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Theorem r19.26-3 3109
Description: Version of r19.26 3108 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 3108 . . 3 (∀𝑥𝐴 ((𝜑𝜓) ∧ 𝜒) ↔ (∀𝑥𝐴 (𝜑𝜓) ∧ ∀𝑥𝐴 𝜒))
2 r19.26 3108 . . 3 (∀𝑥𝐴 (𝜑𝜓) ↔ (∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 𝜓))
31, 2bianbi 625 . 2 (∀𝑥𝐴 ((𝜑𝜓) ∧ 𝜒) ↔ ((∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 𝜓) ∧ ∀𝑥𝐴 𝜒))
4 df-3an 1086 . . 3 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∧ 𝜒))
54ralbii 3090 . 2 (∀𝑥𝐴 (𝜑𝜓𝜒) ↔ ∀𝑥𝐴 ((𝜑𝜓) ∧ 𝜒))
6 df-3an 1086 . 2 ((∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 𝜓 ∧ ∀𝑥𝐴 𝜒) ↔ ((∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 𝜓) ∧ ∀𝑥𝐴 𝜒))
73, 5, 63bitr4i 302 1 (∀𝑥𝐴 (𝜑𝜓𝜒) ↔ (∀𝑥𝐴 𝜑 ∧ ∀𝑥𝐴 𝜓 ∧ ∀𝑥𝐴 𝜒))
Colors of variables: wff setvar class
Syntax hints:  wb 205  wa 394  w3a 1084  wral 3058
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803
This theorem depends on definitions:  df-bi 206  df-an 395  df-3an 1086  df-ral 3059
This theorem is referenced by:  sgrp2rid2ex  18893  axeuclid  28802  axcontlem8  28810  stoweidlem60  45495
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