Wednesday, August 5, 2020

The variety of XOR-s


\[ A\oplus B=\left({A+B}\right)\overline{AB}=^1\]
\[ =^1\overline{\overline{A\overline{AB}}}+\overline{\overline{B\overline{AB}}}=^{21} \]
\[ =^{21}\overline{\overline{A\overline{AB}}\ ·\ \overline{B\overline{AB}}}=^{22} \]
\[ =^1=\left({A+B}\right)\overline{\overline{\overline{A}}\ ·\ \overline{\overline{B}}}=\left({A+B}\right)\overline{\overline{\overline{A}+\overline{B}}}=\left({A+B}\right)\left({\overline{A}+\overline{B}}\right)=^{31} \]
\[ =^{31}\overline{\overline{A+B}}\ ·\ \overline{\overline{\overline{A}+\overline{B}}}=\overline{\overline{A+B}\ +\ \overline{\overline{A}+\overline{B}}}=\overline{\overline{A+B}\ +\ AB}=^{32} \]
\[ =^{31}\cancel{A\overline{A}}+A\overline{B}+B\overline{A}+\cancel{B\overline{B}}=A\overline{B}+B\overline{A}=^{41} \]
\[  =^{41}\overline{\overline{A\overline{B}+B\overline{A}}}=\overline{\overline{A\overline{B}}\ ·\ \overline{B\overline{A}}}=^{42}\overline{\left({A+\overline{B}}\right)\left({\overline{A}+B}\right)}= \]
\[ =\overline{\cancel{A\overline{A}}+AB+\overline{A}\ \overline{B}+\cancel{B\overline{B}}}=\overline{AB+\overline{A}·\overline{B}}=^{32} \]
\[ =^{32}\overline{\overline{\overline{\overline{A+B}+A}}\ ·\ \overline{\overline{\overline{A+B}+B}}}=\overline{\overline{A+B}+A}+\overline{\overline{A+B}+B}=^{33} \]

note : since the Boolean Arithmetic has no priority in between the conjunction and the disjunction -- the following applies (the (2 might be used above ... somewhere) ::

\[ \left({A+B}\right)\left({C+D}\right)=AC+AD+BC+BD\qquad^{(1} \]
\[ A·B+C·D=\left({A+C}\right)·\left({A+D}\right)·\left({B+C}\right)·\left({B+D}\right)=_{"PROOF"}\qquad^{(2} \]
\[ =_{"PROOF"}\left({A+AD+CA+CD}\right)\left({B+BD+CB+CD}\right)= \]
\[ _{=AB+ABD_1+ACB_2+ACD_3+ADB_1+ADCB_4+ACD_3+CAB_2+CABD_4+CAB_2+CAD_3+CDB_5+CD=} \]
\[ =\mathbf{AB}_1+\mathbf{AB}D_2+\mathbf{AB}C_3+A\boxed{CD}+\mathbf{AB}\boxed{CD}_4+B\boxed{CD}+\boxed{CD}= \]
\[ =AB\left({1+D+C+CD}\right)+\left({A+AB+B+1}\right)CD=AB+CD\]
also
\[A+AX=A\left({X+\overline{X}}\right)+AX=AX+A\overline{X}=A=A\left({1+X}\right)=A\left({X+\overline{X}+X}\right) \]
etc. ...


the simulation ::



[EoP]

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