Prove $x \lor (x \land y) = x$ without using distributivity How can one prove the absorption law of Boolean algebra without using distributivity?
I found dozens of pages that proved it using distributivity, but how can one do it without? This is what I want to show.
$$x \lor (x \land y) = x$$
I know that you can describe these as
$$x \lor y :=  x+y - xy \\ x \land y := xy \\ \neg x = 1-x$$
But if I try to calculate it out I get something like:
$$x+xy - x^2y$$
No matter what I try, $x^2$ is hindering me from solving this. I can't get this to disappear, sadly. Does anyone have an idea? Thank you.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$$
\begin{array}{rcl}
\ds{\color{#f00}{\texttt{false}} \lor
\pars{\color{#f00}{\texttt{false}} \land y}}
& \ds{=} & \ds{\color{#f00}{\texttt{false}} \lor \texttt{false} =
\color{#f00}{\texttt{false}}}
\\
\ds{\color{#f00}{\texttt{true}} \lor
\pars{\color{#f00}{\texttt{true}} \land y}}
& \ds{=} & 
\ds{\color{#f00}{\texttt{true}}}\quad \mbox{because the left argument is}\ \texttt{true}.
\end{array}
$$

$$\bbox[15px,#ffd]{%
\begin{array}{c|c|c}
x / y & \texttt{false} & \texttt{true}\\
\hline
\texttt{false} & \texttt{false} & \texttt{false}\\
\hline
\texttt{true} & \texttt{true} & \texttt{true}\\
\hline
\end{array}}
$$
