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Metamath Proof Explorer

Theorem brabidga

Description: The law of concretion for a binary relation. Special case of brabga . Usage of this theorem is discouraged because it depends on ax-13 , see brabidgaw for a weaker version that does not require it. (Contributed by Peter Mazsa, 24-Nov-2018) (New usage is discouraged.)

		Ref	Expression
	Hypothesis	brabidga.1	$⊢ R = \{⟨x, y⟩ \| φ\}$
	Assertion	brabidga	$⊢ x R y \leftrightarrow φ$

Proof

Step	Hyp	Ref	Expression
1		brabidga.1	$⊢ R = \{⟨x, y⟩ \| φ\}$
2	1	breqi	$⊢ x R y \leftrightarrow x \{⟨x, y⟩ \| φ\} y$
3		df-br	$⊢ x \{⟨x, y⟩ \| φ\} y \leftrightarrow ⟨x, y⟩ \in \{⟨x, y⟩ \| φ\}$
4		opabid	$⊢ ⟨x, y⟩ \in \{⟨x, y⟩ \| φ\} \leftrightarrow φ$
5	2 3 4	3bitri	$⊢ x R y \leftrightarrow φ$