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Theorem r19.26-3 3100
Description: Version of r19.26 3099 with three quantifiers. (Contributed by FL, 22-Nov-2010.)
Assertion
Ref Expression
r19.26-3 (∀𝑥𝐴 (𝜑𝜓𝜒) ↔ (∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 𝜓 ∧ ∀𝑥𝐴 𝜒))

Proof of Theorem r19.26-3
StepHypRef Expression
1 r19.26 3099 . . 3 (∀𝑥𝐴 ((𝜑𝜓) ∧ 𝜒) ↔ (∀𝑥𝐴 (𝜑𝜓) ∧ ∀𝑥𝐴 𝜒))
2 r19.26 3099 . . 3 (∀𝑥𝐴 (𝜑𝜓) ↔ (∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 𝜓))
31, 2bianbi 627 . 2 (∀𝑥𝐴 ((𝜑𝜓) ∧ 𝜒) ↔ ((∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 𝜓) ∧ ∀𝑥𝐴 𝜒))
4 df-3an 1088 . . 3 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∧ 𝜒))
54ralbii 3083 . 2 (∀𝑥𝐴 (𝜑𝜓𝜒) ↔ ∀𝑥𝐴 ((𝜑𝜓) ∧ 𝜒))
6 df-3an 1088 . 2 ((∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 𝜓 ∧ ∀𝑥𝐴 𝜒) ↔ ((∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 𝜓) ∧ ∀𝑥𝐴 𝜒))
73, 5, 63bitr4i 303 1 (∀𝑥𝐴 (𝜑𝜓𝜒) ↔ (∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 𝜓 ∧ ∀𝑥𝐴 𝜒))
Colors of variables: wff setvar class
Syntax hints:  wb 206  wa 395  w3a 1086  wral 3052
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809
This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-ral 3053
This theorem is referenced by:  sgrp2rid2ex  18910  axeuclid  28947  axcontlem8  28955  stoweidlem60  46056
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