Proving $(A\rightarrow (B\rightarrow C))\rightarrow ((A\rightarrow B)\rightarrow (A\rightarrow C))$ in Hilbert/Ackermann axiomatic system

Question

I am working upon exercises from "Introduction to Mathematical Logic" by Mendelson chapter 1.6.

Given a formal system L1 : $\vee$ and $\neg$ are the primitive connectives. We use $B\rightarrow C$ as an abbreviation for $\neg B\vee C$. We have four axiom schemas:

(1) $B\vee B\rightarrow B$

(2) $B\rightarrow B\vee C$

(3) $B\vee C\rightarrow C\vee B$

(4) $(C\rightarrow D)\rightarrow (B\vee C\rightarrow B\vee D)$

The only rule of inference is modus ponens. Here and below we use the usual rules for eliminating parentheses. This system is developed in Hilbert and Ackermann (1950).

From the exercises I solved the following:

$B\rightarrow C\vdash D ∨ B\rightarrow D ∨ C$

$\vdash (B \rightarrow C)\rightarrow ((A\rightarrow B)\rightarrow (A\rightarrow C))$

$A\rightarrow B, B\rightarrow C\vdash A\rightarrow C$

$\vdash A\rightarrow A$

$\vdash A ∨ ¬A$

$\vdash A \rightarrow ¬¬A$

For all $\Gamma, \Gamma\vdash B\implies\Gamma\vdash A\rightarrow B$ - this one is meta theorem, not an exercise.

The ultimate goal is to prove the deduction theorem. With deduction theorem other exercises are easy. But I can't prove it because I don't have this formula as an axiom:

$\vdash (A\rightarrow (B\rightarrow C))\rightarrow ((A\rightarrow B)\rightarrow (A\rightarrow C))$

It is given as a theorem by Mendelson:

n. $B\rightarrow (C\rightarrow D), B\rightarrow C\vdash B\rightarrow D$

I think, the flip theorem could be helpful to prove it:

k. $\vdash(B\rightarrow (C\rightarrow D))\rightarrow (C\rightarrow (B\rightarrow D))$

But I can't prove both of them. Is there any method, how I should work? Or I should use trial and error approach or write a computer program to try all the possible combinations?

Answer 1

I shall sketch a mainline basing off David Hilbert and Wilhelm Ackermann's 1950 translation Principles of Mathematical Logic (freely available at the Internet Archive) in the contemporary notation.

I take the formula in question as

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

Otherwise, apparently, there would be multiple questions at once. Also I have corrected the expression, for its previous form was not a theorem.

Here is what I suggest:

1. By Axiom 4, $$((A\wedge B)\rightarrow C)\rightarrow((\neg A\vee(A\wedge B))\rightarrow(\neg A\vee C))$$
2. We note that $$((A\wedge B)\rightarrow C)\equiv (A\rightarrow(B\rightarrow C))$$ which is given as Rule VII on p. 36.
3. We observe that $$(\neg A\vee(A\wedge B))\equiv(\neg A\vee B)$$ by the theorems 19 and 20 on p. 37.
4. We notice that Axiom 4 can be transformed into $$(C\rightarrow D)\rightarrow ((Z\rightarrow C)\rightarrow (Z\rightarrow D))$$

by $(A\rightarrow B)\equiv(\neg A\vee B)$, which also the book gives as Theorem 1 on p. 31.

5. We put together the equivalences above. We may directly substitute the equivalents as Rule VI allows (on p. 33):

Two expressions which stand in a relation of mutual implication may in any theorem be substituted for each other.

Or, if a more stepwise demonstration is demanded, we can use the transitivity of implication given as Rule V on p. 31:

If $A\rightarrow B$ and $B\rightarrow C$ are theorems, then $A\rightarrow C$ is also a theorem.

Now, by (3), we rewrite (1) as $$((A\wedge B)\rightarrow C)\rightarrow((\neg A\vee B)\rightarrow(\neg A\vee C))$$

By (4), we rewrite this as

$$((A\wedge B)\rightarrow C)\rightarrow((A\rightarrow B)\rightarrow (A\rightarrow C))$$

Using (2), we obtain

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

The Hilbert-style systems are notorious for the difficulty they pose in properly manipulating all the lines into a proof if full rigour is demanded (even more, in case that deduction theorem is not to be employed). This is of no exception. However, I hope this much also helps.