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I am reading a document that says "$p$ is stronger than $q$" is another way of saying "$p$ implies $q$".

I'm trying to understand how that maps to generalizations in mathematics. For example, consider the following. Let $p$ be the statement $ax^2 + bx + c = 0 \text{ for } a,b,c \in \mathbb{R}$, $q$ be the statement $bx+c=0 \text{ for } b,c \in \mathbb{R}$. Then I have $p \Rightarrow q$ if $a=0$. I am then left with the conclusion that $p$ is stronger than $q$ if $a=0$ is true. That last statement strikes me as counterintuitive to the way a generalization of something contains that something as a special case ($p$ is a generalization of $q$, intuitively $p$ contains $q$ as a special case when $a=0$ is true).

Can anyone help me understand this?

amrods
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  • What is it exactly that you've problem with? – Parth Bhagwat Oct 28 '22 at 06:45
  • I would say that $p$ is not stronger than $q$ in general, but if also $a=0$, and than if $a=0$ we have that $p$ is equivalent to $q$ i.e. also $q$ implies $p$. Moreover your statements are not valid in general but only for specific values of $x$, so you have to add more restrictions for them to be always true. hope this helps – l4teLearner Oct 28 '22 at 06:48
  • forget what I wrote after "Moreover...", not really useful. – l4teLearner Oct 28 '22 at 07:12
  • I think I got it. $p \Leftrightarrow (q \wedge (a=0))$, which means $p \Rightarrow (q \wedge (a=0))$. Since the last part is a conjunction then it is stronger than simply $q$. – amrods Oct 28 '22 at 11:33

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