Metamath Proof Explorer


Theorem csbopabgALT

Description: Move substitution into a class abstraction. Version of csbopab with a sethood antecedent but depending on fewer axioms. (Contributed by NM, 6-Aug-2007) (Proof shortened by Mario Carneiro, 17-Nov-2016) (New usage is discouraged.) (Proof modification is discouraged.)

Ref Expression
Assertion csbopabgALT ( 𝐴𝑉 𝐴 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝜑 } = { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝐴 / 𝑥 ] 𝜑 } )

Proof

Step Hyp Ref Expression
1 csbeq1 ( 𝑤 = 𝐴 𝑤 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝜑 } = 𝐴 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝜑 } )
2 dfsbcq2 ( 𝑤 = 𝐴 → ( [ 𝑤 / 𝑥 ] 𝜑[ 𝐴 / 𝑥 ] 𝜑 ) )
3 2 opabbidv ( 𝑤 = 𝐴 → { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝑤 / 𝑥 ] 𝜑 } = { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝐴 / 𝑥 ] 𝜑 } )
4 1 3 eqeq12d ( 𝑤 = 𝐴 → ( 𝑤 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝜑 } = { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝑤 / 𝑥 ] 𝜑 } ↔ 𝐴 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝜑 } = { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝐴 / 𝑥 ] 𝜑 } ) )
5 vex 𝑤 ∈ V
6 nfs1v 𝑥 [ 𝑤 / 𝑥 ] 𝜑
7 6 nfopab 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝑤 / 𝑥 ] 𝜑 }
8 sbequ12 ( 𝑥 = 𝑤 → ( 𝜑 ↔ [ 𝑤 / 𝑥 ] 𝜑 ) )
9 8 opabbidv ( 𝑥 = 𝑤 → { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝜑 } = { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝑤 / 𝑥 ] 𝜑 } )
10 5 7 9 csbief 𝑤 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝜑 } = { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝑤 / 𝑥 ] 𝜑 }
11 4 10 vtoclg ( 𝐴𝑉 𝐴 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝜑 } = { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝐴 / 𝑥 ] 𝜑 } )