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Theorem sbabel 3013
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 2282 . . 3 ([𝑦 / 𝑥]∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ ∃𝑣[𝑦 / 𝑥](𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)))
2 sban 2080 . . . . 5 ([𝑦 / 𝑥](𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ ([𝑦 / 𝑥]𝑣𝐴 ∧ [𝑦 / 𝑥]∀𝑧(𝑧𝑣𝜑)))
3 sbabel.1 . . . . . . . 8 𝑥𝐴
43nfcri 2969 . . . . . . 7 𝑥 𝑣𝐴
54sbf 2264 . . . . . 6 ([𝑦 / 𝑥]𝑣𝐴𝑣𝐴)
6 nfv 1909 . . . . . . . . 9 𝑥 𝑧𝑣
76sbf 2264 . . . . . . . 8 ([𝑦 / 𝑥]𝑧𝑣𝑧𝑣)
87sbrbis 2314 . . . . . . 7 ([𝑦 / 𝑥](𝑧𝑣𝜑) ↔ (𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑))
98sbalv 2160 . . . . . 6 ([𝑦 / 𝑥]∀𝑧(𝑧𝑣𝜑) ↔ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑))
105, 9anbi12i 628 . . . . 5 (([𝑦 / 𝑥]𝑣𝐴 ∧ [𝑦 / 𝑥]∀𝑧(𝑧𝑣𝜑)) ↔ (𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
112, 10bitri 277 . . . 4 ([𝑦 / 𝑥](𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ (𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
1211exbii 1842 . . 3 (∃𝑣[𝑦 / 𝑥](𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ ∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
131, 12bitri 277 . 2 ([𝑦 / 𝑥]∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ ∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
14 clabel 2957 . . 3 ({𝑧𝜑} ∈ 𝐴 ↔ ∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)))
1514sbbii 2075 . 2 ([𝑦 / 𝑥]{𝑧𝜑} ∈ 𝐴 ↔ [𝑦 / 𝑥]∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)))
16 clabel 2957 . 2 ({𝑧 ∣ [𝑦 / 𝑥]𝜑} ∈ 𝐴 ↔ ∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
1713, 15, 163bitr4i 305 1 ([𝑦 / 𝑥]{𝑧𝜑} ∈ 𝐴 ↔ {𝑧 ∣ [𝑦 / 𝑥]𝜑} ∈ 𝐴)
Colors of variables: wff setvar class
Syntax hints:  wb 208  wa 398  wal 1529  wex 1774  [wsb 2063  wcel 2108  {cab 2797  wnfc 2959
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1905  ax-6 1964  ax-7 2009  ax-8 2110  ax-9 2118  ax-10 2139  ax-11 2154  ax-12 2170  ax-ext 2791
This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1534  df-ex 1775  df-nf 1779  df-sb 2064  df-clab 2798  df-cleq 2812  df-clel 2891  df-nfc 2961
This theorem is referenced by: (None)
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