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Theorem 19.26-3an 1905
Description: Theorem 19.26 1903 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 1903 . . 3 (∀𝑥(𝜑𝜓) ↔ (∀𝑥𝜑 ∧ ∀𝑥𝜓))
21anbi1i 636 . 2 ((∀𝑥(𝜑𝜓) ∧ ∀𝑥𝜒) ↔ ((∀𝑥𝜑 ∧ ∀𝑥𝜓) ∧ ∀𝑥𝜒))
3 df-3an 1105 . . . 4 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∧ 𝜒))
43albii 1852 . . 3 (∀𝑥(𝜑𝜓𝜒) ↔ ∀𝑥((𝜑𝜓) ∧ 𝜒))
5 19.26 1903 . . 3 (∀𝑥((𝜑𝜓) ∧ 𝜒) ↔ (∀𝑥(𝜑𝜓) ∧ ∀𝑥𝜒))
64, 5bitri 278 . 2 (∀𝑥(𝜑𝜓𝜒) ↔ (∀𝑥(𝜑𝜓) ∧ ∀𝑥𝜒))
7 df-3an 1105 . 2 ((∀𝑥𝜑 ∧ ∀𝑥𝜓 ∧ ∀𝑥𝜒) ↔ ((∀𝑥𝜑 ∧ ∀𝑥𝜓) ∧ ∀𝑥𝜒))
82, 6, 73bitr4i 306 1 (∀𝑥(𝜑𝜓𝜒) ↔ (∀𝑥𝜑 ∧ ∀𝑥𝜓 ∧ ∀𝑥𝜒))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  wb 209  wa 401  w3a 1103  wal 1568
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842
This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105
This theorem is used by:  alrim3con13v  45275  19.21a3con13vVD  45593
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