Logic operations - trying to understand basics in forming I have noted the basics I got from the internet on a sheet of paper to learn but I'm not sure if it's correct because the source isn't really trustable in my eyes, so I'll write it down here what I've written on my paper and I hope this is the correct section for it!
I'm especially not sure if last line is correct. About the others, I think it seems logic and makes sense :)
If possible, please make a little personal task for me and I solve it. You say if I did it right. I would really appreciate that because learning everything myself isn't that easy.
$$A \vee \bar{A} = 1$$

$$A \wedge \bar{A} = 0$$

$$A \wedge 1 = A$$

$$A \vee 1 = 1$$

$$(A\wedge A) \vee (\bar{A} \wedge B) = B \wedge (A \vee \bar{A}) = B \wedge 1 = B$$
 A: The last line is not correct. The correct expression beggins with '(A∧B)'; not with '(A∧A)'. The rest is OK.
I wish you luck in your learning
A: You wrote "$(A∧A)∨(\bar A∧B)=B∧(A∨\bar A)$". From where does it come? You can't just pull it out of thin air. Firstly you can simplify "$A \land A$". Also, use the distributivity rule properly, namely:

$P \lor ( Q \land R ) = ( P \lor Q ) \land ( P \lor R )$.

If you don't like this rule, the other distributivity rule actually suffices, namely:

$P \land ( Q \lor R ) = ( P \land Q ) \lor ( P \land R )$.

The proof will be a little harder, but perhaps make more sense:

$A \lor ( \neg A \land B )$
$\ = ( A \land ( \neg B \lor B ) ) \lor ( \neg A \land B )$
$\ = ( A \land \neg B ) \lor ( A \land B ) \lor ( \neg A \land B )$
$\ = ( A \land \neg B ) \lor ( A \land B ) \lor ( A \land B ) \lor ( \neg A \land B )$
$\ = ( A \land ( \neg B \lor B ) ) \lor ( ( A \lor \neg A ) \land B )$
$\ = A \lor B$.

Draw out the venn diagram to see how each step works.
