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Theorem alsconv 42345
 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 2901 . . 3 (∀𝑥𝐴 𝜑 ↔ ∀𝑥(𝑥𝐴𝜑))
21anbi1i 727 . 2 ((∀𝑥𝐴 𝜑 ∧ ∃𝑥 𝑥𝐴) ↔ (∀𝑥(𝑥𝐴𝜑) ∧ ∃𝑥 𝑥𝐴))
3 df-alsc 42344 . 2 (∀!𝑥𝐴𝜑 ↔ (∀𝑥𝐴 𝜑 ∧ ∃𝑥 𝑥𝐴))
4 df-alsi 42343 . 2 (∀!𝑥(𝑥𝐴𝜑) ↔ (∀𝑥(𝑥𝐴𝜑) ∧ ∃𝑥 𝑥𝐴))
52, 3, 43bitr4ri 292 1 (∀!𝑥(𝑥𝐴𝜑) ↔ ∀!𝑥𝐴𝜑)
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 195   ∧ wa 383  ∀wal 1473  ∃wex 1695   ∈ wcel 1977  ∀wral 2896  ∀!walsi 42341  ∀!walsc 42342 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8 This theorem depends on definitions:  df-bi 196  df-an 385  df-ral 2901  df-alsi 42343  df-alsc 42344 This theorem is referenced by: (None)
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