The basic logical operations are not difficult to understand, but I think that memory is a more efficient way to use it, and I am familiar with it.
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logic operation
- 与 A ⋅ 1 = A A \cdot 1=A A⋅1=A A ⋅ 0 = 0 A\cdot0=0 A⋅0=0
- Or A + 0 = AA + 0 = AA+0=A A + 1 = 1 A+1=1 A+1=1
- non-
- And not (and then not) A ⋅ 1 ‾ = A ˉ \overline{A \cdot 1}=\bar{A}A⋅1=Aˉ A ⋅ 0 ‾ = 1 \overline{A \cdot 0}=1 A⋅0=1
- Or not (present or not) A + 0 ‾ = A ˉ \overline{A + 0}=\bar{A}A+0=Aˉ A + 1 ‾ = 0 \overline{A + 1}=0 A+1=0
- XOR A ⊕ 1 = A ˉ A \oplus 1=\bar{A}A⊕1=Aˉ A ⊕ 0 = A A \oplus 0 =A A⊕0=A
- Same OR A ⊕ 0 = A ˉ A \oplus 0=\bar{A}A⊕0=Aˉ A ⊕ 1 = A A \oplus 1 =A A⊕1=A
The essence of XOR is: A ⊕ B = A ˉ B + BA ˉ A \oplus B = \bar AB+B\bar AA⊕B=Aˉ B+BAˉThe
same or the most essential is:A ⊙ B = A ˉ B ˉ + BAA \odot B = \bar{A}\bar{B}+BAA⊙B=AˉBˉ+B A
The following is a deformed formula, you can pass the most basic formula deformation verification
XOR: A ˉ ⊕ B ˉ \bar{A} \oplus \bar{B}Aˉ⊕Bˉ A ˉ ⊙ B \ bar {A} \ odot BAˉ⊙B A ⊙ B ˉ \quad A \odot \bar{B} A⊙Bˉ
Same or: A ˉ ⊕ B \bar{A} \oplus BAˉ⊕B A ⊕ B ˉ \quad A \oplus \bar{B} A⊕Bˉ A ˉ ⊙ B ˉ \ bar {A} \ odot \ bar {B}Aˉ⊙Bˉ
Boolean algebra
The part of Boolean algebra is repeated with discrete mathematics.
Some theorems of Boolean algebra are basically the same as discrete mathematics, so skip it.