布爾表達式和正規表達式_簡化布爾表達式的執行個體

布爾表達式和正規表達式

Example 1: Simplify the given Boolean Expression to minimum no. of variables or literals. 示例1：将給定的布爾表達式簡化為最小編号。 變量或文字。

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：證明：( 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：減少表達式(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：簡化表達式 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：證明XOR運算的關聯律。 Solution: 解：

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

我們知道XOR運算為： A⊕B = A B + A B

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

為了證明XOR運算的關聯性，我們需要證明：

(A⊕B) ⊕C =A⊕(B⊕C). Therefore,

( A⊕B ) ⊕C = A⊕ ( B⊕C )。 是以，

From equation 1 and 2 we can see

LHS = RHS

, thus associative law holds true for

XOR operation

.

從等式1和2可以看出

LHS = RHS

，是以對于

XOR運算

，關聯律成立。

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

布爾表達式和正規表達式