implication introduction in logic I read that implication like a=>b can be proof using the following steps :
1) suppose a true.
2) Then deduce b from a.
3) Then you can conclude that a=>b is true.
Actually my real problem is to understand why step 1 and 2 are sufficient to prove that a=>b is true. I mean, how can you prove the truth table of a=>b just using 1 and 2 ?
I know that implication a=>b is actually defined as "not(a) or b". How can steps 1 and 2 can prove that a and b are related like "not(a) or b" ?
 A: What you show by steps 1 and 2 is that it is impossible for $a$ to be true and $b$ to be false. So, you could say that what you show is $\neg (a \land \neg b)$, which by DeMorgan is equivalent to $\neg a \lor b$
A: The rule of implication introduction ($\mathord{\to{}}I$) do not alone exhaust the classical meaning of implication as given by the truth tables. You still need a strictly classical principle like the classical rule of reductio ad absurdum or double negation elimination or the principle of excluded middle or Peirce's law or DeMorgan's law (as in the answer from @Bram28) or something of the kind (I am here supposing that you are not allowed to assume classical logic at the outset, otherwise the question does not really make sense, since the classical meaning of implication and, therefore, the truth tables for implication, are embedded as background assumptions).
More precisely, from $a\to{b}$ you can prove $\lnot(a\wedge{\lnot{b}})$ in a logic where implication is solely determined by $\mathord{\to{}}I$ (like intuitionistic or minimal logic), but not the other way around. The formulas $a\to{b}$, $\lnot(a\wedge{\lnot{b}})$ and $\lnot{a}\vee{b}$ are only equivalent classically. So, if you begin only with $\mathord{\to{}}I$, you cannot show the validity of the full classical truth table for implication. On the other hand, if you begin with the truth table for implication, you can show the validity of $\mathord{\to{}}I$, and more. Now, if $b$ is shown to follow classically (in a strict sense) from $a$, then, indeed, you can show that the truth table for implication holds between $a$ and $b$.
