/sci/ - Science & Math

The way I realized this btw, assuming I did not fuck up, is rather straightforward but ugly.
If nothing else is stated, assume that [math]a,b,c[/math] are chosen in a way that neither 0 nor 1
appear in the equations implicitly.

First, you consider [math]a+(b\times 1) = a+b[/math], apply distributivity and find after
some substitution that [math]c\times (a+1) =c[/math].
Heck, if one wants to make no assumptions and case studies this expression is good enough.
Next, consider [math](1+1)\times(b+c)[/math]. Applying distributivity yields
[math](1+1)\times(b+c) = b+c+b+c[/math]
Yet, [math](a\times 1)+(b\times 1) = a+b[/math] but
[eqn](a\times 1)+(b\times 1) = a+b \\
a+b= ((a\times 1)+b)\times((a\times 1)+1)\\
a+b= (a+b)\times((a+1)\times(1+1))\\
a+b= ((a+b)\times(a+1))\times(1+1)\\
a+b= (a+b)\times(1+1)=a+b+a+b[/eqn]

I'll reply in a minute that took a while to type.