Trying to justify each step correctly in proof sequence I am trying Justify each step in the proof sequence below for correctly with [A → (B ∨ C)] ∧ B' ∧ C' → A'
So I justified my steps here but I am not sure at 1 to 3 if I did it correctly. 


*

*A → (B ∨ C) = Conjunctive simplification

*B′ = Conjunctive Simplification

*C′ = Conjunctive Simplification

*B′∧ C′ = Conjunctive Addition

*(B ∨ C)′ = DeMorgan's Law

*A′ = Contrapositive

 A: Lines 4 to 6 of the proof are correct, but getting there raises some question which has more to do with how the conditional was presented to the OP because there is ambiguity around where the parentheses should go in the antecedent of the conditional. 
Should it be like this?
$$([A \to (B \lor C)] \land \lnot B') \land C' \to A'$$
Or should it be like this?
$$[A → (B ∨ C)] ∧ (B' ∧ C') → A'$$
In neither case would the first three lines of the proof be correct. If one were going to separate out all the conjuncts, one would need another line referencing the simplification rule to derive either $[A \to (B \lor C)] \land \lnot B'$ or $B' ∧ C'$. 
Without the parentheses there is an assumption that there exists an inference rule that permits one, given three conjuncts and two $\land$ connectives, to select one of the conjuncts. Note how Wikipedia describes the conjunction simplification inference rule as given two conjuncts connected by $\land$, one can derive one of them.
This also assumes there are parentheses around those conjunctions separating them from the consequent.
Here is a proof using a Fitch-style proof checker which forces me to follow the inference rules and enter only well-formed formulas:

On lines 2 and 3, I used conjunction elimination (simplification) (∧E); on line 4, De Morgan's laws (DeM); and finally modus tollens (MT) on line 5.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/
