Theorem df3nandALT2 36401

Description: The double nand expressed in terms of negation and and not. (Contributed by Anthony Hart, 13-Sep-2011.)

Assertion

Ref	Expression
df3nandALT2	⊢ ((𝜑 ⊼ 𝜓 ⊼ 𝜒) ↔ ¬ (𝜑 ∧ 𝜓 ∧ 𝜒))

Proof of Theorem df3nandALT2

Step	Hyp	Ref	Expression
1		df-3nand 36399	. 2 ⊢ ((𝜑 ⊼ 𝜓 ⊼ 𝜒) ↔ (𝜑 → (𝜓 → ¬ 𝜒)))
2		imnan 399	. . 3 ⊢ ((𝜓 → ¬ 𝜒) ↔ ¬ (𝜓 ∧ 𝜒))
3	2	imbi2i 336	. 2 ⊢ ((𝜑 → (𝜓 → ¬ 𝜒)) ↔ (𝜑 → ¬ (𝜓 ∧ 𝜒)))
4		imnan 399	. . 3 ⊢ ((𝜑 → ¬ (𝜓 ∧ 𝜒)) ↔ ¬ (𝜑 ∧ (𝜓 ∧ 𝜒)))
5		3anass 1095	. . 3 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ (𝜑 ∧ (𝜓 ∧ 𝜒)))
6	4, 5	xchbinxr 335	. 2 ⊢ ((𝜑 → ¬ (𝜓 ∧ 𝜒)) ↔ ¬ (𝜑 ∧ 𝜓 ∧ 𝜒))
7	1, 3, 6	3bitri 297	1 ⊢ ((𝜑 ⊼ 𝜓 ⊼ 𝜒) ↔ ¬ (𝜑 ∧ 𝜓 ∧ 𝜒))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∧ wa 395   ∧ w3a 1087   ⊼ w3nand 36398

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 207  df-an 396  df-3an 1089  df-3nand 36399

This theorem is referenced by: (None)