Theorem andnand1 36940

Description: Double and in terms of double nand. (Contributed by Anthony Hart, 2-Sep-2011.)

Assertion

Ref	Expression
andnand1	⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ⊼ 𝜓 ⊼ 𝜒) ⊼ (𝜑 ⊼ 𝜓 ⊼ 𝜒)))

Proof of Theorem andnand1

Step	Hyp	Ref	Expression
1		3anass 1110	. . 3 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ (𝜑 ∧ (𝜓 ∧ 𝜒)))
2		pm4.63 402	. . . 4 ⊢ (¬ (𝜓 → ¬ 𝜒) ↔ (𝜓 ∧ 𝜒))
3	2	anbi2i 634	. . 3 ⊢ ((𝜑 ∧ ¬ (𝜓 → ¬ 𝜒)) ↔ (𝜑 ∧ (𝜓 ∧ 𝜒)))
4		annim 408	. . 3 ⊢ ((𝜑 ∧ ¬ (𝜓 → ¬ 𝜒)) ↔ ¬ (𝜑 → (𝜓 → ¬ 𝜒)))
5	1, 3, 4	3bitr2i 302	. 2 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ¬ (𝜑 → (𝜓 → ¬ 𝜒)))
6		df-3nand 36937	. . 3 ⊢ ((𝜑 ⊼ 𝜓 ⊼ 𝜒) ↔ (𝜑 → (𝜓 → ¬ 𝜒)))
7	6	notbii 323	. 2 ⊢ (¬ (𝜑 ⊼ 𝜓 ⊼ 𝜒) ↔ ¬ (𝜑 → (𝜓 → ¬ 𝜒)))
8		nannot 1528	. 2 ⊢ (¬ (𝜑 ⊼ 𝜓 ⊼ 𝜒) ↔ ((𝜑 ⊼ 𝜓 ⊼ 𝜒) ⊼ (𝜑 ⊼ 𝜓 ⊼ 𝜒)))
9	5, 7, 8	3bitr2i 302	1 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ⊼ 𝜓 ⊼ 𝜒) ⊼ (𝜑 ⊼ 𝜓 ⊼ 𝜒)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 400   ∧ w3a 1102   ⊼ wnan 1520   ⊼ w3nand 36936

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 401  df-3an 1104  df-nan 1521  df-3nand 36937

This theorem is used by: (None)