digitalLogic

F = A ⊕ B F = A ⊙ B A ⊕ 0 = A A ⊙ 1 = A A ⊕ 1 = A ˉ A ⊙ 0 = A ˉ A ⊕ A = 0 A ⊙ A = 1 A ⊕ A ˉ = 1 A ⊙ A ˉ = 0 A ⊕ B ˉ = A ⊕ B ‾ = A ⊕ B ⊕ 1 A ⊙ B ˉ = A ⊙ B ‾ = A ⊙ B ⊙ 0 A ⊕ B = B ⊕ A A ⊙ B = B ⊙ A A ⊕ ( B ⊕ C ) = ( A ⊕ B ) ⊕ C A ⊙ ( B ⊙ C ) = ( A ⊙ B ) ⊙ C A ( B ⊕ C ) = A B ⊕ A C A + ( B ⊙ C ) = ( A + B ) ⊙ ( A + C ) \begin{array}{|l|l|} \hline {F=A \oplus B} & {F=A \odot B} \\ \hline A \oplus 0=A & A \odot 1=A \\ \hline A \oplus 1=\bar{A} & A \odot 0=\bar{A} \\ \hline A \oplus A=0 & A \odot A=1 \\ \hline A \oplus \bar{A}=1 & A \odot \bar{A}=0 \\ \hline A \oplus \bar{B}=\overline{A \oplus B}=A \oplus B \oplus 1 & A \odot \bar{B}=\overline{A \odot B}=A \odot B \odot 0 \\ \hline A \oplus B=B \oplus A & A \odot B=B \odot A \\ \hline A \oplus(B \oplus C)=(A \oplus B) \oplus C & A \odot(B \odot C)=(A \odot B) \odot C \\ \hline A(B \oplus C)=A B \oplus A C & A+(B \odot C)=(A+B) \odot(A+C) \\ \hline \end{array} F=A⊕BA⊕0=AA⊕1=AˉA⊕A=0A⊕Aˉ=1A⊕Bˉ=A⊕B​=A⊕B⊕1A⊕B=B⊕AA⊕(B⊕C)=(A⊕B)⊕CA(B⊕C)=AB⊕AC​F=A⊙BA⊙1=AA⊙0=AˉA⊙A=1A⊙Aˉ=0A⊙Bˉ=A⊙B​=A⊙B⊙0A⊙B=B⊙AA⊙(B⊙C)=(A⊙B)⊙CA+(B⊙C)=(A+B)⊙(A+C)​​

调换律: 若 A ⊕ B = C , 则必有 A ⊕ C = B , B ⊕ C = A ; 若 A ⊙ B = C , 则必有 A ⊙ C = B , B ⊙ C = A . 若 {A} \oplus \mathrm{B}=\mathrm{C} , 则必有 {A} \oplus \mathrm{C}= {B}, {B} \oplus \mathrm{C}= {A} ; \\ 若 {A} \odot B=\mathrm{C} , 则必有 {A} \odot \mathrm{C}=\mathrm{B}, {B} \odot \mathrm{C}= {A} . 若A⊕B=C,则必有A⊕C=B,B⊕C=A;若A⊙B=C,则必有A⊙C=B,B⊙C=A.