Theorem sseq2d

Description: An equality deduction for the subclass relationship. (Contributed by NM, 14-Aug-1994)

	Ref	Expression
Hypothesis	sseq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	sseq2d	⊢ ( 𝜑 → ( 𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵 ) )

Proof

Step	Hyp	Ref	Expression
1		sseq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		sseq2	⊢ ( 𝐴 = 𝐵 → ( 𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵 ) )
3	1 2	syl	⊢ ( 𝜑 → ( 𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵 ) )