Metamath Proof Explorer


Theorem rexrab2

Description: Existential quantification over a class abstraction. (Contributed by Mario Carneiro, 3-Sep-2015)

Ref Expression
Hypothesis ralab2.1 ⊢ ( 𝑥 = 𝑦 → ( 𝜓 ↔ 𝜒 ) )
Assertion rexrab2 ( ∃ 𝑥 ∈ { 𝑦 ∈ 𝐴 ∣ 𝜑 } 𝜓 ↔ ∃ 𝑦 ∈ 𝐴 ( 𝜑 ∧ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 ralab2.1 ⊢ ( 𝑥 = 𝑦 → ( 𝜓 ↔ 𝜒 ) )
2 df-rab ⊢ { 𝑦 ∈ 𝐴 ∣ 𝜑 } = { 𝑦 ∣ ( 𝑦 ∈ 𝐴 ∧ 𝜑 ) }
3 2 rexeqi ⊢ ( ∃ 𝑥 ∈ { 𝑦 ∈ 𝐴 ∣ 𝜑 } 𝜓 ↔ ∃ 𝑥 ∈ { 𝑦 ∣ ( 𝑦 ∈ 𝐴 ∧ 𝜑 ) } 𝜓 )
4 1 rexab2 ⊢ ( ∃ 𝑥 ∈ { 𝑦 ∣ ( 𝑦 ∈ 𝐴 ∧ 𝜑 ) } 𝜓 ↔ ∃ 𝑦 ( ( 𝑦 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝜒 ) )
5 anass ⊢ ( ( ( 𝑦 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝜒 ) ↔ ( 𝑦 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜒 ) ) )
6 5 exbii ⊢ ( ∃ 𝑦 ( ( 𝑦 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝜒 ) ↔ ∃ 𝑦 ( 𝑦 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜒 ) ) )
7 df-rex ⊢ ( ∃ 𝑦 ∈ 𝐴 ( 𝜑 ∧ 𝜒 ) ↔ ∃ 𝑦 ( 𝑦 ∈ 𝐴 ∧ ( 𝜑 ∧ 𝜒 ) ) )
8 6 7 bitr4i ⊢ ( ∃ 𝑦 ( ( 𝑦 ∈ 𝐴 ∧ 𝜑 ) ∧ 𝜒 ) ↔ ∃ 𝑦 ∈ 𝐴 ( 𝜑 ∧ 𝜒 ) )
9 3 4 8 3bitri ⊢ ( ∃ 𝑥 ∈ { 𝑦 ∈ 𝐴 ∣ 𝜑 } 𝜓 ↔ ∃ 𝑦 ∈ 𝐴 ( 𝜑 ∧ 𝜒 ) )