Prove using a proof sequence and justify each step

Question

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?

1. A-->(B ∨ C) (Given)
2. B’ (Given)
3. C’ (Given) ?
4. B’ ∧ C’ unsure here
5. (B∨C)’ DeMorgan's Law? using 2 and 3.
6. 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.

1. (A-->C)
2. (C-->B’)
3. B
4. A’

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

Answer 1

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

Answer 2

Assuming you're using prime to denote the negation, and that you meant C' instead of C; in the first line of your post, then your first proof is correct.

[EDIT] As pointed out in the comments below, you only really have one given. That is the left side of the initial logic statement: $[A \rightarrow (B\vee C)] \wedge B' \wedge C'$.

Your initial first three statements (now statements 2 through 4) all derive from this given. By specialization, if $A\wedge B$ is true then $A$ is true (as is $B$).

Working from that, your fourth statement does come from the previous 2 - it's called Conjunction. If B' is true and C' is true, then $B'\wedge C'$ is also true. Your statement 5 is an application of DeMorgan's Law on Statement 4 and Statement 6 is because of the contrapositive rule. The contrapositive rule (also known as Modus Tollens) says that if $A \rightarrow B$ is true, and $B'$ is true, then $A'$ is true.

So to recap:

1. $[A \rightarrow (B\vee C)] \wedge B' \wedge C'$ (Given)
2. $B'$ (Specialization)
3. $C'$ (Specialization)
4. $B' \wedge C'$ (Conjunction)
5. $(B \vee C)'$ (DeMorgan's Law)
6. Therefore $A'$ by Modus Tollens

Your second proof will start the same way. State the given. Use Specialization to get the individual statements out. After that, you'll have to to apply the contrapositive rule twice. I'll post how to do it in spoilers below, but see if you can figure it out on your own.

First application:

Statement 4 should be an application of the contrapositive on statements 2 and 3. Because you know that $C \rightarrow B'$ and $B$, that must mean that $C'$ is true.

Second application:

Now that you know that $C'$ is true, combine that with the first statement and apply the contrapositive to reach your conclusion, $A'$

FYI: Here's a good quick reference for most of the basic logic rules.

http://integral-table.com/downloads/logic.pdf