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Theorem alsconv 12828
 Description: There is an equivalence between the two "all some" forms. (Contributed by David A. Wheeler, 22-Oct-2018.)
Assertion
Ref Expression
alsconv (∀!𝑥(𝑥𝐴𝜑) ↔ ∀!𝑥𝐴𝜑)

Proof of Theorem alsconv
StepHypRef Expression
1 df-ral 2380 . . 3 (∀𝑥𝐴 𝜑 ↔ ∀𝑥(𝑥𝐴𝜑))
21anbi1i 449 . 2 ((∀𝑥𝐴 𝜑 ∧ ∃𝑥 𝑥𝐴) ↔ (∀𝑥(𝑥𝐴𝜑) ∧ ∃𝑥 𝑥𝐴))
3 df-alsc 12827 . 2 (∀!𝑥𝐴𝜑 ↔ (∀𝑥𝐴 𝜑 ∧ ∃𝑥 𝑥𝐴))
4 df-alsi 12826 . 2 (∀!𝑥(𝑥𝐴𝜑) ↔ (∀𝑥(𝑥𝐴𝜑) ∧ ∃𝑥 𝑥𝐴))
52, 3, 43bitr4ri 212 1 (∀!𝑥(𝑥𝐴𝜑) ↔ ∀!𝑥𝐴𝜑)
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  ∀wal 1297  ∃wex 1436   ∈ wcel 1448  ∀wral 2375  ∀!walsi 12824  ∀!walsc 12825 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107 This theorem depends on definitions:  df-bi 116  df-ral 2380  df-alsi 12826  df-alsc 12827 This theorem is referenced by: (None)
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