Derive: $(A \implies (B \wedge C)) \models (A\implies B)$ I need help!
I am taking a Math & Truth Course and there are logic and paradox problems on an assignment I don't understand. 
Anyone willing to help me derive the following?

$$
(A \implies (B \wedge C)) \models (A\implies B)
$$

Note:
In the above equation the & sign between B & C is actually an upside down "u" in the problem however I am unable to locate a way to present an upside down "u" which means "and" however I have read that you are to place & sign there on the computer. And the "F" is in place of what appears to be a line with two lines directly from it... 
 A: If $A$ implies both $B$ and $C$, then in particular, $A$ implies $B$.
Thus, in any context where $A\to(B\wedge C)$ is true, then also $A\to B$ is true. So $A\to(B\wedge C)\models A\to B$. 
A: You might write a truth table.
You might also note that if A is false, then (A⟹B) is true also.  Then suppose A true, and (A⟹(B∧C)) true also.  It then follows that (B∧C) is true also.  So, B holds true.  Thus, (A⟹B) holds true additionally.  So, in either case (A⟹(B∧C))⊨(A⟹B)  
A: From your comment you want to show that (A⟹(B∧C))|-(A⟹B) instead of that "(A⟹(B∧C))|=(A⟹B)".  The difference comes as that "|-" means that the left side "yields" or "proves" the right.  If you have "|=", then the left side makes the right side "valid" or "entails" it.  You're using what gets called a "natural deduction" system.
Alright, now note that (A⟹B) consists of a conditional statement.  With conditional statements, a good general method comes as to assume the antecedent (the part on the left side of the "⟹"), derive the consequent (the part on the right), and finally use conditional (⟹) introduction.  So, here you would assume A, and hope to derive B somehow, and then use conditional introduction to get (A⟹B).  You also have (A⟹(B∧C)) and A what can you can you infer from your introduction and elimination rules?  Once you have that, what else can you infer from what you just got along with your rules?  
Does that help?
