Metamath Proof Explorer


Theorem rspesbca

Description: Existence form of rspsbca . (Contributed by NM, 29-Feb-2008) (Proof shortened by Mario Carneiro, 13-Oct-2016)

Ref Expression
Assertion rspesbca ( ( 𝐴 ∈ 𝐵 ∧ [ 𝐴 / 𝑥 ] 𝜑 ) → ∃ 𝑥 ∈ 𝐵 𝜑 )

Proof

Step Hyp Ref Expression
1 dfsbcq2 ⊢ ( 𝑦 = 𝐴 → ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝐴 / 𝑥 ] 𝜑 ) )
2 1 rspcev ⊢ ( ( 𝐴 ∈ 𝐵 ∧ [ 𝐴 / 𝑥 ] 𝜑 ) → ∃ 𝑦 ∈ 𝐵 [ 𝑦 / 𝑥 ] 𝜑 )
3 cbvrexsvw ⊢ ( ∃ 𝑥 ∈ 𝐵 𝜑 ↔ ∃ 𝑦 ∈ 𝐵 [ 𝑦 / 𝑥 ] 𝜑 )
4 2 3 sylibr ⊢ ( ( 𝐴 ∈ 𝐵 ∧ [ 𝐴 / 𝑥 ] 𝜑 ) → ∃ 𝑥 ∈ 𝐵 𝜑 )