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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 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		sseq1	⊢ ( 𝑥 = 𝑧 → ( 𝑥 ⊆ 𝐴 ↔ 𝑧 ⊆ 𝐴 ) )
2		sseq1	⊢ ( 𝑧 = 𝑦 → ( 𝑧 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴 ) )
3	1 2	sbievw2	⊢ ( [ 𝑦 / 𝑥 ] 𝑥 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴 )