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Theorem andnand1 32704
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 1081 . . 3 ((𝜑𝜓𝜒) ↔ (𝜑 ∧ (𝜓𝜒)))
2 pm4.63 436 . . . 4 (¬ (𝜓 → ¬ 𝜒) ↔ (𝜓𝜒))
32anbi2i 732 . . 3 ((𝜑 ∧ ¬ (𝜓 → ¬ 𝜒)) ↔ (𝜑 ∧ (𝜓𝜒)))
4 annim 440 . . 3 ((𝜑 ∧ ¬ (𝜓 → ¬ 𝜒)) ↔ ¬ (𝜑 → (𝜓 → ¬ 𝜒)))
51, 3, 43bitr2i 288 . 2 ((𝜑𝜓𝜒) ↔ ¬ (𝜑 → (𝜓 → ¬ 𝜒)))
6 df-3nand 32701 . . 3 ((𝜑𝜓𝜒) ↔ (𝜑 → (𝜓 → ¬ 𝜒)))
76notbii 309 . 2 (¬ (𝜑𝜓𝜒) ↔ ¬ (𝜑 → (𝜓 → ¬ 𝜒)))
8 nannot 1602 . 2 (¬ (𝜑𝜓𝜒) ↔ ((𝜑𝜓𝜒) ⊼ (𝜑𝜓𝜒)))
95, 7, 83bitr2i 288 1 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓𝜒) ⊼ (𝜑𝜓𝜒)))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 196  wa 383  w3a 1072  wnan 1596  w3nand 32700
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-an 385  df-3an 1074  df-nan 1597  df-3nand 32701
This theorem is referenced by: (None)
  Copyright terms: Public domain W3C validator