Truth values for 2 implications and whether or not they imply each other Let A, B, C, and D be arbitrary statements.
Consider the following implications:


*

*$\text{If $A$ and $B$, then $C$ or $D$}$


*$\text{If $A$, then $D$}$

Question:

*

*Suppose that (1) is true. Is it possible that A is true but both C and D are false?


*Do either of the statements imply the other?
I'm not quite sure how to start off the solution for this problem. I considered a truth table, but I wasn't sure how A and B would form the truth values for C or D. Can anyone explain to me how to approach this problem?
 A: a. Yes. Since C and D are false, the conclusion of Statement 1 is false. But Statement 1 is true, so it must be the case that the hypothesis of Statement 1 is false. Since A is true yet A and B is false, it must be the case that B is false. So the situation is possible, but only when B is false.
b. Case 1: Statement 1 imply Statement 2: No. Let A and B both be $3 < 4$, let C be $2\ge1$, and let D be $2 < 1$. Statement 1 is true, but Statement 2 If $3 < 4$, then $2 < 1$ is false.
Case 2: Statement 2 imply Statement 1: Yes. Assume Statement 2 is true. To prove that Statement 1 is true, we need to assume that the hypothesis of Statement 1 is true, namely A and B . Therefore, both A and B are true. Since A is true and Statement 1 is true, it follows that D is true. Since D is true, C or D is true. Hence Statement 1 is true.
A: 
Statement 1 If A and B, the C or D
Statement 2 If A, then D

This is: $1: (A\wedge B)\to(C\vee D)\\2: (A\to D)$
By Material Implication Equivalence: $1': (\neg A\vee \neg B\vee C\vee D)\\2': (\neg A\vee D)$

a) Suppose that Statement 1 is true. Is it possible that A is true but both C and D are false?

Yes, since either $C$ or $D$ are true whenever both $A$ and $B$ are true, we can have that possiblity happen when $B$ is false.  $(\top\vee \bot)\to(\bot\wedge\bot) = \top$

b) Do either of the statements imply the other?

As above, statement 1 can be satisfied when $D$ is false and $A$ is true, so it does not imply statement 2, which is that $D$ is true whenever $A$ is.
However, every way to satisfy statement 2 satisfies statement 1.  Statement 2 is satisfied when either A is false or D is true.  If A is false then statement 1 is satisfied, and if D is true then statement 1 is satisfied.  Therefore statement 1 is true whenever statement 2 is true.  That's an implication.
( Alternately we can immediately see that statement 2' implies statement 1' by disjunction introduction. )
$$(A\to D) \vdash (A\wedge B)\to (C\vee D) \\\qquad\Box$$
If $D$ is true whenever $A$ is true, then either $C$ or $D$ is true whenever both $A$ and $B$ are true.
