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Theorem reueqd 2515
 Description: Equality deduction for restricted uniqueness quantifier. (Contributed by NM, 5-Apr-2004.)
Hypothesis
Ref Expression
raleqd.1 (𝐴 = 𝐵 → (𝜑𝜓))
Assertion
Ref Expression
reueqd (𝐴 = 𝐵 → (∃!𝑥𝐴 𝜑 ↔ ∃!𝑥𝐵 𝜓))
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵
Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥)

Proof of Theorem reueqd
StepHypRef Expression
1 reueq1 2507 . 2 (𝐴 = 𝐵 → (∃!𝑥𝐴 𝜑 ↔ ∃!𝑥𝐵 𝜑))
2 raleqd.1 . . 3 (𝐴 = 𝐵 → (𝜑𝜓))
32reubidv 2493 . 2 (𝐴 = 𝐵 → (∃!𝑥𝐵 𝜑 ↔ ∃!𝑥𝐵 𝜓))
41, 3bitrd 177 1 (𝐴 = 𝐵 → (∃!𝑥𝐴 𝜑 ↔ ∃!𝑥𝐵 𝜓))
 Colors of variables: wff set class Syntax hints:   → wi 4   ↔ wb 98   = wceq 1243  ∃!wreu 2308 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022 This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-eu 1903  df-cleq 2033  df-clel 2036  df-nfc 2167  df-reu 2313 This theorem is referenced by:  acexmidlemcase  5507
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