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Consider the following system, $N$ which have the soundness property

  1. $a\to (b\to a)$
  2. $(a\to (b\to c)\to ((a\to b)\to (a\to c))$
  3. $(\lnot b\to \lnot a) \to ((\lnot b \to a) \to b)$

We just changed the third axiom of $HPC$. The original one was: $$(\lnot a \to \lnot b) \to (b\to a)$$

I was asked to prove that the two system are equivalent, so I'm trying to show a proof from HPC + the assumption $(\lnot b \to \lnot a)$ to $((\lnot b \to a ) \to b)$ (And the other way around)

I wasn't able to find a proof and I be glad if you could demonstrate it to me.

Thanks

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    For the first part (HPC to HPC+) see: Elliott Mendelson, [Introduction to Mathematical Logic](https://books.google.it/books?id=FS-sCQAAQBAJ&pg=PA31) (6th ed, 2015), **Lemma 1.11(d)**, page 31. – 2017-01-29
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    For the second part (HPC+ to HPC) see: Geoffrey Hunter, [Metalogic: An Introduction to the Metatheory of Standard First Order Logic](https://books.google.it/books?id=oHpMtskGcv0C&pg=PA107) (1971), **32.5**, page 107. – 2017-01-29
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    Thanks, I'll check it out. – 2017-01-29
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    I'm actually interested as to why all of these axioms are Tautologies. It kind of makes sense that an axiom would be one; but its surprising because, in propositional logic they would all be logically equivalent by virtue of been Tautologies. – 2017-01-30
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    @user400188: The system used here has no **other** inference rules besides modus ponens, so in fact it may be **surprising** that these three axioms are sufficient to prove **every** tautology that use just implication and negation. Note also that conjunction and disjunction can be written in terms of implication and negation. However, if you allow deducing any classical propositional tautology, then you don't need any of these three axioms. – 2017-02-05
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    @blueplusgreen: You should specify clearly in your question that the only inference rule available in this system is modus ponens, to prevent any misunderstanding. – 2017-02-05

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