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Metamath Proof Explorer

Theorem brcnvssrid

Description: Any set is a converse subset of itself. (Contributed by Peter Mazsa, 9-Jun-2021)

		Ref	Expression
	Assertion	brcnvssrid	$⊢ A \in V \to A S^{-1} A$

Step	Hyp	Ref	Expression
1		ssid	$⊢ A \subseteq A$
2		brcnvssr	$⊢ A \in V \to (A S^{-1} A \leftrightarrow A \subseteq A)$
3	1 2	mpbiri	$⊢ A \in V \to A S^{-1} A$