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Theorem repizf2lem 3988
Description: Lemma for repizf2 3989. If we have a function-like proposition which provides at most one value of 𝑦 for each 𝑥 in a set 𝑤, we can change "at most one" to "exactly one" by restricting the values of 𝑥 to those values for which the proposition provides a value of 𝑦. (Contributed by Jim Kingdon, 7-Sep-2018.)
Assertion
Ref Expression
repizf2lem (∀𝑥𝑤 ∃*𝑦𝜑 ↔ ∀𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑}∃!𝑦𝜑)

Proof of Theorem repizf2lem
StepHypRef Expression
1 df-mo 1952 . . . 4 (∃*𝑦𝜑 ↔ (∃𝑦𝜑 → ∃!𝑦𝜑))
21imbi2i 224 . . 3 ((𝑥𝑤 → ∃*𝑦𝜑) ↔ (𝑥𝑤 → (∃𝑦𝜑 → ∃!𝑦𝜑)))
32albii 1404 . 2 (∀𝑥(𝑥𝑤 → ∃*𝑦𝜑) ↔ ∀𝑥(𝑥𝑤 → (∃𝑦𝜑 → ∃!𝑦𝜑)))
4 df-ral 2364 . 2 (∀𝑥𝑤 ∃*𝑦𝜑 ↔ ∀𝑥(𝑥𝑤 → ∃*𝑦𝜑))
5 df-ral 2364 . . 3 (∀𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑}∃!𝑦𝜑 ↔ ∀𝑥(𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑} → ∃!𝑦𝜑))
6 rabid 2542 . . . . . 6 (𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑} ↔ (𝑥𝑤 ∧ ∃𝑦𝜑))
76imbi1i 236 . . . . 5 ((𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑} → ∃!𝑦𝜑) ↔ ((𝑥𝑤 ∧ ∃𝑦𝜑) → ∃!𝑦𝜑))
8 impexp 259 . . . . 5 (((𝑥𝑤 ∧ ∃𝑦𝜑) → ∃!𝑦𝜑) ↔ (𝑥𝑤 → (∃𝑦𝜑 → ∃!𝑦𝜑)))
97, 8bitri 182 . . . 4 ((𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑} → ∃!𝑦𝜑) ↔ (𝑥𝑤 → (∃𝑦𝜑 → ∃!𝑦𝜑)))
109albii 1404 . . 3 (∀𝑥(𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑} → ∃!𝑦𝜑) ↔ ∀𝑥(𝑥𝑤 → (∃𝑦𝜑 → ∃!𝑦𝜑)))
115, 10bitri 182 . 2 (∀𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑}∃!𝑦𝜑 ↔ ∀𝑥(𝑥𝑤 → (∃𝑦𝜑 → ∃!𝑦𝜑)))
123, 4, 113bitr4i 210 1 (∀𝑥𝑤 ∃*𝑦𝜑 ↔ ∀𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑}∃!𝑦𝜑)
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wb 103  wal 1287  wex 1426  wcel 1438  ∃!weu 1948  ∃*wmo 1949  wral 2359  {crab 2363
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1381  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-ext 2070
This theorem depends on definitions:  df-bi 115  df-sb 1693  df-mo 1952  df-clab 2075  df-cleq 2081  df-clel 2084  df-ral 2364  df-rab 2368
This theorem is referenced by:  repizf2  3989
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