boolean-logic - 简化布尔表达式：X + X'Y'Z

Question

我知道以下是相等的： X + X'Y'Z = X + Y'Z 如何使用基本布尔身份简化左侧到达右侧？提前致谢。

Answer 1

Expression                            Justification
---------------------------------     -------------------------
X + X'Y'Z                             initial expression
(XY'Z + X(Y'Z)') + X'Y'Z              r  = rs + rs'
(XY'Z + XY'Z + X(Y'Z)') + X'Y'Z       r = r + r
(XY'Z + X(Y'Z)' + XY'Z) + X'Y'Z       r + s = s + r
(XY'Z + X(Y'Z)') + (XY'Z + X'Y'Z)     (r + s) + t = r + (s + t)
X(Y'Z + (Y'Z)') + (Y'Z)(X + X')       rs + rt = r(s + t)
X(1) + (Y'Z)(1)                       r + r' = 1
X + Y'Z                               r(1) = r

Answer 2

证明这个表达式的最快方法是添加一个将丢弃 X' 的冗余项

X + X'Y'Z = X(1+Y'Z) + X'Y'Z
          = X + XY'Z + X'Y'Z
          = X + (X+X')Y'Z
          = X + Y'Z