Theorem ne3anior 3050

Description: A De Morgan's law for inequality. (Contributed by NM, 30-Sep-2013.)

Assertion

Ref	Expression
ne3anior	⊢ ((𝐴 ≠ 𝐵 ∧ 𝐶 ≠ 𝐷 ∧ 𝐸 ≠ 𝐹) ↔ ¬ (𝐴 = 𝐵 ∨ 𝐶 = 𝐷 ∨ 𝐸 = 𝐹))

Proof of Theorem ne3anior

Step	Hyp	Ref	Expression
1		3anor 1125	. 2 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐶 ≠ 𝐷 ∧ 𝐸 ≠ 𝐹) ↔ ¬ (¬ 𝐴 ≠ 𝐵 ∨ ¬ 𝐶 ≠ 𝐷 ∨ ¬ 𝐸 ≠ 𝐹))
2		nne 2960	. . 3 ⊢ (¬ 𝐴 ≠ 𝐵 ↔ 𝐴 = 𝐵)
3		nne 2960	. . 3 ⊢ (¬ 𝐶 ≠ 𝐷 ↔ 𝐶 = 𝐷)
4		nne 2960	. . 3 ⊢ (¬ 𝐸 ≠ 𝐹 ↔ 𝐸 = 𝐹)
5	2, 3, 4	3orbi123i 1174	. 2 ⊢ ((¬ 𝐴 ≠ 𝐵 ∨ ¬ 𝐶 ≠ 𝐷 ∨ ¬ 𝐸 ≠ 𝐹) ↔ (𝐴 = 𝐵 ∨ 𝐶 = 𝐷 ∨ 𝐸 = 𝐹))
6	1, 5	xchbinx 337	1 ⊢ ((𝐴 ≠ 𝐵 ∧ 𝐶 ≠ 𝐷 ∧ 𝐸 ≠ 𝐹) ↔ ¬ (𝐴 = 𝐵 ∨ 𝐶 = 𝐷 ∨ 𝐸 = 𝐹))

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ∨ w3o 1102   ∧ w3a 1103   = wceq 1570   ≠ wne 2956

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 402  df-or 862  df-3or 1104  df-3an 1105  df-ne 2957

This theorem is used by:  eldiftp  4648