Subido por leonardo correa loaiza

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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
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