Theorem rabrabi

Description: Abstract builder restricted to another restricted abstract builder with implicit substitution. (Contributed by AV, 2-Aug-2022) Avoid ax-10 and ax-11 . (Revised by Gino Giotto, 20-Aug-2023)

	Ref	Expression
Hypothesis	rabrabi.1	⊢ ( 𝑥 = 𝑦 → ( 𝜒 ↔ 𝜑 ) )
Assertion	rabrabi	⊢ { 𝑥 ∈ { 𝑦 ∈ 𝐴 ∣ 𝜑 } ∣ 𝜓 } = { 𝑥 ∈ 𝐴 ∣ ( 𝜒 ∧ 𝜓 ) }

Proof

Step	Hyp	Ref	Expression
1		rabrabi.1	⊢ ( 𝑥 = 𝑦 → ( 𝜒 ↔ 𝜑 ) )
2	1	cbvrabv	⊢ { 𝑥 ∈ 𝐴 ∣ 𝜒 } = { 𝑦 ∈ 𝐴 ∣ 𝜑 }
3	2	rabeqi	⊢ { 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜒 } ∣ 𝜓 } = { 𝑥 ∈ { 𝑦 ∈ 𝐴 ∣ 𝜑 } ∣ 𝜓 }
4		rabrab	⊢ { 𝑥 ∈ { 𝑥 ∈ 𝐴 ∣ 𝜒 } ∣ 𝜓 } = { 𝑥 ∈ 𝐴 ∣ ( 𝜒 ∧ 𝜓 ) }
5	3 4	eqtr3i	⊢ { 𝑥 ∈ { 𝑦 ∈ 𝐴 ∣ 𝜑 } ∣ 𝜓 } = { 𝑥 ∈ 𝐴 ∣ ( 𝜒 ∧ 𝜓 ) }