proposition logic problem below is proposition 
"When A holds,  if B holds C holds, if B does not hold C also does not holds.  Regardless of B,C  if A does not hold, then D does not hold. "
Write the above sentence with a propositional formula. 
$A \implies (B \implies  C)$
$A \implies ( \sim B \implies  \sim C)$
$\sim A \implies  \sim D$
Suppose that the sentence with proposition above is holds.  At this time, determine the truth of "If both C and D hold, then B holds."
is this mean i have to find : 
$(A \implies  (B \implies C) \land (A \implies  ( \sim B \implies  \sim C)) \land (\sim A \implies  \sim D)) \implies  ((C \land D) \implies  B)$
is tautology or not?
I draw the table , but there is false so it means the proposition does not holds?
 A: $$C\implies B\equiv \neg C\lor B$$
$$D\implies(C\implies B)\equiv \neg D\lor\neg C\lor B\equiv \neg(C\land D)\lor B\equiv(C\land D)\implies B$$
We have two true sentences in your conjunction. $Q.E.D.$
Edited:
This is the final step following my last statement in comments, but the core of the problem is here:
Looking at $A$ I wrote has to hold necessarily (transitivity of the implication)
@kili asked for an explanation when rewriting the implication using $\lor\;\&\;\land$. In that case, it appears, the conjunction has to be true (read more on CNF & DNF), it can be checked by plugging into all the truth values $1,0$ for each of the subsentences.
A: We have to check whether the premises entail the conclusion, i.e. 

$A ⟹ (B ⟹ C), A ⟹ (\lnot B ⟹ \lnot C), \lnot A ⟹ \lnot D \vDash (C \land D) ⟹ B$

which is the same as checking whether:

$(A ⟹ (B ⟹ C)) \land (A ⟹ (\lnot B ⟹ \lnot C)) \land( \lnot A ⟹ \lnot D) ⟹ ((C \land D) ⟹ B)$ 

is a tautology.
We can avoid the full Truth Table development considering the following approach: assume not, i.e. assume that the conclusion is FALSE and the premises are all TRUE.
In order for the conclusion: $(C \land D) ⟹ B$ to be false, we must have a line in the TT where:

$C$ and $D$ are both TRUE and $B$ is FALSE.

What happens to the premises with this line of the TT ? 
With $D$ TRUE we must have also $A$ TRUE, in order to satisfy the third premise: $\lnot A ⟹ \lnot D$.
Finally, what happens to 1st and 2nd premises with $A$ TRUE ? 
Consider 1st one: $A ⟹ (B ⟹ C)$. We have that $A$ and $B$ are TRUE while $C$ is FALSE. This means: $\top ⟹ (\top ⟹ \bot)$ which is FALSE, and this contradicts our assumption that all premises are TRUE.

Conclusion: the assumption that the formula is not a tautology led us to a contradiction, and thus the formula must be a tautology.

A: You have : 
(1) A --> ( B --> C) 
(2) A --> ( ~B --> ~C) 
(3) ~ A --> ~ D 
The goal is : ( C & D) --> B 
So, let's suppose (C & D) is true, and let's try to derive that B is true. 


*

*If ( C& D) is true, then, in particular, D is true. But, by contraposition, (3) is equivalent to (D --> A) ( once one has applied double negation). From D and ( D -->A) one can derive that A is true ( by modus ponens). 

*Now, since A is true, using (2) , it follows that ( ~B --> ~C) is true. Using contraposition ( and double negation) this is equivalent to ( C --> B) . 

*But from our initial hypothesis ( namely: C&D) it follows that C is true. 

*From (C -->B) and C, it follows that B is true, as desired. 

*The conditional proof rule allows us to say that : (C&D) implies B. 
