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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 2290 . . 3 ([𝑦 / 𝑥]∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ ∃𝑣[𝑦 / 𝑥](𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)))
2 sban 2087 . . . . 5 ([𝑦 / 𝑥](𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ ([𝑦 / 𝑥]𝑣𝐴 ∧ [𝑦 / 𝑥]∀𝑧(𝑧𝑣𝜑)))
3 sbabel.1 . . . . . . . 8 𝑥𝐴
43nfcri 2969 . . . . . . 7 𝑥 𝑣𝐴
54sbf 2273 . . . . . 6 ([𝑦 / 𝑥]𝑣𝐴𝑣𝐴)
6 nfv 1916 . . . . . . . . 9 𝑥 𝑧𝑣
76sbf 2273 . . . . . . . 8 ([𝑦 / 𝑥]𝑧𝑣𝑧𝑣)
87sbrbis 2322 . . . . . . 7 ([𝑦 / 𝑥](𝑧𝑣𝜑) ↔ (𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑))
98sbalv 2168 . . . . . 6 ([𝑦 / 𝑥]∀𝑧(𝑧𝑣𝜑) ↔ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑))
105, 9anbi12i 629 . . . . 5 (([𝑦 / 𝑥]𝑣𝐴 ∧ [𝑦 / 𝑥]∀𝑧(𝑧𝑣𝜑)) ↔ (𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
112, 10bitri 278 . . . 4 ([𝑦 / 𝑥](𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ (𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
1211exbii 1849 . . 3 (∃𝑣[𝑦 / 𝑥](𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ ∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
131, 12bitri 278 . 2 ([𝑦 / 𝑥]∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)) ↔ ∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
14 clabel 2960 . . 3 ({𝑧𝜑} ∈ 𝐴 ↔ ∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)))
1514sbbii 2082 . 2 ([𝑦 / 𝑥]{𝑧𝜑} ∈ 𝐴 ↔ [𝑦 / 𝑥]∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣𝜑)))
16 clabel 2960 . 2 ({𝑧 ∣ [𝑦 / 𝑥]𝜑} ∈ 𝐴 ↔ ∃𝑣(𝑣𝐴 ∧ ∀𝑧(𝑧𝑣 ↔ [𝑦 / 𝑥]𝜑)))
1713, 15, 163bitr4i 306 1 ([𝑦 / 𝑥]{𝑧𝜑} ∈ 𝐴 ↔ {𝑧 ∣ [𝑦 / 𝑥]𝜑} ∈ 𝐴)
Colors of variables: wff setvar class
Syntax hints:  wb 209  wa 399  wal 1536  wex 1781  [wsb 2070  wcel 2115  {cab 2802  wnfc 2962
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-10 2146  ax-11 2162  ax-12 2179  ax-ext 2796
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2071  df-clab 2803  df-cleq 2817  df-clel 2896  df-nfc 2964
This theorem is referenced by: (None)
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