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Aren't A,C, and D transitive? Also, I thought E, F, and B also transitive with other elements due to vacuous truth? Thanks!

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    It is not transitive for the reason that there is no shortcut to go from $A$ to $D$ (*notice the direction on the arrow*). You have $A\sim C$ and $C\sim D$ but you do not have $A\sim D$. – 2017-02-22
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    To be formal, elements of the set can't be transitive; it's the relation that's transitive. You can say that $A, B$ and $C$ satisfy transitivity for that relation. – 2017-02-22

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No, because of the fact that we are given $A\to C$ and $C\to D$, but $A\not\to D$, the relation is not transitive. If it were a transitive relation, we would need an arrow from $A$ to $D$.

I think you are confused about transitivity: it is a property of a relation, not of particular elements in the set on which the relation is defined.

So it makes no sense to describe elements that are transitive, etc. That is, the relation on the set $\{A, B, C, D, E, F\}$ can be summarized as $R = \{(a, c), (c, d), (d, a)\}.$

So we see $a\sim c$ and $c \sim d$ but not $a \sim d$.

A closer look at the set shows us, also, that we have $(c, d), (d, a) \in R$, but $(c, a) \notin R$, and we therefore can conclude the relation is not transitive, even if we overlooked the other example showing that transitivity fails in this relation.