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The following proposition in (1) is taken as an axiom in intuitionistic propositional logic.

$$(A\rightarrow(B\rightarrow C))\rightarrow((A\rightarrow B)\rightarrow(A\rightarrow C))\quad\quad(1)$$

What about its converse in (2)?

$$((A\rightarrow B)\rightarrow(A\rightarrow C))\rightarrow(A\rightarrow(B\rightarrow C))\quad\quad (2)$$

It's clear that (2) is also valid in intuitionistic propositional logic. But why it's less mentioned in the literature compared to (1)?

Kelly
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  • Do you mean contraposition? – Wuestenfux Jul 08 '20 at 16:40
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    @Wuestenfux Apparently not, since (2) is the converse, not the contrapositive, of (1). – Andreas Blass Jul 08 '20 at 16:46
  • It is easily provable. – Mauro ALLEGRANZA Jul 08 '20 at 16:53
  • According to some extensive notes I wrote several years, basically cataloging as much as I could 1-variable, 2-variable, 3-variable (and a few higher variable numbers) well formed formulas in implication logic systems -- which can be proved in the positive implicational fragment of intuitionistic propositional logic, which additional formulas can be proved in minimal negation logic, which additional formulas can be proved in intuitionistic implicational logic --- this can be proved in the positive implicational fragment of intuitionistic propositional logic. – Dave L. Renfro Jul 08 '20 at 16:55
  • @DaveL.Renfro Thanks! Do you have more references about this? – Kelly Jul 08 '20 at 17:05
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    Eric Schecter's 2005 book Classical and Nonclassical Logics is the best place I know for discussions of this sort (i.e. conversations to the reader about how to prove propositional statements in different ways, various interpretations of them and connections between them, etc.). Possibly this technical answer and this literature list answer might have something of interest to you. – Dave L. Renfro Jul 08 '20 at 17:16
  • @DaveL.Renfro Thanks! I just took a look at the contents of Schecter's 2005 book. It looks very interesting and it is very different from van Dalen's, which I am using now. Thanks again! – Kelly Jul 08 '20 at 18:47
  • As far as I'm concerned, the primary reason to assume $(P \rightarrow (Q \rightarrow R)) \rightarrow ((P \rightarrow Q) \rightarrow (P \rightarrow R))$ as an axiom is that it plays a central role in proving the abstraction theorem for Hilbert-type proof systems - and on the other hand the converse doesn't appear in that proof. (Not sure whether this would be suitable as an answer to the original question, though.) – Daniel Schepler Jul 08 '20 at 19:41

1 Answers1

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Formula (2) is indeed valid intuitionistically. The likely reason for its not being an axiom is that it follows easily from other intuitionistically valid formulas. The details of that would depend on the particular deductive system, but the idea is that $(B\to(A\to C))\to(A\to (B\to C))$ and $B\to(A\to B)$ together imply (2).

Andreas Blass
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  • Yes, thanks! I know how to prove (2). I am just curious about why the direction in (1) is considered an axiom but not the other direction in (2). But that's just a non-technical question. Thanks again! – Kelly Jul 08 '20 at 17:00
  • @Kelly I don't think the last statement is correct, that (2) is provable from those two. But, I don't have a model at the moment. – Doug Spoonwood Jul 09 '20 at 06:21
  • Alright, I have a model now. This part: "but the idea is that (B→(A→C))→(A→(B→C)) and B→(A→B) together imply (2)." is not correct. – Doug Spoonwood Jul 09 '20 at 06:38
  • @DougSpoonwood Does your model violate monotonicity of implication? I ask because this monotonicity seems to be all that's needed as a foundation for the "idea" in my previous comment. – Andreas Blass Jul 09 '20 at 13:30
  • @AndreasBlass I don't know. Using Cxy instead of (x$\rightarrow$y), the model is C01 = 3, C02 = 2, C03 = 1, C12 = 1, and Cxy = 0 for all other values of x and y in {0, 1, 2, 3}. That CxCyx = 0, follows from C1a = 0, C2a = 0, C3a = 0, and Ca0 = 0. If x = 0, y = 1, and z = 2, then CCCxyCxzCxCyz = CCC01C02C0C12 = CC32C01 = C03 = 1, but CCC00C00C0C00 = CC00C00 = C00 = 0. That CCxCyzCyCxz = 0 appears to take a little more work. – Doug Spoonwood Jul 09 '20 at 20:21