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Theorem alsconv 13719
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 2440 . . 3 (∀𝑥𝐴 𝜑 ↔ ∀𝑥(𝑥𝐴𝜑))
21anbi1i 454 . 2 ((∀𝑥𝐴 𝜑 ∧ ∃𝑥 𝑥𝐴) ↔ (∀𝑥(𝑥𝐴𝜑) ∧ ∃𝑥 𝑥𝐴))
3 df-alsc 13718 . 2 (∀!𝑥𝐴𝜑 ↔ (∀𝑥𝐴 𝜑 ∧ ∃𝑥 𝑥𝐴))
4 df-alsi 13717 . 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 1333  wex 1472  wcel 2128  wral 2435  ∀!walsi 13715  ∀!walsc 13716
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107
This theorem depends on definitions:  df-bi 116  df-ral 2440  df-alsi 13717  df-alsc 13718
This theorem is referenced by: (None)
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