Theorem psseq1d

Description: An equality deduction for the proper subclass relationship. (Contributed by NM, 9-Jun-2004)

	Ref	Expression
Hypothesis	psseq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
Assertion	psseq1d	⊢ ( 𝜑 → ( 𝐴 ⊊ 𝐶 ↔ 𝐵 ⊊ 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		psseq1d.1	⊢ ( 𝜑 → 𝐴 = 𝐵 )
2		psseq1	⊢ ( 𝐴 = 𝐵 → ( 𝐴 ⊊ 𝐶 ↔ 𝐵 ⊊ 𝐶 ) )
3	1 2	syl	⊢ ( 𝜑 → ( 𝐴 ⊊ 𝐶 ↔ 𝐵 ⊊ 𝐶 ) )