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Theorem ordir 1020
Description: Distributive law for disjunction. (Contributed by NM, 12-Aug-1994.)
Assertion
Ref Expression
ordir (((𝜑𝜓) ∨ 𝜒) ↔ ((𝜑𝜒) ∧ (𝜓𝜒)))

Proof of Theorem ordir
StepHypRef Expression
1 ordi 1019 . 2 ((𝜒 ∨ (𝜑𝜓)) ↔ ((𝜒𝜑) ∧ (𝜒𝜓)))
2 orcom 888 . 2 (((𝜑𝜓) ∨ 𝜒) ↔ (𝜒 ∨ (𝜑𝜓)))
3 orcom 888 . . 3 ((𝜑𝜒) ↔ (𝜒𝜑))
4 orcom 888 . . 3 ((𝜓𝜒) ↔ (𝜒𝜓))
53, 4anbi12i 614 . 2 (((𝜑𝜒) ∧ (𝜓𝜒)) ↔ ((𝜒𝜑) ∧ (𝜒𝜓)))
61, 2, 53bitr4i 294 1 (((𝜑𝜓) ∨ 𝜒) ↔ ((𝜑𝜒) ∧ (𝜓𝜒)))
Colors of variables: wff setvar class
Syntax hints:  wb 197  wa 384  wo 865
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866
This theorem is referenced by:  orddi  1023  pm5.62  1033  dn1  1073  cadan  1703  elnn0z  11652  ifpim123g  38342  rp-fakeanorass  38355  fvmptrabdm  41880
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