Metamath Proof Explorer


Theorem neanior

Description: A De Morgan's law for inequality. (Contributed by NM, 18-May-2007)

Ref Expression
Assertion neanior ( ( 𝐴 ≠ 𝐵 ∧ 𝐶 ≠ 𝐷 ) ↔ ¬ ( 𝐴 = 𝐵 ∨ 𝐶 = 𝐷 ) )

Proof

Step Hyp Ref Expression
1 df-ne ⊢ ( 𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵 )
2 df-ne ⊢ ( 𝐶 ≠ 𝐷 ↔ ¬ 𝐶 = 𝐷 )
3 1 2 anbi12i ⊢ ( ( 𝐴 ≠ 𝐵 ∧ 𝐶 ≠ 𝐷 ) ↔ ( ¬ 𝐴 = 𝐵 ∧ ¬ 𝐶 = 𝐷 ) )
4 pm4.56 ⊢ ( ( ¬ 𝐴 = 𝐵 ∧ ¬ 𝐶 = 𝐷 ) ↔ ¬ ( 𝐴 = 𝐵 ∨ 𝐶 = 𝐷 ) )
5 3 4 bitri ⊢ ( ( 𝐴 ≠ 𝐵 ∧ 𝐶 ≠ 𝐷 ) ↔ ¬ ( 𝐴 = 𝐵 ∨ 𝐶 = 𝐷 ) )