Prove using a proof sequence and justify each step Prove using a proof sequence that the argument is valid [ A --> (B ∨ C) ] ∧ B' ∧ C' --> A'
I'm having some trouble figuring the proof out here.  Here is what I have so far.  Is this on the right track?


*

*A-->(B ∨ C)     (Given)

*B’               (Given)

*C’               (Given) ?

*B’ ∧ C’          unsure here

*(B∨C)’           DeMorgan's Law? using 2 and 3.

*A’               Contradiction using 1 and 5


Prove using a proof sequence that the argument is valid (hint: the last A’ has to be
inferred). Justify each step. 
(A-->C) ∧ (C-->B’) ∧ B-->A’
This one I'm not really sure where to go with it, any help would be appreciated.


*

*(A-->C)

*(C-->B’)

*B

*A’


Edit: Apologies there were some typos in the original question, ' is used for negation in this case.

This proof is similar to what the OP and Duncan presented. I am using a proof checker to make sure I follow the rules:

To enter a statement into this proof checker I have to use parentheses if there is any ambiguity, so line 1 is the premise with some added parentheses.
On lines 2 and 3, I used conjunction elimination (∧E) to derive the two conjuncts from line 1.
I used De Morgan's laws (DeM) referencing line 3 to derive line 4.  
Finally, I used Modus Tollens (MT) referencing lines 2 and 4 to derive $\lnot A$ on line 5.

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