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Theorem ifpdfnan 38148
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 1488 . 2 ((𝜑𝜓) ↔ ¬ (𝜑𝜓))
2 ifpdfan 38127 . . 3 ((𝜑𝜓) ↔ if-(𝜑, 𝜓, ⊥))
32notbii 309 . 2 (¬ (𝜑𝜓) ↔ ¬ if-(𝜑, 𝜓, ⊥))
4 ifpnot23 38140 . . 3 (¬ if-(𝜑, 𝜓, ⊥) ↔ if-(𝜑, ¬ 𝜓, ¬ ⊥))
5 notfal 1559 . . . 4 (¬ ⊥ ↔ ⊤)
6 ifpbi3 38129 . . . 4 ((¬ ⊥ ↔ ⊤) → (if-(𝜑, ¬ 𝜓, ¬ ⊥) ↔ if-(𝜑, ¬ 𝜓, ⊤)))
75, 6ax-mp 5 . . 3 (if-(𝜑, ¬ 𝜓, ¬ ⊥) ↔ if-(𝜑, ¬ 𝜓, ⊤))
84, 7bitri 264 . 2 (¬ if-(𝜑, 𝜓, ⊥) ↔ if-(𝜑, ¬ 𝜓, ⊤))
91, 3, 83bitri 286 1 ((𝜑𝜓) ↔ if-(𝜑, ¬ 𝜓, ⊤))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 196  wa 383  if-wif 1032  wnan 1487  wtru 1524  wfal 1528
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 1033  df-nan 1488  df-tru 1526  df-fal 1529
This theorem is referenced by: (None)
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