Metamath Proof Explorer

Theorem dfnul4

Description: Alternate definition of the empty class/set. (Contributed by BJ, 30-Nov-2019) Avoid ax-8 , df-clel . (Revised by GG, 3-Sep-2024) Prove directly from definition to allow shortening dfnul2 . (Revised by BJ, 23-Sep-2024)

		Ref	Expression
	Assertion	dfnul4	$⊢ \emptyset = \{x \| ⊥\}$

Proof

Step	Hyp	Ref	Expression
1		df-nul	$⊢ \emptyset = V ∖ V$
2		df-dif	$⊢ V ∖ V = \{x \| (x \in V \land \neg x \in V)\}$
3		pm3.24	$⊢ \neg (x \in V \land \neg x \in V)$
4	3	bifal	$⊢ (x \in V \land \neg x \in V) \leftrightarrow ⊥$
5	4	abbii	$⊢ \{x \| (x \in V \land \neg x \in V)\} = \{x \| ⊥\}$
6	1 2 5	3eqtri	$⊢ \emptyset = \{x \| ⊥\}$