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Theorem an6 1299
Description: Rearrangement of 6 conjuncts. (Contributed by NM, 13-Mar-1995.)
Assertion
Ref Expression
an6 (((𝜑𝜓𝜒) ∧ (𝜃𝜏𝜂)) ↔ ((𝜑𝜃) ∧ (𝜓𝜏) ∧ (𝜒𝜂)))

Proof of Theorem an6
StepHypRef Expression
1 df-3an 964 . . . 4 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∧ 𝜒))
2 df-3an 964 . . . 4 ((𝜃𝜏𝜂) ↔ ((𝜃𝜏) ∧ 𝜂))
31, 2anbi12i 455 . . 3 (((𝜑𝜓𝜒) ∧ (𝜃𝜏𝜂)) ↔ (((𝜑𝜓) ∧ 𝜒) ∧ ((𝜃𝜏) ∧ 𝜂)))
4 an4 575 . . 3 ((((𝜑𝜓) ∧ 𝜒) ∧ ((𝜃𝜏) ∧ 𝜂)) ↔ (((𝜑𝜓) ∧ (𝜃𝜏)) ∧ (𝜒𝜂)))
5 an4 575 . . . 4 (((𝜑𝜓) ∧ (𝜃𝜏)) ↔ ((𝜑𝜃) ∧ (𝜓𝜏)))
65anbi1i 453 . . 3 ((((𝜑𝜓) ∧ (𝜃𝜏)) ∧ (𝜒𝜂)) ↔ (((𝜑𝜃) ∧ (𝜓𝜏)) ∧ (𝜒𝜂)))
73, 4, 63bitri 205 . 2 (((𝜑𝜓𝜒) ∧ (𝜃𝜏𝜂)) ↔ (((𝜑𝜃) ∧ (𝜓𝜏)) ∧ (𝜒𝜂)))
8 df-3an 964 . 2 (((𝜑𝜃) ∧ (𝜓𝜏) ∧ (𝜒𝜂)) ↔ (((𝜑𝜃) ∧ (𝜓𝜏)) ∧ (𝜒𝜂)))
97, 8bitr4i 186 1 (((𝜑𝜓𝜒) ∧ (𝜃𝜏𝜂)) ↔ ((𝜑𝜃) ∧ (𝜓𝜏) ∧ (𝜒𝜂)))
Colors of variables: wff set class
Syntax hints:  wa 103  wb 104  w3a 962
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107
This theorem depends on definitions:  df-bi 116  df-3an 964
This theorem is referenced by:  3an6  1300  elfzuzb  9814
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