Why does A -> A or B

Question

Here's the problem. I have to prove that ((A or B') and (B' or C) -> ((B' or A) and (B' or C)) is always true no matter what A, B or C are. Which is easy, because I know (A -> A) is the same as (A' or A) and since the above statement is equal to (A -> A) (if you changed the long statements into A's) so you can change the above statement into (((A or B') and (B' or C)))' or ((B' or A) and (B' or C)) . And I know that (A' or A) is always true so that means that the above statement is always true.

So why am I here? well, to get to ((A or B') and (B' or C) -> ((B' or A) and (B' or C)) the thing I have to prove is always true, I first have to get (A and (B' or C) -> ((B' or A) and (B' or C)) to be equivalent to that above statement. And the only way to make those to statement equivalent is if you add an B' and an or to the A. Which there just so happens to be a proof that states, A -> A or B.

Long story short, can someone give me the proof that states why A -> A or B?

Answer 1

Here's a copout. If $A$ is true, then no matter what $B$ happens to be, $A \ \vee B$ is true. Why? Because the disjunction is true when one of the component propositions is true.

Answer 2

$A \to (A\vee B)$ is equivalent to $A' \vee (A\vee B)$. By associativity of $\vee$, this is equivalent to $(A' \vee A) \vee B$. This is equivalent to $T \vee B$, which in turn is equivalent to $T$.