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Is there any meaningful mathematical connection between these two things: a) proving an implication and then its converse, and b) showing that two sets $P$, $Q$ are equal by establishing $P ⊆ Q$ and $Q ⊆ P$, or proving that $a = b$ by proving $a ≤ b$ and $b ≤ a$. There's some structural similarity, but is there more to know about this structure? Maybe some category theory?

Gary
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Theo H
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    In the case of $\subset$ and $\leq$ you use that the relationship is antisimmetrical since both are order relationships. For more general relationships it may not be true. Consider for example the relationship $x \leq y$ iff $x$ is younger or has the same age than $y$. – Zanzag Apr 09 '22 at 23:18
  • The sets P and Q does not need to have the same size. For example modelling voting uses different sized sets where there's more people who vote than there are representatives to vote for. Still voting can be modelled with P=>Q and Q=>P, where => means votes for some representative. – tp1 Apr 09 '22 at 23:36
  • @tp1 I'm talking specifically about things like https://en.wikipedia.org/wiki/Combinatorial_proof and https://en.wikipedia.org/wiki/Bijective_proof. Is there any connection to the example you give? I can't see it. – Theo H Apr 10 '22 at 00:44
  • My answer is here. – ryang Apr 10 '22 at 05:27
  • @ryang Your answer observes the situation I described but it doesn't go beyond that. I'm looking for something deeper, along the lines of, for example, "implication in propositional logic is adjoint to conjunction". In other words a great answer would say something like e.g. "these are all examples of the following categorical structure: mutual right adjoints". I understand why the sets are equinumerous. – Theo H Apr 10 '22 at 13:12

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If $P(x)$ is a logical formula in the variable $x$, which ranges over some universe set $U$, then one may form the set $P = \{x \in U : P(x)\}$. Similarly, let $Q = \{x \in U : Q(x)\}$ be the set of points $x$ for which $Q(x)$ is true.

$P(x) \implies Q(x)$ is a logical formula equivalent to $\neg P(x) \vee Q(x)$, and so the set of points $x\in U$ for which this implication is true is the same as the set $\{x \in U : \neg P(x) \vee Q(x)\}$, which is also the same as the set $P^c \cup Q$ using set operations.

If $P (x) \implies Q(x)$ were true for every $x \in U$, then $P^c \cup Q = U$ or equivalently $P \subset Q$. Here is one connection between implication and the subset relation illustrated.

Moreover, if $P(x) \iff Q(x)$ were true for every $x \in U$, then of course $P = Q$ and so $P$ and $Q$ would be equinumerous in that case. However, if $P$ and $Q$ were known from the start to be equinumerous, then of course it is not necessarily entailed that $P = Q$.

Rob
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