A⊕B=(A+B)¯AB=1
=1¯¯A¯AB+¯¯B¯AB=21
=21¯¯A¯AB · ¯B¯AB=22
=1=(A+B)¯¯¯A · ¯¯B=(A+B)¯¯¯A+¯B=(A+B)(¯A+¯B)=31
=31¯¯A+B · ¯¯¯A+¯B=¯¯A+B + ¯¯A+¯B=¯¯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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