Prove:
$$\frac{(a \land b) \rightarrow (b \leftrightarrow c)}{\therefore a \rightarrow(b \rightarrow c)}$$
My conclusion:
$1.\space(a \land b) \rightarrow (b \leftrightarrow c) \qquad Premise.$
$\boxed{ 2. \space a \qquad \qquad Assumption. \\ \boxed{3. \space b \qquad \qquad Assumption. \\ 4. \space a \land b \qquad \land-intro\space(2,3)\\ 5. \space b \leftrightarrow c \qquad \rightarrow-elim(1,4) \\ 6. \space c \qquad \qquad \leftrightarrow-elim(3,5)} \\7. \space b \rightarrow c \qquad \rightarrow -intro (3-6)
}$
$8. \space a \rightarrow (b\rightarrow c) \quad \rightarrow-intro(2-7)$
Here I am asking to see if my steps to the conclusion are right. I am not super clear on the "correct" way to go about this problem. I know that I am assuming things to be true and drawing conclusions based on what I have assumed and what comes out with the statements I deduct from.
Some insight on this problem and if someone could validate my work would me very helpful.
Your derivation is correct. The assumtions are perfect, since your conlusion has only (→) you have to use the "practical rule" of I→ that states: "assume the antecedent of the formula you want to derivate and try to derivate the consequent". In this case you had to do it twice since your conclusion is (a →(b → c)). You have to assume (a) and try to derivate (b → c) and in order to get (b → c) you have to assume (b) and try to get (c).