Relations between statements involving universal quantifier, conditional and biconditional If we consider two predicates:

$b(x)$: x is a boy

$c(x)$: x is clever

Then, there are four statements involving $∀, b(x), c(x), →$ and  $↔$ . These are below along with my interpretation of their meaning. Correct me if I am wrong in any of these interpretations.

(a) $∀x(b(x)→c(x))$ - All boys are clever

(b) $∀x(b(x))→∀x(c(x))$ - If all are boys, then all are clever

(c) $∀x(b(x)↔c(x))$ - All boys are clever &amp; all those who are clever are boys

(d) $∀x(b(x))↔∀x(c(x))$ - If all are boys, then all are clever &amp; if all are clever then all are boys

I am trying to determine all conditional $(→,↛)$ / bi-conditional $(↔)$ relations between each two of above four. Between any two of these, if there is $↔$, then there is only one relation. However, if there is conditional relation in one direction, then conditional relation in other direction will be invalid. That is there can be $(→,↛)$. So at max there will be $4C_2\times 2=12$ such relations.

Now to understand these relations, I prepared below table. I considered that there are only two person in the universe $x_1,x_2$. $\{(b,c),(¬b,¬c)\}$ means $x_1$ is boy and is clever and $x_2$ is not a boy and is not clever. 
The truth values (in table) in boldface (all other than vi, viii and xiv) are one which I feel are correct, though I am not sure. The non bold-faced truth values (vi, viii and xiv) are ones which I am unsure. So, I am more likely make mistake with them.



Based on the truth values in above table, I prepared following relations:



Q. Are relations 1 to 12 correct?

Q. Also is there some fault in the example predicates $b(x)$ and $c(x)$ themselves?

Actually, I must doing similar for existential quantifier. But afterwards. But am I correct with all this thinking or am just over thinking? Or am severly screwed with my logics?
Reasoning based on examples is fine for disproving implications, as you have done. In order to prove that an implication holds, however, you need something more elaborate. One possibility, as mentioned in the comments, is to use resolution to prove the implications. For example, suppose you want to show that (a) implies (b), i.e. that

$\varphi:=(\forall x.(b(c)\to c(x)))\to((\forall x.b(x))\to(\forall x.c(x)))$

is valid. We will do this by deriving a contradiction from $\neg\varphi$, which (using standard equivalences) is equivalent to

$\forall x.(\neg b(x)\vee c(x)))\wedge\forall x.b(x)\wedge\exists x.\neg c(x)$.

Skolemization results in 

$(\neg b(x)\vee c(x))\wedge b(x)\wedge\neg c(a)$,

from which resolution derives first $\neg b(a)$ and then an empty clause, i.e. false. Therefore the original implication is valid.

As for your table, note that the example I gave also shows that (d) does not imply (a); the remaining 5 implications look ok.