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.

$\color{gray}{\boxed{\color{black}{
\because\quad \text{Assuming } A 
\\[2ex]
\qquad A ,\; [ (A\to B\vee C)\wedge B' \wedge C' ]
\\ \quad \vdash (\text{ conjunction elimination: } S\wedge T \vdash S, \text{ and } S\wedge T\vdash T)
\\ \qquad A ,\; (A\to B\vee C) ,\; B' ,\; C'
\\ \quad \vdash (\text{ implication elimination: } S\to T, S \vdash T)
\\ \qquad (B\vee C),\; B',\; C'
\\ \quad \vdash (\text{ disjunctive syllogism: } S', S\vee T \vdash T)
\\ \qquad C,\; C'
\\ \quad \vdash (\text{ contradiction: } S', S\vee S' \vdash \bot)
\\ \qquad \bot
\\[2ex]
\therefore \quad (A\to B\vee C)\wedge B' \wedge C' \;\vdash\; A'
}}}$
