I was asked to prove that the following is TRUE by "multiplying it out" and simplifying:
$(A+BC)(A+DE)=A+BCDE$
I am already familiar with the theorem
$A+BC=(A+B)(A+C)$
which would be enough proof to prove this true in standard environments, but I was specifically told to "multiply it out". So I tried and this is how far I got:
$(A+BC)(A+DE)$
$=AA+BCDE+ABC+ADE$
$=A+BCDE+ABC+ADE$
$=A(BC+DE)+BCDE$
How does that look so far? Am I headed in the right direction, or do I need to take a different route for simplification?
You need to take a different route at your second-to-last line: $$A+BCDE+ABC+ADE$$ $$=A(BC+DE+1)+BCDE$$ $$=A+BCDE$$