How to prove that $A\bar B+(\bar A + B)C=A\bar B+C$?
I don't even know how to start. The distribution law doesn't help. All other laws are not applicable.
Similarly, how to prove that
$AD+B\bar D+C\bar D+A\bar C+\bar A\bar D=A+\bar D$?
Boolean algebra simplification proof without truth table
boolean-algebra
Related Solutions
Essentially, we manipulate the equation in whatever way possible, trying to make the left-hand side look like the right-hand side. There's no real magic method here, we just keep trying until we succeed.
In the second example, we can do:
$\small \begin{align*} x\cdot y+y\cdot z+x'\cdot z &= x\cdot y + 1 \cdot (y\cdot z)+x'\cdot z & \text{identity for } \cdot \\ &= x\cdot y + (x+x') \cdot (y\cdot z)+x'\cdot z & \text{complementation } x+x'=1\\ &=x\cdot y + x\cdot (y\cdot z)+x'\cdot (y\cdot z)+x'\cdot z & \text{distributivity} \\ &=x\cdot y + (x\cdot y)\cdot z+x'\cdot (y\cdot z)+x'\cdot z & \text{associativity} \\ &=(x\cdot y) \cdot 1 + (x\cdot y)\cdot z+x'\cdot (y\cdot z)+x'\cdot z & \text{identity for } \cdot \\ &=(x\cdot y) \cdot (1 + z)+x'\cdot (y\cdot z)+x'\cdot z & \text{associativity} \\ &=(x\cdot y) \cdot (1 + z)+x'\cdot (y\cdot z)+x'\cdot z & \text{distributivity} \\ &=(x\cdot y) \cdot 1+x'\cdot (y\cdot z)+x'\cdot z & \text{annihilator for } + \\ &=x\cdot y+x'\cdot (y\cdot z)+x'\cdot z & \text{identity for } \cdot \\ &=x\cdot y+(x'\cdot z) \cdot y+x'\cdot z & \text{assoc. and comm. of } \cdot \\ &=x\cdot y+(x'\cdot z) \cdot y+(x'\cdot z) \cdot 1 & \text{identity for } \cdot \\ &=x\cdot y+(x'\cdot z) \cdot (y+1) & \text{distributivity} \\ &=x\cdot y+x'\cdot z & \text{annihilator for } +, \\ \end{align*} $
as desired.
$$\begin{array}{ccc|cc} x & y & z & a & b \\ \hline 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0\end{array}$$ $$\implies \begin{cases}a= \overline x \cdot\overline y\cdot \overline z\\ b= \overline x \cdot\overline y\cdot \overline z+\overline x \cdot \overline y \cdot z + x\cdot\overline y \cdot \overline z \\ a+b = \overline x \cdot\overline y\cdot \overline z+\overline x \cdot \overline y \cdot z + x\cdot\overline y \cdot \overline z = b \\ a\cdot b = \overline x \cdot\overline y\cdot \overline z=a\end{cases}$$
Let's try to simplify $b$ a little:
$$\begin{align}b &= \overline x \cdot\overline y\cdot \overline z+\overline x \cdot \overline y \cdot z + x\cdot\overline y \cdot \overline z \\ &= \overline y \cdot (\overline x\cdot \overline z+\overline x \cdot z + x \cdot \overline z) \\ &= \overline y \cdot [\overline x\cdot (\overline z + z) + x \cdot \overline z] \\ &= \overline y \cdot (\overline x + x\cdot \overline z) \\ &= \overline y \cdot (\overline x\cdot (\overline z+1) + x\cdot \overline z) \\ &= \overline y \cdot (\overline x\cdot \overline z+\overline x + x\cdot \overline z) \\ &= \overline y \cdot ((\overline x +x)\cdot \overline z+\overline x) \\ &= \overline y \cdot (\overline z+\overline x) \\ &= \overline y \cdot \overline {z\cdot x}\end{align}$$
There's probably a slicker way of reaching that result, but it's been a while since I took electronics. But this way works! ;)
Best Answer
The easies way is to compare the truth tables.
\begin{align} A\bar B+(\bar A + B)C&=A\bar B+C \tag{1}\label{1} \\ AD+B\bar D+C\bar D+A\bar C+\bar A\bar D &=A+\bar D \tag{2}\label{2} \end{align}
Things can also be simplified by checking if eqns hold for both values of one variable.
For example, let's check \eqref{1} for $C=0$
\begin{align} A\bar B+(\bar A + B)0&=A\bar B+0 \tag{3}\label{3} \\ A\bar B&=A\bar B \tag{4}\label{4} \end{align}
and $C=1$ separately:
\begin{align} A\bar B+(\bar A + B)1&=A\bar B+1 \tag{5}\label{5} ,\\ A\bar B+(\bar A + B)(A+\bar A)&=1 \tag{6}\label{6} ,\\ A\bar B+A\bar A + AB+\bar A\bar A+\bar AB&=1 \tag{7}\label{7} ,\\ A\bar B+0 + AB+\bar A+\bar AB&=1 \tag{8}\label{8} ,\\ A\bar B+0 + AB+\bar A+\bar AB&=1 \tag{9}\label{9} ,\\ A(\bar B+B)+\bar A(1+B)&=1 \tag{10}\label{10} ,\\ A+\bar A&=1 \tag{11}\label{11} . \end{align}
Similarly, for \eqref{2}, check $A=0$
\begin{align} 0D+B\bar D+C\bar D+0\bar C+1\bar D &=0+\bar D \tag{12}\label{12} ,\\ B\bar D+(C+1)\bar D &=\bar D \tag{13}\label{13} ,\\ B\bar D+\bar D &=\bar D \tag{14}\label{14} ,\\ (B+1)\bar D &=\bar D \tag{15}\label{15} , \end{align}
and $A=1$:
\begin{align} 1D+B\bar D+C\bar D+1\bar C+0\bar D &=1+\bar D \tag{16}\label{16} ,\\ D+B\bar D+C\bar D+\bar C &=1 \tag{17}\label{17} ,\\ D+B\bar D+C\bar D+\bar C(D+\bar D) &=1 \tag{18}\label{18} ,\\ D+B\bar D+C\bar D+\bar CD+\bar C\bar D &=1 \tag{19}\label{19} ,\\ (1+\bar C)D+(B+C+\bar C)\bar D &=1 \tag{20}\label{20} ,\\ D+\bar D &=1 \tag{21}\label{21} . \end{align}