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Theorem ifpdfnan 38510
Description: Define nand as conditional logic operator. (Contributed by RP, 20-Apr-2020.)
Assertion
Ref Expression
ifpdfnan ((𝜑𝜓) ↔ if-(𝜑, ¬ 𝜓, ⊤))

Proof of Theorem ifpdfnan
StepHypRef Expression
1 df-nan 1609 . 2 ((𝜑𝜓) ↔ ¬ (𝜑𝜓))
2 ifpdfan 38489 . . 3 ((𝜑𝜓) ↔ if-(𝜑, 𝜓, ⊥))
32notbii 311 . 2 (¬ (𝜑𝜓) ↔ ¬ if-(𝜑, 𝜓, ⊥))
4 ifpnot23 38502 . . 3 (¬ if-(𝜑, 𝜓, ⊥) ↔ if-(𝜑, ¬ 𝜓, ¬ ⊥))
5 notfal 1681 . . . 4 (¬ ⊥ ↔ ⊤)
6 ifpbi3 38491 . . . 4 ((¬ ⊥ ↔ ⊤) → (if-(𝜑, ¬ 𝜓, ¬ ⊥) ↔ if-(𝜑, ¬ 𝜓, ⊤)))
75, 6ax-mp 5 . . 3 (if-(𝜑, ¬ 𝜓, ¬ ⊥) ↔ if-(𝜑, ¬ 𝜓, ⊤))
84, 7bitri 266 . 2 (¬ if-(𝜑, 𝜓, ⊥) ↔ if-(𝜑, ¬ 𝜓, ⊤))
91, 3, 83bitri 288 1 ((𝜑𝜓) ↔ if-(𝜑, ¬ 𝜓, ⊤))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 197  wa 384  if-wif 1085  wnan 1608  wtru 1653  wfal 1665
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 874  df-ifp 1086  df-nan 1609  df-tru 1656  df-fal 1666
This theorem is referenced by: (None)
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