Theorem sseq1d

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

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

Proof

Step	Hyp	Ref	Expression
1		sseq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		sseq1	⊢ ( 𝐴 = 𝐵 → ( 𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶 ) )
3	1 2	syl	⊢ ( 𝜑 → ( 𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐶 ) )