布尔表达式和正则表达式_简化布尔表达式的实例

布尔表达式和正则表达式

Example 1: Simplify the given Boolean Expression to minimum no. of variables or literals.

示例1：将给定的布尔表达式简化为最小编号。 变量或文字。

1. (A+B). (A+B)

   (A + B)。 (A + B )

2. ABC + AB + ABC

   ABC + A B + AB C

Answer:

回答：

1) (A+B). (A+B)

1)(A + B)。 (A + B )

    = A.A + A. B + B.A + B.B
    = A + A. B + B.A + 0
    = A (1 + B + B) = A

2) ABC + AB + ABC

2)ABC + A B + AB C

    = ABC + ABC + AB
    = AB (C + C) +  AB
    = AB +  AB
    = B (A + A) = B. 1 = B

Example 2: Prove that: (A + C) (A.B +C) = 0

示例2：证明：( A + C)( A .B + C )= 0

Solution:

解：

    LHS     = A.(A + C) (A.B + C)
            = (A. A + AC) (A.B + C)
            = AC (A.B + C)		[Since, A.A = 0]
            = AC. A.B + AC. C
            = 0 = RHS			[Since, A. A = 0 and C. C = 0]
    Hence, our result is proved.

Example 3: Reduce the expression (B +BC) (B + BC) (B + D)

示例3：减少表达式(B + BC)(B + B C)(B + D)

Solution:

解：

    = (B +BC) (B + BC) (B + D)
    = (BB +BBC) (B + D)
    = (B +0) (B +D)		[Since, B.B =0]
    = BB +BD			[Since, B.B =1]
    = B + BD
    = B (1+D) =B 		[Since, 1+D =1].

Example 4: Reduce the expression

示例4：简化表达式

Solution:

解：

    = A.BC. (AB +ABC)		        [De-Morgan's Law]
    = A.BC (AB+ABC)
    = A.BC. AB + A.BC. ABC
    = A.A.B.B.C + A.A.B.B.C.C		[Manipulation]
    = 0 + 0 = 0

Example 4: Prove the Associative Law for XOR operation.

示例4：证明XOR运算的关联律。

Solution:

解：

We know that XOR operation is given as: A⊕B = AB + AB

我们知道XOR运算为： A⊕B = A B + A B

To prove associativity for XOR operation we are required to prove:

为了证明XOR运算的关联性，我们需要证明：

(A⊕B) ⊕C =A⊕(B⊕C). Therefore,

( A⊕B ) ⊕C = A⊕ ( B⊕C )。 因此，

From equation 1 and 2 we can see LHS = RHS, thus associative law holds true for XOR operation.

从等式1和2可以看出LHS = RHS ，因此对于XOR运算 ，关联律成立。

翻译自: https://www.includehelp.com/basics/solved-examples-on-reduction-of-boolean-expression.aspx

布尔表达式和正则表达式