# Stronger condition lead to weaker result?

Suppose there are two theorems $A \Rightarrow B$ and $C \Rightarrow A$. Then we have $C \Rightarrow B$.

Now comparing $A \Rightarrow B$ and $C \Rightarrow B$, we know that $C \Rightarrow A$ means C is a stronger condition than A. Is it to say $A \Rightarrow B$ is stronger or weaker than $C \Rightarrow B$? I personally think $A \Rightarrow B$ is a stronger result/theorem than $C \Rightarrow B$, but I also saw $A \Rightarrow B$ is said to be weaker than $C \Rightarrow B$.

Thanks!

• Where did you see "$A\Rightarrow B$ is said to be weaker than $C \Rightarrow B$"? – Ben Millwood Jul 5 '12 at 12:57
• I looked back, but forgot where I saw it. – Tim Jul 5 '12 at 13:02
• Surely $\mathcal{P}$ is only stronger (more general) than $\mathcal{Q}$ if $\mathcal{P}\Rightarrow \mathcal{Q}$. In your example, $A\Rightarrow B$ does not imply $C\Rightarrow B$, and $C\Rightarrow B$ does not imply $A\Rightarrow B$. So...I'd say they were incomparable. Neither is stronger than the other. – user1729 Jul 5 '12 at 13:05
• @user1729: in this case, we already know $C\Rightarrow A$, so one of those does imply the other. – Ben Millwood Jul 5 '12 at 13:08
• @BenMillwood: Is the OP not asking about the raw implications? If we throw away our knowledge that $C\Rightarrow A$ then which is stronger? I mean, ($A\Rightarrow B$ and $C\Rightarrow A$) is stronger than $C\Rightarrow B$, but we need the fact that $C\Rightarrow A$ to make this so... – user1729 Jul 5 '12 at 13:13

Suppose we later find that $D\Rightarrow A$. Then we can use $A\Rightarrow B$ to show $D\Rightarrow B$, but we can't use $C\Rightarrow B$ to do that.
Conversely, if we find that $E\Rightarrow C$, then we can use either $A\Rightarrow B$ or $C\Rightarrow B$ to prove $E\Rightarrow B$, since the latter can be recovered from the former and $C\Rightarrow A$.
So, in the presence of $C\Rightarrow A$, the statement $A\Rightarrow B$ is stronger than $C\Rightarrow B$.
You are correct, $A \Rightarrow B$ is stronger than $C \Rightarrow B$ as it is the more direct path.