Informal Proof for tautology I proved the following tautology formally using a truth table:
$$((A \to B) \land (A \to C)) \to (A \to (B \land C)).$$

What would an Informal proof (non-symbolic) in the English Language to show this is a tautology look like ? 
 A: General comment: If I had a pound for every exam answer I've marked over the years where someone has "proved" that $X$ is/isn't a tautology or is/isn't valid, when a moment's informal argumentation would have shown them that their answer can't be right, then ... well, I'd not be rich, but at least could have bought a few more drinks along the way! 
So, yes, faced with a claim that so-and-so is or isn't a tautology (or that such-and-such is or isn't a tautologically valid inference), it is often a very good instinct to ask yourself "what's an informal argument for this?" 

Specific response: In this case something as informal as the following is enough to make it clear that the proposition given should indeed come out as  a tautology:

Suppose we get B if A is true, and also get C if A is true; then if A
  is true we get both B and C.

Of course, this sort of very informal reasoning can only take us so far, and can sometimes lead us astray. But that's why we have formal logical systems, to regiment and systematize argumentation! 
A: Well, I am not sure if this is what you are looking for but here is another way to express it:
If we use the fact that $X \implies Y = \lnot X \lor Y$ here, we have:
$$((A \implies B) \land (A \implies C)) \implies (A \implies (B \land C))$$ $$= ((\lnot A \lor B) \land (\lnot A \lor C)) \implies (\lnot A \lor (B \land C))$$ $$= ((\lnot A \lor B) \land (\lnot A \lor C)) \implies ((\lnot A \lor B) \land (\lnot A \lor C))$$
which is $1$ because the expression before the $\implies$ is same as the expression after $\implies$. So we can write it as $D \implies D = 1$ where $D =(\lnot A \lor B) \land (\lnot A \lor C) $.
A: If $A$ implies $B$, and $A$ implies $C$, then $A$ implies both $B$ and $C$.
