The variety of XOR-s

\[ A\oplus B=\left({A+B}\right)\overline{AB}=^1\]

\[ =^1\overline{\overline{A\overline{AB}}}+\overline{\overline{B\overline{AB}}}=^{21} \]

\[ =^{21}\overline{\overline{A\overline{AB}}\ ·\ \overline{B\overline{AB}}}=^{22} \]

\[ =^1=\left({A+B}\right)\overline{\overline{\overline{A}}\ ·\ \overline{\overline{B}}}=\left({A+B}\right)\overline{\overline{\overline{A}+\overline{B}}}=\left({A+B}\right)\left({\overline{A}+\overline{B}}\right)=^{31} \]

\[ =^{31}\overline{\overline{A+B}}\ ·\ \overline{\overline{\overline{A}+\overline{B}}}=\overline{\overline{A+B}\ +\ \overline{\overline{A}+\overline{B}}}=\overline{\overline{A+B}\ +\ AB}=^{32} \]

\[ =^{31}\cancel{A\overline{A}}+A\overline{B}+B\overline{A}+\cancel{B\overline{B}}=A\overline{B}+B\overline{A}=^{41} \]

\[  =^{41}\overline{\overline{A\overline{B}+B\overline{A}}}=\overline{\overline{A\overline{B}}\ ·\ \overline{B\overline{A}}}=^{42}\overline{\left({A+\overline{B}}\right)\left({\overline{A}+B}\right)}= \]

\[ =\overline{\cancel{A\overline{A}}+AB+\overline{A}\ \overline{B}+\cancel{B\overline{B}}}=\overline{AB+\overline{A}·\overline{B}}=^{32} \]

\[ =^{32}\overline{\overline{\overline{\overline{A+B}+A}}\ ·\ \overline{\overline{\overline{A+B}+B}}}=\overline{\overline{A+B}+A}+\overline{\overline{A+B}+B}=^{33} \]

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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. ...

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the simulation ::

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[EoP]

Swapping Display Memory on Desk Calculator

v.1
\[\begin{array}{llll}
math.&{key\\presses}&{DISP.\\B}&{MEM\\A}\\
M=M+D & [\ M+\ ] &B&A+B\\
D=-D & [\ ±\ ] &-B&A+B\\
D=D+M & [\ +\ ]\ [\ MR\ ]\ [\ =\ ] &-B+A+B&A+B\\
M=M-D & [\ M-\ ] &A&A+B-A\\
M ↔ D &done&A&B
\end{array}
\]

v.2
\[\begin{array}{llll}
math.&{key\\presses}&{DISP.\\B}&{MEM\\A}\\
D=-D & [\ ±\ ] &-B&A\\
D=D+M & [\ +\ ]\ [\ MR\ ]\ [\ =\ ] &-B+A&A\\
M=M-D & [\ M-\ ] &A-B&A-A+B\\
D="0,=" & [\ 0\ ]\ [\ =\ ] &0(+A)&B\\
M ↔ D &done&A&B
\end{array}
\]

v.3
\[\begin{array}{llll}
math.&{key\\presses}&{DISP.\\B}&{MEM\\A}\\
D=D-M & [\ -\ ]\ [\ MR\ ]\ [\ =\ ] &B-A&A\\
M=M+D & [\ M+\ ] &B-A&A+B-A\\
D=D-M & [\ -\ ]\ [\ MR\ ]\ [\ =\ ] &-A+B-B&B\\
D=-D & [\ ×\ ]\ [\ 1\ ]\ [\ -\ ]\ [\ =\ ]\ \left({\ _{[\ =\ ]}\ }\right)^{\ (1} &(-A)·1-((-A)·1)&B\\
M ↔ D &done&A&B
\end{array}\\
_{NOTE^{\ (1}\ :\ DEPENDING\ ON\ THE\ SPECIFIC\ CALCULATOR\ THE\ ALTERNATE\ SEQUENCE\\ FOR\ THE\ ARITHMETIC\ NEGATION\ [\ ±\ ]\ MAY\ REQUIRE\ DOUBLE\ [\ =\ ]\ PRESS} 
\]

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[Eop]

Resistor Bridge

Simulation / Schematic :

\[U_R=\frac{U_LR_2R_4+U_UR_0R_4+U_DR_0R_2}{R_2R_4+R_0R_4+R_0R_2}=
\frac{U_L\left({R_1R_3+R_0R_3+R_0R_1}\right)-U_UR_0R_3-U_DR_0R_1}{R_1R_3}\]
\[U_L=\frac{U_RR_1R_3+U_UR_0R_3+U_DR_0R_1}{R_1R_3+R_0R_3+R_0R_1}=
\frac{U_R\left({R_2R_4+R_0R_4+R_0R_2}\right)-U_UR_0R_4-U_DR_0R_2}{R_2R_4}\]
\(U_RR_1R_3R_2R_4+U_UR_0R_3R_2R_4+U_DR_0R_1R_2R_4=U_R\sum_E^3\sum_O^3-U_UR_0R_4\sum_O^3-U_DR_0R_2\sum_O^3\)
\[U_R=R_0·\frac{\left({U_UR_3+U_DR_1}\right)R_2R_4+\left({U_UR_4+U_DR_2}\right)\sum_O^3}{\sum_E^3\sum_O^3-R_1R_2R_3R_4}\]
\(U_LR_1R_3R_2R_4+U_UR_0R_4R_1R_3+U_DR_0R_2R_1R_3=U_L\sum_O^3\sum_E^3-U_UR_0R_3\sum_E^3-U_DR_0R_1\sum_E^3\)
\[U_L=R_0·\frac{\left({U_UR_4+U_DR_2}\right)R_1R_3+\left({U_UR_3+U_DR_1}\right)\sum_E^3}{\sum_E^3\sum_O^3-R_1R_2R_3R_4}\]

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[Eop]

3 - Resistor Voltage Divider

\[I_0+I_1+I_2=0\qquad,\qquad\text{! Notice the directions of the currents !}\]
\[{\large \cases{I_0=\frac{V_0-V_X}{R_0}\\ {\ }\\
I_1=\frac{V_1-V_X}{R_1}\\ {\ }\\
I_2=\frac{V_2-V_X}{R_2}}}\]
\[\frac{V_0-V_X}{R_0}+\frac{V_1-V_X}{R_1}+\frac{V_2-V_X}{R_2}=0\]
\[\frac{V_0R_1R_2-V_XR_1R_2+V_1R_0R_2-V_XR_0R_2+V_2R_0R_1-V_XR_0R_1}{R_0R_1R_2}=0\qquad|\qquad×R_0R_1R_2\]
\[\frac{V_0R_1R_2+V_1R_0R_2+V_2R_0R_1}{R_1R_2+R_0R_2+R_0R_1}=V_X\]
Note: the \(V_X\) is the unmarked voltage point on the schematic !

It is easy to see the calculation scheme easily adjusts for N voltages all connecting to the single \(V_X\)
\[V_{X_N}=\frac{\displaystyle{\sum_{i=0}^{N-1}V_i\frac{\displaystyle{\prod_{k=0}^{N-1}R_k}}{R_i}}}{\displaystyle{\sum_{i=0}^{N-1}\frac{\displaystyle{\prod_{k=0}^{N-1}R_k}}{R_i}}}\]
thus BWD chek ::
for N=2 :
\[V_X=\frac{V_1R_0+V_0R_1}{R_0+R_1}\]
incase of the \(V_0=0\) it reduces to a trivial 2 resistor divider formula :
\[V_X=V_1\frac{R_0}{R_0+R_1}\]

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Simulation example/chk. for N=4 :

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[Eop]