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 𝜑 → 𝐴 ∈ V )
2 sbcex ⊢ ( [ 𝐴 / 𝑦 ] 𝜑 → 𝐴 ∈ V )
3 2 spsv ⊢ ( ∀ 𝑥 [ 𝐴 / 𝑦 ] 𝜑 → 𝐴 ∈ 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 )