Theorem elssabg

Description: Membership in a class abstraction involving a subset. Unlike elabg , A does not have to be a set. (Contributed by NM, 29-Aug-2006)

	Ref	Expression
Hypothesis	elssabg.1	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
Assertion	elssabg	⊢ ( 𝐵 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ ( 𝑥 ⊆ 𝐵 ∧ 𝜑 ) } ↔ ( 𝐴 ⊆ 𝐵 ∧ 𝜓 ) ) )

Proof

Step	Hyp	Ref	Expression
1		elssabg.1	⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2		ssexg	⊢ ( ( 𝐴 ⊆ 𝐵 ∧ 𝐵 ∈ 𝑉 ) → 𝐴 ∈ V )
3	2	expcom	⊢ ( 𝐵 ∈ 𝑉 → ( 𝐴 ⊆ 𝐵 → 𝐴 ∈ V ) )
4	3	adantrd	⊢ ( 𝐵 ∈ 𝑉 → ( ( 𝐴 ⊆ 𝐵 ∧ 𝜓 ) → 𝐴 ∈ V ) )
5		sseq1	⊢ ( 𝑥 = 𝐴 → ( 𝑥 ⊆ 𝐵 ↔ 𝐴 ⊆ 𝐵 ) )
6	5 1	anbi12d	⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 ⊆ 𝐵 ∧ 𝜑 ) ↔ ( 𝐴 ⊆ 𝐵 ∧ 𝜓 ) ) )
7	6	elab3g	⊢ ( ( ( 𝐴 ⊆ 𝐵 ∧ 𝜓 ) → 𝐴 ∈ V ) → ( 𝐴 ∈ { 𝑥 ∣ ( 𝑥 ⊆ 𝐵 ∧ 𝜑 ) } ↔ ( 𝐴 ⊆ 𝐵 ∧ 𝜓 ) ) )
8	4 7	syl	⊢ ( 𝐵 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ ( 𝑥 ⊆ 𝐵 ∧ 𝜑 ) } ↔ ( 𝐴 ⊆ 𝐵 ∧ 𝜓 ) ) )