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Theorem ifpananb 38372
Description: Factor conditional logic operator over conjunction in terms 2 and 3. (Contributed by RP, 21-Apr-2020.)
Assertion
Ref Expression
ifpananb (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ∧ if-(𝜑, 𝜒, 𝜏)))

Proof of Theorem ifpananb
StepHypRef Expression
1 anor 511 . . 3 ((𝜓𝜒) ↔ ¬ (¬ 𝜓 ∨ ¬ 𝜒))
2 anor 511 . . 3 ((𝜃𝜏) ↔ ¬ (¬ 𝜃 ∨ ¬ 𝜏))
3 ifpbi23 38338 . . 3 ((((𝜓𝜒) ↔ ¬ (¬ 𝜓 ∨ ¬ 𝜒)) ∧ ((𝜃𝜏) ↔ ¬ (¬ 𝜃 ∨ ¬ 𝜏))) → (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ if-(𝜑, ¬ (¬ 𝜓 ∨ ¬ 𝜒), ¬ (¬ 𝜃 ∨ ¬ 𝜏))))
41, 2, 3mp2an 710 . 2 (if-(𝜑, (𝜓𝜒), (𝜃𝜏)) ↔ if-(𝜑, ¬ (¬ 𝜓 ∨ ¬ 𝜒), ¬ (¬ 𝜃 ∨ ¬ 𝜏)))
5 ifpororb 38371 . . . . 5 (if-(𝜑, (¬ 𝜓 ∨ ¬ 𝜒), (¬ 𝜃 ∨ ¬ 𝜏)) ↔ (if-(𝜑, ¬ 𝜓, ¬ 𝜃) ∨ if-(𝜑, ¬ 𝜒, ¬ 𝜏)))
6 ifpnotnotb 38345 . . . . . 6 (if-(𝜑, ¬ 𝜓, ¬ 𝜃) ↔ ¬ if-(𝜑, 𝜓, 𝜃))
7 ifpnotnotb 38345 . . . . . 6 (if-(𝜑, ¬ 𝜒, ¬ 𝜏) ↔ ¬ if-(𝜑, 𝜒, 𝜏))
86, 7orbi12i 544 . . . . 5 ((if-(𝜑, ¬ 𝜓, ¬ 𝜃) ∨ if-(𝜑, ¬ 𝜒, ¬ 𝜏)) ↔ (¬ if-(𝜑, 𝜓, 𝜃) ∨ ¬ if-(𝜑, 𝜒, 𝜏)))
95, 8bitri 264 . . . 4 (if-(𝜑, (¬ 𝜓 ∨ ¬ 𝜒), (¬ 𝜃 ∨ ¬ 𝜏)) ↔ (¬ if-(𝜑, 𝜓, 𝜃) ∨ ¬ if-(𝜑, 𝜒, 𝜏)))
109notbii 309 . . 3 (¬ if-(𝜑, (¬ 𝜓 ∨ ¬ 𝜒), (¬ 𝜃 ∨ ¬ 𝜏)) ↔ ¬ (¬ if-(𝜑, 𝜓, 𝜃) ∨ ¬ if-(𝜑, 𝜒, 𝜏)))
11 ifpnotnotb 38345 . . 3 (if-(𝜑, ¬ (¬ 𝜓 ∨ ¬ 𝜒), ¬ (¬ 𝜃 ∨ ¬ 𝜏)) ↔ ¬ if-(𝜑, (¬ 𝜓 ∨ ¬ 𝜒), (¬ 𝜃 ∨ ¬ 𝜏)))
12 anor 511 . . 3 ((if-(𝜑, 𝜓, 𝜃) ∧ if-(𝜑, 𝜒, 𝜏)) ↔ ¬ (¬ if-(𝜑, 𝜓, 𝜃) ∨ ¬ if-(𝜑, 𝜒, 𝜏)))
1310, 11, 123bitr4i 292 . 2 (if-(𝜑, ¬ (¬ 𝜓 ∨ ¬ 𝜒), ¬ (¬ 𝜃 ∨ ¬ 𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ∧ if-(𝜑, 𝜒, 𝜏)))
144, 13bitri 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  wo 382  wa 383  if-wif 1050
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 384  df-an 385  df-ifp 1051
This theorem is referenced by:  ifpnannanb  38373
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