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Theorem 19.26-3an 1961
Description: Theorem 19.26 1959 with triple conjunction. (Contributed by NM, 13-Sep-2011.)
Assertion
Ref Expression
19.26-3an (∀𝑥(𝜑𝜓𝜒) ↔ (∀𝑥𝜑 ∧ ∀𝑥𝜓 ∧ ∀𝑥𝜒))

Proof of Theorem 19.26-3an
StepHypRef Expression
1 19.26 1959 . . 3 (∀𝑥(𝜑𝜓) ↔ (∀𝑥𝜑 ∧ ∀𝑥𝜓))
21anbi1i 612 . 2 ((∀𝑥(𝜑𝜓) ∧ ∀𝑥𝜒) ↔ ((∀𝑥𝜑 ∧ ∀𝑥𝜓) ∧ ∀𝑥𝜒))
3 df-3an 1102 . . . 4 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∧ 𝜒))
43albii 1904 . . 3 (∀𝑥(𝜑𝜓𝜒) ↔ ∀𝑥((𝜑𝜓) ∧ 𝜒))
5 19.26 1959 . . 3 (∀𝑥((𝜑𝜓) ∧ 𝜒) ↔ (∀𝑥(𝜑𝜓) ∧ ∀𝑥𝜒))
64, 5bitri 266 . 2 (∀𝑥(𝜑𝜓𝜒) ↔ (∀𝑥(𝜑𝜓) ∧ ∀𝑥𝜒))
7 df-3an 1102 . 2 ((∀𝑥𝜑 ∧ ∀𝑥𝜓 ∧ ∀𝑥𝜒) ↔ ((∀𝑥𝜑 ∧ ∀𝑥𝜓) ∧ ∀𝑥𝜒))
82, 6, 73bitr4i 294 1 (∀𝑥(𝜑𝜓𝜒) ↔ (∀𝑥𝜑 ∧ ∀𝑥𝜓 ∧ ∀𝑥𝜒))
Colors of variables: wff setvar class
Syntax hints:  wb 197  wa 384  w3a 1100  wal 1635
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894
This theorem depends on definitions:  df-bi 198  df-an 385  df-3an 1102
This theorem is referenced by:  alrim3con13v  39235  19.21a3con13vVD  39575
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