In the book, Tu states that the inverse function theorem is equivalent to the implicit function theorem. Also I have read that the Constant Rank Theorem contains the inverse function theorem as a special cases. From a logic point of view, if theorem $T_1$ is a special case of theorem $T_2$, does that mean that $T_1 \Rightarrow T_2$?
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1$\begingroup$ That's...not really how it works. From a logical point of view, once you've held your axioms fixed, all your inferences are set in stone. Such "one theorem is contained in another" statements are not really logical statements, strictly speaking. They are more like "with $T_1$ in hand, $T_2$ has a trivial proof". $\endgroup$– IanCommented Jan 15, 2018 at 17:01
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No. If theorem $T_1$ is a special case of theorem $T_2$, this means that $T_2 \Rightarrow T_1$: from $T_2$ (the most general theorem), the special case $T_1$ follows.
answered Jan 15, 2018 at 17:02
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$\begingroup$ I used to always "imagine" the statement $A\Rightarrow B$ as two Venn diagrams where $A$ is a subset of $B$. This intuition has served me well in all cases except this one. $\endgroup$– EEEBCommented Jan 15, 2018 at 17:06
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$\begingroup$ Your intuition could be correct if you think as follows: often, a theorem $T_2$ is more general than theorem $T_1$ because the hypothesis of $T_2$ are satisfied by a larger number of "cases" than the hypothesis of $T_1$, and among the cases satisfying the hypothesis of $T_2$ there are all the cases where the hypothesis of $T_1$ is satisfied. $\endgroup$ Commented Jan 15, 2018 at 17:13
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$\begingroup$ I see, so you're saying I should view a theorem $T_1$ as being in fact $H_1 \Rightarrow C_1$ and say $T_2$ as being $H_2\Rightarrow C_2$. Then indeed if $H_1 \Rightarrow H_2$, ($H_2$ being "weaker" and hence satisfying more cases), then the theorem with the weakest hypotheses, namely $T_2$ implies the one with the stronger hypothesis, namely $T_1$, so $T_2$ is the more general one, as you stated. $\endgroup$– EEEBCommented Jan 15, 2018 at 17:17
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$\begingroup$ This of course provided the consequences are unchanged.. $\endgroup$– EEEBCommented Jan 15, 2018 at 17:18
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$\begingroup$ That's exactly how it is. $\endgroup$ Commented Jan 15, 2018 at 17:21