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Theorem andnand1 33757
Description: Double and in terms of double nand. (Contributed by Anthony Hart, 2-Sep-2011.)
Assertion
Ref Expression
andnand1 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓𝜒) ⊼ (𝜑𝜓𝜒)))

Proof of Theorem andnand1
StepHypRef Expression
1 3anass 1092 . . 3 ((𝜑𝜓𝜒) ↔ (𝜑 ∧ (𝜓𝜒)))
2 pm4.63 401 . . . 4 (¬ (𝜓 → ¬ 𝜒) ↔ (𝜓𝜒))
32anbi2i 625 . . 3 ((𝜑 ∧ ¬ (𝜓 → ¬ 𝜒)) ↔ (𝜑 ∧ (𝜓𝜒)))
4 annim 407 . . 3 ((𝜑 ∧ ¬ (𝜓 → ¬ 𝜒)) ↔ ¬ (𝜑 → (𝜓 → ¬ 𝜒)))
51, 3, 43bitr2i 302 . 2 ((𝜑𝜓𝜒) ↔ ¬ (𝜑 → (𝜓 → ¬ 𝜒)))
6 df-3nand 33754 . . 3 ((𝜑𝜓𝜒) ↔ (𝜑 → (𝜓 → ¬ 𝜒)))
76notbii 323 . 2 (¬ (𝜑𝜓𝜒) ↔ ¬ (𝜑 → (𝜓 → ¬ 𝜒)))
8 nannot 1490 . 2 (¬ (𝜑𝜓𝜒) ↔ ((𝜑𝜓𝜒) ⊼ (𝜑𝜓𝜒)))
95, 7, 83bitr2i 302 1 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓𝜒) ⊼ (𝜑𝜓𝜒)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 209  wa 399  w3a 1084  wnan 1482  w3nand 33753
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-3an 1086  df-nan 1483  df-3nand 33754
This theorem is referenced by: (None)
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