逻辑运算

基本运算

逻辑运算	含义	逻辑表达式	运算符
与	`A AND B`	$`F = A \cdot B`$	`a & b`
或	`A OR B`	$`F = A + B`$	`a \| b`
非	`NOT A`	$`F = \bar{A}`$	`!a`
异或	`A XOR B`	$`F = \bar{A}B + A\bar{B} = A \oplus B`$	`a ^ b`
同或	`A XNOR B`	$`F = AB + \bar{A}\bar{B} = A \odot B =\overline{A\oplus B}`$

基本公式

运算法则	或	与
交换律	$`A+B=B+A`$	$`AB=BA`$
结合律	$`A+(B+C)=(A+B)+C`$	$`A(BC)=(AB)C`$
分配律	$`A(B+C)=AB+AC`$	$`A+(BC)=(A+B)(A+C)`$
吸收律	$`A+AB=A`$	$`A(A+B)=A`$
`0-1` 律	$`A+1=1,A+0=A`$	$`A\cdot1=A,A\cdot0=0`$
互补律	$`A+\bar{A}=1`$	$`A\cdot\bar{A}=0`$
重叠律	$`A+A=A`$	$`AA=A`$
反演律	$`\overline{A+B}=\bar{A}\cdot\bar{B}`$	$`\overline{AB}=\bar{A}+\bar{B}`$
包含律	$`AB+\bar{A}C+BC=AB+\bar{A}C`$	$`(A+B)(\bar{A}+C)(B+C)=(A+B)(\bar{A}+C)`$

分配律

与运算对异或运算的分配律

$$ \begin{aligned} (AB) \oplus (AC) &= \overline{AB}AC + AB\overline{AC} \\ &=(\bar{A}+\bar{B})AC + AB(\bar{A}+\bar{C}) \\ &=A\bar{B}C+AB\bar{C}\\ &=A(\bar{B}C+B\bar{C}) \\ &=A(B \oplus C) \end{aligned} $$

或运算对异或运算的分配律 (不满足)

$$ \begin{aligned} (A+B)\oplus(A+C) &= \overline{A+B}(A+C) + (A+B)\overline{A+C} \\ &=\bar{A}\bar{B}(A+C) + (A+B)\bar{A}\bar{C} \\ &=\bar{A}\bar{B}C+\bar{A}B\bar{C} \\ &=\bar{A}(\bar{B}C+B\bar{C}) \\ &=\bar{A}(B \oplus C) \neq A + (B \oplus C) \end{aligned} $$

$$ \begin{aligned} \overline{\bar{A}(B \oplus C)} &= A + \overline{B \oplus C}=A+ B\odot C \end{aligned} $$

真值表

$$ \begin{array}{|c|cccccccc|} \hline & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline A & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ B & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ C & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ \hline \bar{A} & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ \bar{B} & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ \bar{C} & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 \\ \hline AB & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ AC & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ BC & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ \hline A + B & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ A + C & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \\ B + C & 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 \\ \hline A\oplus B & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 \\ A\oplus C & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 \\ B\oplus C & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 \\ \hline A(B+C) & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\ AB+AC & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 \\ \hline A(B\oplus C) & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ (AB)\oplus(AC) & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ \hline A+(BC) & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\ (A+B)(A+C) & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\ \hline A+(B\oplus C) & 0 & 1 & 1 & 0 & 1 & 1 & 1 & 1 \\ (A+B)\oplus(A+C) & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ \hline A\oplus(BC) & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 \\ (A\oplus B)(A\oplus C) & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ \hline A\oplus(B+C) & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ (A\oplus B)+(A\oplus C) & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ \hline \end{array} $$

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p = [[0] * 8 for _ in range(3)]
for i in range(8):
    for j in range(3):
        p[2 - j][i] = (i >> j) & 1

A, B, C = p


def S(A):
    return " & ".join(map(str, A))


def Not(A):
    return list(map(lambda x: x ^ 1, A))


def And(A, B):
    return list(map(lambda x: x[0] & x[1], zip(A, B)))


def Or(A, B):
    return list(map(lambda x: x[0] | x[1], zip(A, B)))


def Xor(A, B):
    return list(map(lambda x: x[0] ^ x[1], zip(A, B)))


s = f"""
$$
\\begin{{array}}{{|c|cccccccc|}}
\\hline
  & {S(range(8))} \\\\
\\hline
A & {S(A)} \\\\
B & {S(B)} \\\\
C & {S(C)} \\\\
\\hline
\\bar{{A}} & {S(Not(A))} \\\\
\\bar{{B}} & {S(Not(B))} \\\\
\\bar{{C}} & {S(Not(C))} \\\\
\\hline
AB & {S(And(A, B))} \\\\
AC & {S(And(A, C))} \\\\
BC & {S(And(B, C))} \\\\
\\hline
A + B & {S(Or(A, B))} \\\\
A + C & {S(Or(A, C))} \\\\
B + C & {S(Or(B, C))} \\\\
\\hline
A\\oplus B & {S(Xor(A, B))} \\\\
A\\oplus C & {S(Xor(A, C))} \\\\
B\\oplus C & {S(Xor(B, C))} \\\\
\\hline
A(B+C) & {S(And(A, Or(B, C)))} \\\\
AB+AC & {S(Or(And(A, B), And(A, C)))} \\\\
\\hline
A(B\\oplus C) & {S(And(A, Xor(B, C)))} \\\\
(AB)\\oplus(AC) & {S(Xor(And(A, B), And(A, C)))} \\\\
\\hline
A+(BC) & {S(Or(A, And(B, C)))} \\\\
(A+B)(A+C) & {S(And(Or(A, B), Or(A, C)))} \\\\
\\hline
A+(B\\oplus C) & {S(Or(A, Xor(B, C)))} \\\\
(A+B)\\oplus(A+C) & {S(Xor(Or(A, B), Or(A, C)))} \\\\
\\hline
A\\oplus(BC) & {S(Xor(A, And(B, C)))} \\\\
(A\\oplus B)(A\\oplus C) & {S(And(Xor(A, B), Xor(A, C)))} \\\\
\\hline
A\\oplus(B+C) & {S(Xor(A, Or(B, C)))} \\\\
(A\\oplus B)+(A\\oplus C) & {S(Or(Xor(A, B), Xor(B, C)))} \\\\
\\hline
\\end{{array}}
$$
"""

print(s)