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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?

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2 Answers 2

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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.

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

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