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Theorem 19.26-3an 1459
Description: Theorem 19.26 of [Margaris] p. 90 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 1457 . . 3 (∀𝑥((𝜑𝜓) ∧ 𝜒) ↔ (∀𝑥(𝜑𝜓) ∧ ∀𝑥𝜒))
2 19.26 1457 . . . 4 (∀𝑥(𝜑𝜓) ↔ (∀𝑥𝜑 ∧ ∀𝑥𝜓))
32anbi1i 453 . . 3 ((∀𝑥(𝜑𝜓) ∧ ∀𝑥𝜒) ↔ ((∀𝑥𝜑 ∧ ∀𝑥𝜓) ∧ ∀𝑥𝜒))
41, 3bitri 183 . 2 (∀𝑥((𝜑𝜓) ∧ 𝜒) ↔ ((∀𝑥𝜑 ∧ ∀𝑥𝜓) ∧ ∀𝑥𝜒))
5 df-3an 964 . . 3 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∧ 𝜒))
65albii 1446 . 2 (∀𝑥(𝜑𝜓𝜒) ↔ ∀𝑥((𝜑𝜓) ∧ 𝜒))
7 df-3an 964 . 2 ((∀𝑥𝜑 ∧ ∀𝑥𝜓 ∧ ∀𝑥𝜒) ↔ ((∀𝑥𝜑 ∧ ∀𝑥𝜓) ∧ ∀𝑥𝜒))
84, 6, 73bitr4i 211 1 (∀𝑥(𝜑𝜓𝜒) ↔ (∀𝑥𝜑 ∧ ∀𝑥𝜓 ∧ ∀𝑥𝜒))
Colors of variables: wff set class
Syntax hints:  wa 103  wb 104  w3a 962  wal 1329
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1423  ax-gen 1425
This theorem depends on definitions:  df-bi 116  df-3an 964
This theorem is referenced by:  hb3and  1466
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