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Theorem sbabel 2966
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, 26-Dec-2019.)
Hypothesis
Ref Expression
sbabel.1 𝑥𝐴
Assertion
Ref Expression
sbabel ([𝑦 / 𝑥]{𝑧𝜑} ∈ 𝐴 ↔ {𝑧 ∣ [𝑦 / 𝑥]𝜑} ∈ 𝐴)
Distinct variable groups:   𝑥,𝑧   𝑦,𝑧
Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧)   𝐴(𝑥,𝑦,𝑧)

Proof of Theorem sbabel
Dummy variable 𝑣 is distinct from all other variables.
StepHypRef Expression
1 sbex 2501 . . 3 ([𝑦 / 𝑥]∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ ∃𝑣[𝑦 / 𝑥](𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)))
2 sban 2031 . . . . 5 ([𝑦 / 𝑥](𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ ([𝑦 / 𝑥]𝑣𝐴 ∧ [𝑦 / 𝑥]∀𝑧(𝑧𝑣𝜑)))
3 sbabel.1 . . . . . . . 8 𝑥𝐴
43nfcri 2926 . . . . . . 7 𝑥 𝑣𝐴
54sbf 2199 . . . . . 6 ([𝑦 / 𝑥]𝑣𝐴𝑣𝐴)
6 nfv 1873 . . . . . . . . 9 𝑥 𝑧𝑣
76sbf 2199 . . . . . . . 8 ([𝑦 / 𝑥]𝑧𝑣𝑧𝑣)
87sbrbis 2243 . . . . . . 7 ([𝑦 / 𝑥](𝑧𝑣𝜑) ↔ (𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑))
98sbalv 2502 . . . . . 6 ([𝑦 / 𝑥]∀𝑧(𝑧𝑣𝜑) ↔ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑))
105, 9anbi12i 617 . . . . 5 (([𝑦 / 𝑥]𝑣𝐴 ∧ [𝑦 / 𝑥]∀𝑧(𝑧𝑣𝜑)) ↔ (𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
112, 10bitri 267 . . . 4 ([𝑦 / 𝑥](𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ (𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
1211exbii 1810 . . 3 (∃𝑣[𝑦 / 𝑥](𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ ∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
131, 12bitri 267 . 2 ([𝑦 / 𝑥]∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ ∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
14 clabel 2914 . . 3 ({𝑧𝜑} ∈ 𝐴 ↔ ∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)))
1514sbbii 2027 . 2 ([𝑦 / 𝑥]{𝑧𝜑} ∈ 𝐴 ↔ [𝑦 / 𝑥]∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)))
16 clabel 2914 . 2 ({𝑧 ∣ [𝑦 / 𝑥]𝜑} ∈ 𝐴 ↔ ∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
1713, 15, 163bitr4i 295 1 ([𝑦 / 𝑥]{𝑧𝜑} ∈ 𝐴 ↔ {𝑧 ∣ [𝑦 / 𝑥]𝜑} ∈ 𝐴)
Colors of variables: wff setvar class
Syntax hints:  wb 198  wa 387  wal 1505  wex 1742  [wsb 2015  wcel 2050  {cab 2758  wnfc 2916
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772  ax-5 1869  ax-6 1928  ax-7 1965  ax-8 2052  ax-9 2059  ax-10 2079  ax-11 2093  ax-12 2106  ax-ext 2750
This theorem depends on definitions:  df-bi 199  df-an 388  df-or 834  df-tru 1510  df-ex 1743  df-nf 1747  df-sb 2016  df-clab 2759  df-cleq 2771  df-clel 2846  df-nfc 2918
This theorem is referenced by: (None)
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