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Metamath Proof Explorer

Theorem sbss

Description: Set substitution into the first argument of a subset relation. (Contributed by Rodolfo Medina, 7-Jul-2010) (Proof shortened by Mario Carneiro, 14-Nov-2016)

		Ref	Expression
	Assertion	sbss	$⊢ [y/ x] x \subseteq A \leftrightarrow y \subseteq A$

Proof

Step	Hyp	Ref	Expression
1		sseq1	$⊢ x = z \to (x \subseteq A \leftrightarrow z \subseteq A)$
2		sseq1	$⊢ z = y \to (z \subseteq A \leftrightarrow y \subseteq A)$
3	1 2	sbievw2	$⊢ [y/ x] x \subseteq A \leftrightarrow y \subseteq A$