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.
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$ \begin{array}{rcl} \ds{\color{#f00}{\texttt{false}} \lor \pars{\color{#f00}{\texttt{false}} \land y}} & \ds{=} & \ds{\color{#f00}{\texttt{false}} \lor \texttt{false} = \color{#f00}{\texttt{false}}} \\ \ds{\color{#f00}{\texttt{true}} \lor \pars{\color{#f00}{\texttt{true}} \land y}} & \ds{=} & \ds{\color{#f00}{\texttt{true}}}\quad \mbox{because the left argument is}\ \texttt{true}. \end{array} $$
$$\bbox[15px,#ffd]{% \begin{array}{c|c|c} x / y & \texttt{false} & \texttt{true}\\ \hline \texttt{false} & \texttt{false} & \texttt{false}\\ \hline \texttt{true} & \texttt{true} & \texttt{true}\\ \hline \end{array}} $$