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Metamath Proof Explorer

Theorem elrabd

Description: Membership in a restricted class abstraction, using implicit substitution. Deduction version of elrab . (Contributed by Glauco Siliprandi, 23-Oct-2021)

	Ref	Expression
Hypotheses	elrabd.1	⊢ ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) )
	elrabd.2	⊢ ( 𝜑 → 𝐴 ∈ 𝐵 )
	elrabd.3	⊢ ( 𝜑 → 𝜒 )
Assertion	elrabd	⊢ ( 𝜑 → 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜓 } )

Proof

Step	Hyp	Ref	Expression
1		elrabd.1	⊢ ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) )
2		elrabd.2	⊢ ( 𝜑 → 𝐴 ∈ 𝐵 )
3		elrabd.3	⊢ ( 𝜑 → 𝜒 )
4	1	elrab	⊢ ( 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜓 } ↔ ( 𝐴 ∈ 𝐵 ∧ 𝜒 ) )
5	2 3 4	sylanbrc	⊢ ( 𝜑 → 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜓 } )