User:IssaRice/Strength of a mathematical statement

Negation

Negating a strong statement produces a weak statement, and negating a weak statement produces a strong statement. If a statement has strong and weak components, then the flip occurs at each stage. For example, in $\mathrm{\forall }xW(x)$ with $W(x)$ a weak statement, negating it produces $\mathrm{\exists }x\mathrm{\neg }W(x)$, where the strong $\mathrm{\forall }x$ has become the weak $\mathrm{\exists }x$, and the weak $W(x)$ has become a strong $\mathrm{\neg }W(x)$. See Gowers's posts for more discussion on this.

Strong vs subset

A puzzle: why do we say P is stronger than Q if P is a subset of Q, but we also say that a theorem is stronger if it is more general (so bigger)?

• One reply/intuition uses something like possible world semantics, e.g. see Wei Dai's post on Aumann's agreement theorem. There is just one possible world (a single $\omega \in \mathrm{\Omega }$), but our information state is the set of all possible worlds that we cannot distinguish, so the less we know, the more possible worlds we think we could be in.
• One visualization is to use a Venn diagram. The stronger the statement, the more our movement is restricted, as we are forced to be in more and more sets.
• When we say a strong statement like $\mathrm{\forall }xP(x)$, we are saying $P(x_1)\wedge P(x_2)\wedge \cdots \wedge P(x_n)$. When we say a weak statement like $\mathrm{\exists }xP(x)$, we are saying $P(x_1)\vee P(x_2)\vee \cdots \vee P(x_n)$. It seems like in both cases we are accumulating more and more things.
• But if we're working in a proof system, $\mathrm{\forall }xP(x)$ means we have all of $P(x_1),\dots ,P(x_n)$ separately, whereas with $\mathrm{\exists }xP(x)$ we only have one long statement $P(x_1)\vee P(x_2)\vee \cdots \vee P(x_n)$.
• In causal inference, I think $X\perp \perp Y\cup W$ is stronger than $(X\perp \perp Y)\vee (X\perp \perp W)$, even though both seem to use a single "or"-type operation. But if $Y$ and $W$ are disjoint, then I think the former is true while the latter may be false. I think this is similar to how $\mathrm{\forall }x\in X(P(x))$ is usually stronger than $\mathrm{\exists }x\in X(P(x))$, unless $X=\mathrm{\varnothing }$.
• Maybe another way to state the puzzle is this: "P is stronger than Q" ↔ "P implies Q" ↔ "Q is at least as true as P" ↔ "Q ≥ P" ↔ "Q is 'more powerful' than P"!

External links

• https://gowers.wordpress.com/2008/12/28/how-can-one-equivalent-statement-be-stronger-than-another/ (haven't read this yet)
• https://gowers.wordpress.com/2011/09/26/basic-logic-connectives-not/ (search "strong")
• https://gowers.wordpress.com/2011/10/02/basic-logic-relationships-between-statements-negation/ (search "strong")