Mediante manipulaciones y utilizando los teoremas se comprobó que. ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ ̅̅̅̅ 𝐴 ⋅ 𝐵 + 𝐴 ⋅ 𝐵̅ = 𝐴 ⋅ 𝐵 + 𝐴̅ ⋅ 𝐵̅ Teorema de Morgan 𝐵̅ + 𝐴̅ = 𝐵̅ ⋅ 𝐴̅ Teorema ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ ̅̅̅̅ 𝐴 ⋅ 𝐵 + 𝐴 ⋅ 𝐵̅ = ̅̅̅̅ 𝐴̅𝐵 ⋅ ̅̅̅̅ 𝐴𝐵̅ 𝐵̅ ⋅ 𝐴̅ = 𝐵̅ + 𝐴̅ (Teorema) y 𝐴̅ = 𝐴 ̅̅̅̅ 𝐴̅𝐵 ⋅ ̅̅̅̅ 𝐴𝐵̅ = (𝐴 + 𝐵̅) ⋅ (𝐴̅ + 𝐵) Luego aplicamos distributiva (𝐴 + 𝐵̅) ⋅ (𝐴̅ + 𝐵) = (𝐴 ⋅ 𝐴̅) + (𝐴 ⋅ 𝐵) + (𝐵̅ ⋅ 𝐴̅) + (𝐵̅ ⋅ 𝐵) Sabiendo que (𝐴 ⋅ 𝐴̅) = 0 y (𝐵̅ ⋅ 𝐵) = 0 (𝐴 ⋅ 𝐴̅) + (𝐴 ⋅ 𝐵) + (𝐵̅ ⋅ 𝐴̅) + (𝐵̅ ⋅ 𝐵) = 𝐴 ⋅ 𝐵 + 𝐴̅ ⋅ 𝐵̅