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Theorem ifpororb 38634
Description: Factor conditional logic operator over disjunction in terms 2 and 3. (Contributed by RP, 21-Apr-2020.)
Assertion
Ref Expression
ifpororb (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)))

Proof of Theorem ifpororb
StepHypRef Expression
1 dfifp2 1088 . 2 (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ ((𝜑 → (𝜓𝜒)) ∧ (¬ 𝜑 → (𝜃𝜏))))
2 df-or 875 . . . 4 ((𝜓𝜒) ↔ (¬ 𝜓𝜒))
32imbi2i 328 . . 3 ((𝜑 → (𝜓𝜒)) ↔ (𝜑 → (¬ 𝜓𝜒)))
4 df-or 875 . . . 4 ((𝜃𝜏) ↔ (¬ 𝜃𝜏))
54imbi2i 328 . . 3 ((¬ 𝜑 → (𝜃𝜏)) ↔ (¬ 𝜑 → (¬ 𝜃𝜏)))
63, 5anbi12i 621 . 2 (((𝜑 → (𝜓𝜒)) ∧ (¬ 𝜑 → (𝜃𝜏))) ↔ ((𝜑 → (¬ 𝜓𝜒)) ∧ (¬ 𝜑 → (¬ 𝜃𝜏))))
7 ifpimimb 38633 . . 3 (if-(𝜑, (¬ 𝜓𝜒), (¬ 𝜃𝜏)) ↔ (if-(𝜑, ¬ 𝜓, ¬ 𝜃) → if-(𝜑, 𝜒, 𝜏)))
8 dfifp2 1088 . . 3 (if-(𝜑, (¬ 𝜓𝜒), (¬ 𝜃𝜏)) ↔ ((𝜑 → (¬ 𝜓𝜒)) ∧ (¬ 𝜑 → (¬ 𝜃𝜏))))
9 imor 880 . . . 4 ((if-(𝜑, ¬ 𝜓, ¬ 𝜃) → if-(𝜑, 𝜒, 𝜏)) ↔ (¬ if-(𝜑, ¬ 𝜓, ¬ 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)))
10 ifpnot23d 38614 . . . . 5 (¬ if-(𝜑, ¬ 𝜓, ¬ 𝜃) ↔ if-(𝜑, 𝜓, 𝜃))
1110orbi1i 938 . . . 4 ((¬ if-(𝜑, ¬ 𝜓, ¬ 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)))
129, 11bitri 267 . . 3 ((if-(𝜑, ¬ 𝜓, ¬ 𝜃) → if-(𝜑, 𝜒, 𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)))
137, 8, 123bitr3i 293 . 2 (((𝜑 → (¬ 𝜓𝜒)) ∧ (¬ 𝜑 → (¬ 𝜃𝜏))) ↔ (if-(𝜑, 𝜓, 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)))
141, 6, 133bitri 289 1 (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ∨ if-(𝜑, 𝜒, 𝜏)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 198  wa 385  wo 874  if-wif 1086
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 199  df-an 386  df-or 875  df-ifp 1087
This theorem is referenced by:  ifpananb  38635
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