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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!

Git Gud
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Tim
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    Where did you see "$A\Rightarrow B$ is said to be weaker than $C \Rightarrow B$"? – Ben Millwood Jul 05 '12 at 12:57
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    I looked back, but forgot where I saw it. – Tim Jul 05 '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 05 '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 05 '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 05 '12 at 13:13
  • @user1729: I'm sort of taking $C\Rightarrow A$ as if it were an axiom (or at least, an assumption). Notice that the last line of my answer below makes that explicit, because as you say, without that assumption neither is stronger than the other. – Ben Millwood Jul 05 '12 at 13:17

2 Answers2

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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 $C\Rightarrow B$.

Ben Millwood
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You are correct, $A \Rightarrow B$ is stronger than $C \Rightarrow B$ as it is the more direct path.

user1729
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BYS2
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