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Consider the following system, $S$ above $\{\lnot, \to, \lor \}$:

Axioms (1-3 are HPC's original ones):

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

Deduction Rules:

  1. MP
  2. $\frac{a}{a\lor b}$
  3. $\frac{b}{a\lor b}$

It's given that $S$ is sound - Prove that $S$ isn't complete.

My Try:
Let's look at $\varphi = p\lor \lnot p$. $\varphi$ is a tautology. Therefore, if $S$ is complete then $\vdash_S p\lor\lnot p$. Therefore, there's a proof for $\varphi$: $l_1, \ldots, l_n\equiv p\lor\lnot p$. Hence, WLOG for some $k

Since $S$ is sound, $\vDash p$ - but obviously that is not true that $p$ is a tautology (considering the interpretation $v(p) = f$)

I'd be glad to get a proof-verification

Thanks!

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    It seems to be broadly correct, but that "WLOG" hides a lot of important details. – 2017-01-31
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    Basically, this WLOG means $l_k$ could be $p$ or $\lnot p$. Either way, one could prove $p\lor\lnot p$ from it (using the deduction rule). So that was my intention here. – 2017-01-31
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    and obviously, if $l_k\equiv \lnot p$ then the proof stays the same since $\lnot p$ isn't a tautology either. – 2017-01-31
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    Ah, then I misunderstood. My point is that you have to rule out that you can derive $\phi$ by MP somehow. – 2017-01-31
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    Oh I see.. That's quite a challenge – 2017-01-31

2 Answers 2

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One approach is is as follows. First, note that all your axioms are tautologies under the normal (boolean) interpretation of propositional logic. Hence, just from the axioms it is impossible to prove, for a propositional letter $p$, either $p$ or $\neg p$, since tautologies need to hold regardless of the truth value of $p$.

Now, fix your formula $\phi$. We recursively define what it means for $\psi$ to be good (for $\phi$) as follows:

  • $\phi$ itself is good,
  • if $\psi$ and $\chi$ are good, then so is $\psi \lor \chi$,
  • if $\psi$ is good, and $\chi$ is not good, then $\chi \to \psi$ is good,
  • if $\psi$ is not good, then $\neg \psi$ is good,
  • all formulas whose goodness is not set by the above rules are not good.

Now, suppose towards a contradiction that we have a proof of $\phi$. Since $\phi$ is good, there must be a first good formula in the proof -- let us say that it is $\psi$. Now, $\psi$ cannot be an axiom, since none of the axioms are good. It also cannot follow by modus ponens, since if $\psi$ is good, then at least one of $\chi$ and $\chi \to \psi$ must be good. Finally, it cannot follow by your rules 2 or 3: if $\psi$ is $\phi$ itself, then it must follow from $p$ or $\neg p$, but we had established that your axioms cannot prove those. If $\psi = \psi_0 \lor \psi_1$ is good but not equal to $\phi$, then both $\psi_0$ and $\psi_1$ must be good; but $\psi$ must be derived from one of them.

All in all, this gives a contradiction. Hence your axioms do not prove $\phi$.

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    Why did you set "$\psi$ is not good, then $\lnot\psi$ is good"? Is that because semantically, $\lnot (p\lor \lnot p) \equiv p\lor \lnot p$? – 2017-02-05
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    Also, if $a\equiv p$ then you can derive $p\lor \lnot p$ so it isn't true that "the rules of inference can only prove good formulas". – 2017-02-05
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    The useless answer is "I use this definition because it works". More to the point: I designed the "good" formulas to be a class including those formulas from which you can prove $\phi$, and so you can't prove any good formulas. Hence it might be important to have formulas like $\neg\neg\phi$ be "good". Note that you crucially use the goodness clause for negated formulas when showing that the third axiom is not good. – 2017-02-05
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    To your second point, you're quite right. I made an error in my answer. The clause for disjunctions should be that both formulas in it are good. Will correct it now. – 2017-02-05
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    @blueplusgreen, I'm sorry, I didn't initially read your comment correctly. Your objection is very much right. I have edited my answer *yet again*, and am fairly confident that it is now correct. This question was much more subtle than I thought. – 2017-02-06
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There is a standard, but tedious way to prove that the law of excluded middle does not hold in some system. Let us have three truth values: $T,F$ and $U$, standing for "true", "false" and "unknown". Then we can define $\vee$ as maximum of the values ($F

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    Actually, on second thought, I don't think this works: axiom 3 does not always evaluate to T (and it shouldn't, as under a normal axiom system, it is equivalent to LEM). – 2017-02-05
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    @MeesdeVries On which input it doesn't? – 2017-02-05
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    On $b=T, a=U $. – 2017-02-05