I am working with the book of Professor Liebeck 'A Concise Introduction to Pure Mathematics'. Exercise 4, Chapter 1. (Even numbered exercises have neither hints nor solutions.)
($\bar{A}$ denotes the nagation of statement $A$.)
Statement (d) implies (e) and vice versa. I do not see other possibilities. Am I right? Thanks a lot.
Thanks to the useful comments, I post here an answer.
First, the negation of $\bar{A}\Rightarrow B$ is $\bar{B}\Rightarrow A$ and vice versa. That is,
$\bar{A}\Rightarrow B\iff \bar{B}\Rightarrow A$
so that (d) implies (e) and (e) implies (d).
Furthermore, (a) implies (d) for (a) says that either $A$ is true or $B$ is true. Thus, if $A$ is not true (so $\bar{A}$ is true), then $B$ must be true. But, this is exactly statement (d). In a similar manner, (a) implies (e), since if $B$ is not true (so $\bar{B}$ is true), then $A$ must be true. But, this is exactly statement (e).
Since (d) and (e) have been shown to be equivalent, one concludes that (a),(d),(e) are equivalent to one another.
Next, we shall prove that neither (b) nor (c) imply (a), (d) and (e). We shall do it by a counterexample. Let $x$ be some real number. Let $A$ be the statement $x>6$, and let $B$ be the statement $x>2$.
Then (b) is is true, since $A\Rightarrow B$ means that if $x>6$ then $x>2$ which is clearly true.
But, (a) is wrong, since it means that $x>2$ (either $A$ is true or $B$ is true). But all we know about $x$ is that it is a real number.
Similarly, (c) is wrong, since it says that if $x>2$, then $x>6$, which is not true in general.
In addition, (d) is wrong, for it says that if $x\leq 6$, then $x>2$, which is not true in general.
Finally, (e) is wrong, for it says that if $x\leq 2$, then $x>6$, which is absurd.
Therefore, we have demonstrate, via a a counterexample, that (b) does not imply any of the other statements. Working in a similar manner, one can show that (c) does not imply any of the other statements as well.
To sum up, the only implications are that statements (a),(d),(e) are equivalent to one another.