Metamath Proof Explorer


Theorem sbcal

Description: Move universal quantifier in and out of class substitution. (Contributed by NM, 31-Dec-2016) (Revised by NM, 18-Aug-2018)

Ref Expression
Assertion sbcal ( [ 𝐴 / 𝑦 ]𝑥 𝜑 ↔ ∀ 𝑥 [ 𝐴 / 𝑦 ] 𝜑 )

Proof

Step Hyp Ref Expression
1 sbcex ( [ 𝐴 / 𝑦 ]𝑥 𝜑𝐴 ∈ V )
2 sbcex ( [ 𝐴 / 𝑦 ] 𝜑𝐴 ∈ V )
3 2 sps ( ∀ 𝑥 [ 𝐴 / 𝑦 ] 𝜑𝐴 ∈ V )
4 dfsbcq2 ( 𝑧 = 𝐴 → ( [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑[ 𝐴 / 𝑦 ]𝑥 𝜑 ) )
5 dfsbcq2 ( 𝑧 = 𝐴 → ( [ 𝑧 / 𝑦 ] 𝜑[ 𝐴 / 𝑦 ] 𝜑 ) )
6 5 albidv ( 𝑧 = 𝐴 → ( ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 ↔ ∀ 𝑥 [ 𝐴 / 𝑦 ] 𝜑 ) )
7 sbal ( [ 𝑧 / 𝑦 ] ∀ 𝑥 𝜑 ↔ ∀ 𝑥 [ 𝑧 / 𝑦 ] 𝜑 )
8 4 6 7 vtoclbg ( 𝐴 ∈ V → ( [ 𝐴 / 𝑦 ]𝑥 𝜑 ↔ ∀ 𝑥 [ 𝐴 / 𝑦 ] 𝜑 ) )
9 1 3 8 pm5.21nii ( [ 𝐴 / 𝑦 ]𝑥 𝜑 ↔ ∀ 𝑥 [ 𝐴 / 𝑦 ] 𝜑 )