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Theorem ifpxorxorb 37376
Description: Factor conditional logic operator over xor in terms 2 and 3. (Contributed by RP, 21-Apr-2020.)
Assertion
Ref Expression
ifpxorxorb (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ⊻ if-(𝜑, 𝜒, 𝜏)))

Proof of Theorem ifpxorxorb
StepHypRef Expression
1 df-xor 1462 . . 3 ((𝜓𝜒) ↔ ¬ (𝜓𝜒))
2 df-xor 1462 . . 3 ((𝜃𝜏) ↔ ¬ (𝜃𝜏))
3 ifpbi23 37337 . . 3 ((((𝜓𝜒) ↔ ¬ (𝜓𝜒)) ∧ ((𝜃𝜏) ↔ ¬ (𝜃𝜏))) → (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ if-(𝜑, ¬ (𝜓𝜒), ¬ (𝜃𝜏))))
41, 2, 3mp2an 707 . 2 (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ if-(𝜑, ¬ (𝜓𝜒), ¬ (𝜃𝜏)))
5 ifpbibib 37375 . . . 4 (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ↔ if-(𝜑, 𝜒, 𝜏)))
65notbii 310 . . 3 (¬ if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ ¬ (if-(𝜑, 𝜓, 𝜃) ↔ if-(𝜑, 𝜒, 𝜏)))
7 ifpnotnotb 37344 . . 3 (if-(𝜑, ¬ (𝜓𝜒), ¬ (𝜃𝜏)) ↔ ¬ if-(𝜑, (𝜓𝜒), (𝜃𝜏)))
8 df-xor 1462 . . 3 ((if-(𝜑, 𝜓, 𝜃) ⊻ if-(𝜑, 𝜒, 𝜏)) ↔ ¬ (if-(𝜑, 𝜓, 𝜃) ↔ if-(𝜑, 𝜒, 𝜏)))
96, 7, 83bitr4i 292 . 2 (if-(𝜑, ¬ (𝜓𝜒), ¬ (𝜃𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ⊻ if-(𝜑, 𝜒, 𝜏)))
104, 9bitri 264 1 (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ⊻ if-(𝜑, 𝜒, 𝜏)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 196  if-wif 1011  wxo 1461
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 197  df-or 385  df-an 386  df-ifp 1012  df-xor 1462
This theorem is referenced by: (None)
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