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Theorem ifpnannanb 40212
 Description: Factor conditional logic operator over nand in terms 2 and 3. (Contributed by RP, 21-Apr-2020.)
Assertion
Ref Expression
ifpnannanb (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ⊼ if-(𝜑, 𝜒, 𝜏)))

Proof of Theorem ifpnannanb
StepHypRef Expression
1 df-nan 1483 . . 3 ((𝜓𝜒) ↔ ¬ (𝜓𝜒))
2 df-nan 1483 . . 3 ((𝜃𝜏) ↔ ¬ (𝜃𝜏))
3 ifpbi23 40178 . . 3 ((((𝜓𝜒) ↔ ¬ (𝜓𝜒)) ∧ ((𝜃𝜏) ↔ ¬ (𝜃𝜏))) → (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ if-(𝜑, ¬ (𝜓𝜒), ¬ (𝜃𝜏))))
41, 2, 3mp2an 691 . 2 (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ if-(𝜑, ¬ (𝜓𝜒), ¬ (𝜃𝜏)))
5 ifpananb 40211 . . . 4 (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ∧ if-(𝜑, 𝜒, 𝜏)))
65notbii 323 . . 3 (¬ if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ ¬ (if-(𝜑, 𝜓, 𝜃) ∧ if-(𝜑, 𝜒, 𝜏)))
7 ifpnotnotb 40184 . . 3 (if-(𝜑, ¬ (𝜓𝜒), ¬ (𝜃𝜏)) ↔ ¬ if-(𝜑, (𝜓𝜒), (𝜃𝜏)))
8 df-nan 1483 . . 3 ((if-(𝜑, 𝜓, 𝜃) ⊼ if-(𝜑, 𝜒, 𝜏)) ↔ ¬ (if-(𝜑, 𝜓, 𝜃) ∧ if-(𝜑, 𝜒, 𝜏)))
96, 7, 83bitr4i 306 . 2 (if-(𝜑, ¬ (𝜓𝜒), ¬ (𝜃𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ⊼ if-(𝜑, 𝜒, 𝜏)))
104, 9bitri 278 1 (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ⊼ if-(𝜑, 𝜒, 𝜏)))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   ↔ wb 209   ∧ wa 399  if-wif 1058   ⊼ wnan 1482 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 210  df-an 400  df-or 845  df-ifp 1059  df-nan 1483 This theorem is referenced by: (None)
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