简化布尔表达式:X + X'Y'Z
Simplify Boolean Expression: X + X'Y'Z
我知道下面是相等的:
X + X'Y'Z = X + Y'Z
如何使用基本的布尔恒等式化简左边到右边?
提前致谢。
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
证明这个表达式的最快方法是添加一个冗余项来丢弃 X'
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
我知道下面是相等的: X + X'Y'Z = X + Y'Z 如何使用基本的布尔恒等式化简左边到右边? 提前致谢。
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
证明这个表达式的最快方法是添加一个冗余项来丢弃 X'
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