to prove with the help of a truth table.
If $X \not \subset A$ then there is some $x \in X$ and $x \not \in A$
So, is there any $x \in \emptyset$ and $x \not \in A$ ? No there isn't (because there is no $x \in \emptyset$ ), so $\emptyset \subset A$.