Metamath Proof Explorer


Theorem ne3anior

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

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

Proof

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