# 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:

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

2. $$\text{If A, then D}$$

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. 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.

• Is there any way to show part a in a truth table? Commented May 12, 2015 at 18:39
• Yes there is. It is very simple from what I have shown. But I will leave that to you.
– user174622
Commented May 12, 2015 at 18:44

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.