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Metamath Proof Explorer

Theorem sbabel

Description: Theorem to move a substitution in and out of a class abstraction. (Contributed by NM, 27-Sep-2003) (Revised by Mario Carneiro, 7-Oct-2016) (Proof shortened by Wolf Lammen, 28-Oct-2024)

Ref Expression
Hypothesis sbabel.1 ⊢ Ⅎ 𝑥 𝐴
Assertion sbabel ( [ 𝑦 / 𝑥 ] { 𝑧 ∣ 𝜑 } ∈ 𝐴 ↔ { 𝑧 ∣ [ 𝑦 / 𝑥 ] 𝜑 } ∈ 𝐴 )

Proof

Step Hyp Ref Expression
1 sbabel.1 ⊢ Ⅎ 𝑥 𝐴
2 clabel ⊢ ( { 𝑧 ∣ 𝜑 } ∈ 𝐴 ↔ ∃ 𝑣 ( 𝑣 ∈ 𝐴 ∧ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ 𝜑 ) ) )
3 2 sbbii ⊢ ( [ 𝑦 / 𝑥 ] { 𝑧 ∣ 𝜑 } ∈ 𝐴 ↔ [ 𝑦 / 𝑥 ] ∃ 𝑣 ( 𝑣 ∈ 𝐴 ∧ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ 𝜑 ) ) )
4 sbex ⊢ ( [ 𝑦 / 𝑥 ] ∃ 𝑣 ( 𝑣 ∈ 𝐴 ∧ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ 𝜑 ) ) ↔ ∃ 𝑣 [ 𝑦 / 𝑥 ] ( 𝑣 ∈ 𝐴 ∧ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ 𝜑 ) ) )
5 sban ⊢ ( [ 𝑦 / 𝑥 ] ( 𝑣 ∈ 𝐴 ∧ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ 𝜑 ) ) ↔ ( [ 𝑦 / 𝑥 ] 𝑣 ∈ 𝐴 ∧ [ 𝑦 / 𝑥 ] ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ 𝜑 ) ) )
6 1 nfcri ⊢ Ⅎ 𝑥 𝑣 ∈ 𝐴
7 6 sbf ⊢ ( [ 𝑦 / 𝑥 ] 𝑣 ∈ 𝐴 ↔ 𝑣 ∈ 𝐴 )
8 sbv ⊢ ( [ 𝑦 / 𝑥 ] 𝑧 ∈ 𝑣 ↔ 𝑧 ∈ 𝑣 )
9 8 sbrbis ⊢ ( [ 𝑦 / 𝑥 ] ( 𝑧 ∈ 𝑣 ↔ 𝜑 ) ↔ ( 𝑧 ∈ 𝑣 ↔ [ 𝑦 / 𝑥 ] 𝜑 ) )
10 9 sbalv ⊢ ( [ 𝑦 / 𝑥 ] ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ 𝜑 ) ↔ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ [ 𝑦 / 𝑥 ] 𝜑 ) )
11 7 10 anbi12i ⊢ ( ( [ 𝑦 / 𝑥 ] 𝑣 ∈ 𝐴 ∧ [ 𝑦 / 𝑥 ] ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ 𝜑 ) ) ↔ ( 𝑣 ∈ 𝐴 ∧ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ [ 𝑦 / 𝑥 ] 𝜑 ) ) )
12 5 11 bitri ⊢ ( [ 𝑦 / 𝑥 ] ( 𝑣 ∈ 𝐴 ∧ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ 𝜑 ) ) ↔ ( 𝑣 ∈ 𝐴 ∧ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ [ 𝑦 / 𝑥 ] 𝜑 ) ) )
13 12 exbii ⊢ ( ∃ 𝑣 [ 𝑦 / 𝑥 ] ( 𝑣 ∈ 𝐴 ∧ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ 𝜑 ) ) ↔ ∃ 𝑣 ( 𝑣 ∈ 𝐴 ∧ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ [ 𝑦 / 𝑥 ] 𝜑 ) ) )
14 3 4 13 3bitri ⊢ ( [ 𝑦 / 𝑥 ] { 𝑧 ∣ 𝜑 } ∈ 𝐴 ↔ ∃ 𝑣 ( 𝑣 ∈ 𝐴 ∧ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ [ 𝑦 / 𝑥 ] 𝜑 ) ) )
15 clabel ⊢ ( { 𝑧 ∣ [ 𝑦 / 𝑥 ] 𝜑 } ∈ 𝐴 ↔ ∃ 𝑣 ( 𝑣 ∈ 𝐴 ∧ ∀ 𝑧 ( 𝑧 ∈ 𝑣 ↔ [ 𝑦 / 𝑥 ] 𝜑 ) ) )
16 14 15 bitr4i ⊢ ( [ 𝑦 / 𝑥 ] { 𝑧 ∣ 𝜑 } ∈ 𝐴 ↔ { 𝑧 ∣ [ 𝑦 / 𝑥 ] 𝜑 } ∈ 𝐴 )