Well, this argument isn't valid... An argument is invalid when there is a case where all it's premises are true and the conclusion is false. This is because the entire argument can be written as the conjunction of all it's premises implying in the conclusion. So, if the conjunction of the premises is true and the conclusion is false, then the implication is false, and the argument is invalid, as you can see in this reducted truth table:

About your proof, you can't just assume that a variable if true or false if you don't have this data. Actually, you can do this when both true and false values for the variable in question leads you to the same answer, meaning that the answer does not depend on this variable, like the following exemple:

P ⊃ Q
¬P ⊃ Q
∴ Q

In this exemple, both P and ¬P implies in Q, which mean that anything implies in Q. This is equivalent to (P^¬P) ⊃ Q, which is always true. But this is not the case, because if A if true, it's conclusion isn't the same as it would be if A was false. If you choose values for a variable, you will only know what would be the result of that if this were to be really the truth value, but you don't know what it is.
I don't know if it answers your question. In any case, one thing I like to do is always begin by checking if the argument is invalid. If I can't make the premises true and the conclusion false, then I start using inference rules to prove the argument validity.

*Sorry for any grammar error, i haven't been practicing writing in english.

Ferreira is generally correct. The problem with your "proof" is in line 1. There is no rule of replacement or inference that allows the move from (D ->A) ->T to D->(A->T). in fact, "exposition" is not even a term used in prop logic. If you meant "exportation," that is different rule from the one you use. Since Ferreira has provided a value assignment showing invalidity, there should be no way to manipulate the rules of inference to produce a valid argument from the premises you cite.

Consider it on a more intuitive level: "If it is the case that if the moon is blue, then cats bark, then roger is happy" creates much different conditions from "if the moon is blue, then it is the case that if cats bark, roger is happy".

More formally you can see the transformation you use does not preserve truth value under the following value assignments:

A D T | (A→D)→T | A→(D→T)
F F F F T

A rule of inference must preserve truth value regardless of truth assignment, so "exposition" can not be a rule of inference.