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Theorem repizf2lem 4085
Description: Lemma for repizf2 4086. 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 2003 . . . 4 (∃*𝑦𝜑 ↔ (∃𝑦𝜑 → ∃!𝑦𝜑))
21imbi2i 225 . . 3 ((𝑥𝑤 → ∃*𝑦𝜑) ↔ (𝑥𝑤 → (∃𝑦𝜑 → ∃!𝑦𝜑)))
32albii 1446 . 2 (∀𝑥(𝑥𝑤 → ∃*𝑦𝜑) ↔ ∀𝑥(𝑥𝑤 → (∃𝑦𝜑 → ∃!𝑦𝜑)))
4 df-ral 2421 . 2 (∀𝑥𝑤 ∃*𝑦𝜑 ↔ ∀𝑥(𝑥𝑤 → ∃*𝑦𝜑))
5 df-ral 2421 . . 3 (∀𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑}∃!𝑦𝜑 ↔ ∀𝑥(𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑} → ∃!𝑦𝜑))
6 rabid 2606 . . . . . 6 (𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑} ↔ (𝑥𝑤 ∧ ∃𝑦𝜑))
76imbi1i 237 . . . . 5 ((𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑} → ∃!𝑦𝜑) ↔ ((𝑥𝑤 ∧ ∃𝑦𝜑) → ∃!𝑦𝜑))
8 impexp 261 . . . . 5 (((𝑥𝑤 ∧ ∃𝑦𝜑) → ∃!𝑦𝜑) ↔ (𝑥𝑤 → (∃𝑦𝜑 → ∃!𝑦𝜑)))
97, 8bitri 183 . . . 4 ((𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑} → ∃!𝑦𝜑) ↔ (𝑥𝑤 → (∃𝑦𝜑 → ∃!𝑦𝜑)))
109albii 1446 . . 3 (∀𝑥(𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑} → ∃!𝑦𝜑) ↔ ∀𝑥(𝑥𝑤 → (∃𝑦𝜑 → ∃!𝑦𝜑)))
115, 10bitri 183 . 2 (∀𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑}∃!𝑦𝜑 ↔ ∀𝑥(𝑥𝑤 → (∃𝑦𝜑 → ∃!𝑦𝜑)))
123, 4, 113bitr4i 211 1 (∀𝑥𝑤 ∃*𝑦𝜑 ↔ ∀𝑥 ∈ {𝑥𝑤 ∣ ∃𝑦𝜑}∃!𝑦𝜑)
Colors of variables: wff set class
Syntax hints:  wi 4  wa 103  wb 104  wal 1329  wex 1468  wcel 1480  ∃!weu 1999  ∃*wmo 2000  wral 2416  {crab 2420
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1423  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-ext 2121
This theorem depends on definitions:  df-bi 116  df-sb 1736  df-mo 2003  df-clab 2126  df-cleq 2132  df-clel 2135  df-ral 2421  df-rab 2425
This theorem is referenced by:  repizf2  4086
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