Metamath Proof Explorer

Theorem injust

Description: Soundness justification theorem for df-in . (Contributed by Rodolfo Medina, 28-Apr-2010) (Proof shortened by Andrew Salmon, 9-Jul-2011)

		Ref	Expression
	Assertion	injust	$⊢ \{x \| (x \in A \land x \in B)\} = \{y \| (y \in A \land y \in B)\}$

Proof

Step	Hyp	Ref	Expression
1		eleq1w	$⊢ x = z \to (x \in A \leftrightarrow z \in A)$
2		eleq1w	$⊢ x = z \to (x \in B \leftrightarrow z \in B)$
3	1 2	anbi12d	$⊢ x = z \to ((x \in A \land x \in B) \leftrightarrow (z \in A \land z \in B))$
4	3	cbvabv	$⊢ \{x \| (x \in A \land x \in B)\} = \{z \| (z \in A \land z \in B)\}$
5		eleq1w	$⊢ z = y \to (z \in A \leftrightarrow y \in A)$
6		eleq1w	$⊢ z = y \to (z \in B \leftrightarrow y \in B)$
7	5 6	anbi12d	$⊢ z = y \to ((z \in A \land z \in B) \leftrightarrow (y \in A \land y \in B))$
8	7	cbvabv	$⊢ \{z \| (z \in A \land z \in B)\} = \{y \| (y \in A \land y \in B)\}$
9	4 8	eqtri	$⊢ \{x \| (x \in A \land x \in B)\} = \{y \| (y \in A \land y \in B)\}$