# Question:

I am developing the proof for the following exercise from An Introduction to Mathematical Logic and Type Theory: To Truth Through Proof by Peter B. Andrews:

X1102. Let $\mathscr{M}$ be the system which has the same wffs and rule of inference as $\mathscr{P}$, and the single axiom schema

$$\left[\mathbf{A} \supset \mathbf{B} \supset { }_\blacksquare \mathbf{C} \vee { }_\blacksquare \mathbf{D} \vee \mathbf{E} \right] \supset { }_\blacksquare \mathbf{D} \supset \mathbf{A} \supset { }_\blacksquare \mathbf{C} \vee { }_\blacksquare \mathbf{E} \vee \mathbf{A}$$

Show that each theorem of $\mathscr{P}$ is a theorem of $\mathscr{M}$.

# Beginning of a Proof:

Start with a theorem $\mathbf{X}$ of $\mathscr{P}$. I want to prove that $\mathbf{X}$ is also a theorem of $\mathscr{M}$. To do so, look at the proof $\mathbf{X}_1 \ldots \mathbf{X}_n$ in $\mathscr{P}$ of $\mathbf{X}$ from the empty set. Consider $\mathbf{X}_i$ for some $i$ with $1 \leq n$. We need to prove by induction on $i$ that $\mathbf{X}_i$ has a proof in $\mathscr{M}$.

From the definition of a proof in $\mathscr{P}$, there are three cases to consider: (1) $\mathbf{X}_i$ is an axiom, (2) $\mathbf{X}_i$ is a member of the empty set, and (3) $\mathbf{X}_i$ inferred by modus ponens from $\mathbf{X}_j$ and $\mathbf{X}_k$ where $j < i$ and $k < i$. Condition (2) is never true for any $\mathbf{X}_i$ and (3) follows from a trivial inductive argument.

Condition (1) is where I am struggling. Since $\mathscr{P}$ has three axiom schemata (see details below), I need to find a proof of each from the single axiom of $\mathscr{M}$. I'm not certain how to proceed from here. Am I even heading in the right direction? Should I look at proving any intermediary lemmas?

# Definitions:

For clarification, here are the definitions with which I am working from the text. First, the syntactic and axiomatic structure of $\mathscr{P}$:

Definition. The set of wffs is the intersection of all sets $\mathscr{S}$ of formulas such that:

(i) $\mathbf{p} \in \mathscr{S}$ for each propositional variable $\mathbf{p}$.

(ii) For each formula $\mathbf{A}$ if $\mathbf{A} \in \mathscr{S}$, then $\mathord{\sim} \mathbf{A} \in \mathscr{S}$.

(iii) For all formulas $\mathbf{A}$ and $\mathscr{B}$, if $\mathbf{A} \in \mathscr{S}$ and $\mathbf{B} \in \mathscr{S}$, then $\left[\mathbf{A} \lor \mathbf{B} \right] \in \mathscr{S}$.

Axioms.

(1) $\mathord{\sim} \left[ \mathbf{A} \vee \mathbf{A} \right] \vee \mathbf{A}$

(2) $\mathord{\sim} \mathbf{A} \vee {}_\blacksquare \mathbf{B} \vee \mathbf{A}$

(3) $\mathord{\sim} \left[ \mathord{\sim} \mathbf{A} \vee \mathbf{B} \right] \vee {}_\blacksquare \mathord{\sim} \left[ \mathbf{C} \vee \mathbf{A} \right] \vee {}_\blacksquare \mathbf{B} \vee \mathbf{C}$

Defining $\supset$ as $\mathbf{A} \supset \mathbf{B}$ stands for $\mathord{\sim} \mathbf{A} \vee \mathbf{B}$, we can write these axioms as

(1) $\mathbf{A} \vee \mathbf{A} \supset \mathbf{A}$

(2) $\mathbf{A} \supset {}_\blacksquare \mathbf{B} \vee \mathbf{A}$

(3) $\mathbf{A} \supset \mathbf{B} \supset {}_\blacksquare \mathbf{C} \vee \mathbf{A} \supset {}_\blacksquare \mathbf{B} \vee \mathbf{C}$

$\mathscr{P}$ has one rule of inference:

Modus Ponens (MP). From $\mathbf{A}$ and $\mathord{\sim} \mathbf{A} \vee \mathbf{B}$ to infer $\mathbf{B}$.

More definitions needed to solve the exercise:

Def1. A proof of a wff $\mathbf{B}$ from the set $\mathscr{H}$ of hypotheses is a finite sequence $\mathbf{B}_1,\ldots,\mathbf{B}_m$ of wffs such that $\mathbf{B}_m$ is $\mathbf{B}$ and for each $j$ ($1 \leq j \leq m$) at least one of the following conditions is satisfied:

(1) $\mathbf{B}_j$ is an axiom.

(2) $\mathbf{B}_j$ is an member of $\mathscr{H}$.

(3) $\mathbf{B}_j$ is inferred by modus ponents from wffs $\mathbf{B}_i$ and $\mathbf{B}_k$, where $i < j$ and $k < j$.

Def2. A proof of a wff $\mathbf{B}$ is a proof of $\mathbf{B}$ from the emtpy set of hypotheses.

Def3. A theorem is a wff which has a proof.

• Can you refresh us (me, anyway) on the meaning of the the $\otimes_\blacksquare$ notation? I know it indicates the grouping, but what, e.g., would be the parenthesized version of $\mathscr M$'s axiom schema? – Joshua Taylor Jun 14 '13 at 11:19
• @JoshuaTaylor The ${}_\blacksquare$ is the left parenthesis. The matching right parenthesis goes as far right as possible without breaking other existing groupings, including the ones implied by order of operations. So $\mathbf{A} \supset \mathbf{B} \supset {}_\blacksquare \mathbf{C} \vee \mathbf{A} \supset {}_\blacksquare \mathbf{B} \vee \mathbf{C}$ is equivalent to $\mathbf{A} \supset \mathbf{B} \supset [\mathbf{C} \vee \mathbf{A}] \supset [\mathbf{B} \vee \mathbf{C}]$ – Code-Guru Jun 14 '13 at 15:58
• What are the order of operations in use here? (Sorry, I always tend to fully parenthesize.) Is this the fully parenthesized axiom of $\mathscr{M}$? $(A \to (B \to (C \lor (D \lor E)))) \to (D \to (A \to (C \lor (E \lor A))))$ – Joshua Taylor Jun 14 '13 at 16:18
• @JoshuaTaylor Andrews associates all connectives, including $\supset$, from left to right. And unless otherwise specified, $\supset$ has higher precedence than $\vee$. – Code-Guru Jun 14 '13 at 16:28
• So the axiom is $((A \to B) \to (C \lor (D \lor E))) \to ((D \to A) \to (C \lor (A \lor E)))$? – Joshua Taylor Jun 14 '13 at 16:39

I'm sure that this sort of single axiom schema for propositional calculus was found by C. Meredith about 50 years ago. Google tells me that his original paper was C. Meredith, Single axioms for the systems (C, N), (C, 0) and (A, N) of the two-valued propositional calculus, Journal of Computing Systems, p. 155-164, 1954. That paper will presumably give proofs that the axioms of some more familiar axiom systems can be derived from the single schema.

However, this is a mere curiosity, and I've never seen any interest in this sort of brainteaser: what's the point? I'd just ignore this exercise in Andrews!

• The completionist in me wants to solve this problem. Beyond that, you are probably right. As for the paper, I have been unable to find a copy of it with my googling. Did you find a link? If so, will you please include it in your answer? – Code-Guru Dec 18 '12 at 19:09
• @Code-Guru A logical system consists of a set of theorems (or axioms and theorems together). The axioms under rule(s) of inference can get thought of as a way of encoding the information of a logical system. Now say we have two axiom sets "A" and "B" with A having multiple axioms and "B" having a single axiom. So both A and B under the same set of rule(s) can get said to contain the same information. Or they both allow the same information. But A and B aren't of the same length. Consequently, this sort of problem might have a relationship with some form information compression. – Doug Spoonwood May 18 '14 at 13:06
• This wasn't an answer before. It also doesn't seem very useful in light of actually having an answer. – Doug Spoonwood May 21 '14 at 1:55
• What you mean not an answer before? It's not been changed in five months. But I agree it isn't an answer in one sense: though an important thing for budding logicians to learn is when an exercise elucidates some significant idea, and when it is just a brain-teaser. – Peter Smith May 21 '14 at 9:17
• @Petersmith Before what you wrote here might have come as useful. Also, I have to disagree that this exercise is just a brain-teaser. Not only does it show a way to compress the information of a logical system, it gives us a formula which has more power than any of the axioms of Andrews. From what I can tell it's an open question as to whether or not there exists any shorter (A-N) axiom than the Meredith axiom of the OP. If we do have a metalogical proof that the Meredith axiom has the shortest length, it's not known how many axioms of that length (up to re-lettering of variables) exist. 1/2 – Doug Spoonwood May 22 '14 at 17:21

So in Polish notation the single axiom becomes

The three axiom basis becomes:

1. CAaaa
2. CaAba
3. CCabCAcaAbc

Let's use the definition C:=AN, and rewrite M1 in its definition-free form:

We'll rewrite the sought after formulas as:

1. ANAaaa
2. ANaAba
3. ANANabANAcaAbc

Now, I've used those formulas to produce an input for Prover9 [1] for this problem. And thus:

After 32,449.76 seconds Prover9 completed three proofs which give us the required theorems

1. ANAxxx
2. ANxAyx
3. ANANxyANAzxAyz

from the axiom ANANANxyAzAuwANANuxAzAwx with the rule {AN$\alpha$$\beta, \alpha \vdash \beta}, and then condensed those proofs into one proof. I've managed to shorten things down to around a 70 step proof (using certain theorems from the initial run as "hints"). Notice how much the axiom "3" gets used here. I'll work through it a little by hand to make it more comprehensible. First, we'll start with 3 ANANANxyAzAuwANANuxAzAwx. CCCxyAzAuxCCuxAxAwx. So we can draw: ANANAN x y A z Au w ANANuxAzAwx |||| | | | | | | |||| ---- ----- | ----- | --- ANAN ANxy AzAuw A NANux Az Awx  Thus (see previous versions for an explanation of the notation for proofs) 3 x/ANxy, y/AzAuw, z/NANux, u/z, w/Awx * AN 3-5  5 ANANzANxyANANuxAAwxANxy. CCzCxCCuxAAwxCxy. Now we'll draw 3 ANANAN x y A z A u w ANANuxAzAwx | | | | | | | | ---- | ----- | --- ---- 5 ANAN z ANxy A NANux A Awx ANxy  Thus, 3 x/z, y/ANxy, z/NANux, u/Awx, w/ANxy * AN 5-6 6 ANANAwxzANANuxAANxyz. CCAwxzCCuxACxyz. 3 ANANAN x y A z A u w ANANuxAzAwx | | | | | --- | | ----- ---- | 6 ANAN Awx z A NANux A ANxy z  Thus, 3 x/Awx, y/z, z/NANux, u/ANxy, w/z * AN 6-7 7 ANANANxyAwxANANuxAzAwx. CCCxyAwxCCuxAzAwx. Now the path to the next formula comes as more involved, from my viewpoint at least, than the previous paths. To avoid confusion, we'll re-letter all the variables in M7 one time, and not re-letter them another time. In fact, in general, the safest way to do a condensed detachment comes as to re-letter all the variables of the major premise (schema) and the minor premise (schema). 7 AN ANAN x y A w x ANANuxAzAwx 7 AN ANAN a b A c a ANANdaAeAca | | | | ---- --- ----- ----- 7 ANAN ANxy Awx A NANux AzAwx  So, we at first glance it might look like we have a/ANxy, b/Awx, c/NANux, a/AzAwx. Of course though we can't substitute two things for "a". So, can we find a common form (or "unifier") for ANxy and AzAwx? Well, if z/Nx, y/Awx, then ANxy becomes ANxAwx, and AzAwx becomes ANxAwx also. Thus, we can substitute a/ANxAwx, b/Awx, c/NANux. But, we'll also need to apply the same substitutions we made when re-lettering 7. And we have "u" and "w" appearing on the right hand side, which did not appear on the left hand side. So, to make sure that this "u" and "w" don't match any other letters that result because of a detachment, we'll change them also to get a more general formula. This means that in 7 we apply x/ANxAwx, y/Awx, w/NANux, u/a, z/b for the major, and we retain the substitutions we made in the minor z/Nx, y/Awx. Thus, 7 x/ANxAwx, y/Awx, w/NANux, u/a, z/b * AN 7 z/Nx, y/Awx - 8 8 ANANaANxAwxAbANANuxANxAwx. CCaCxAwxAbCCuxCxAwx. This step also shows something interesting. Suppose we can know what substitutions need to get made in the major. Suppose that we can also know that we need not make any more substitutions in the major on the left hand side of the formula for the left hand side of the major and the minor to have a common form (or "unifier"). Suppose we also change all variables on the right hand side of the major if they do not appear on the left hand side of the major. Then, we can effectively ignore the substitutions in the minor, since the rule of inference here tells us that we infer to a substitution instance of the right hand side of the major. 3 AN ANAN x y A z A u w ANANuxAzAwx | | | | | | ------ | ----- ------ 8 ANAN a ANxAwx A b A NANux ANxAwx  Thus, 3 x/a, y/ANxAwx, z/b, u/NANux, w/ANxAwx * AN 8-9 9 ANANNANuxaAbAANxAwxa. CCNCuxaAbACxAwxa. 8 ANAN a A Nx A w x AbANANuxANxAwx 8 ANAN c A Nd A e d AfANANgdANdAed | -- | | -------- | ------ | 9 AN ANNANuxa A b A ANxAwx a  Now this diagram suggests that in the lower part we substitute b with Nd. But notice that b does not appear anywhere else in 9 (aside... there's a beautiful relationship to "contraction" or "CCpCpqCpq" here), and thus the above comment implies that we can ignore that substitution. "f" and "g" only appear on the right hand side of 9. We could substitute d in 8 with a, or we could substitute a in 9 with d in 8 here. But since d only appears once in 8, it seems simpler to substitute d with a. Taking that all together we can infer: 10 AfANANgaANaAANxAwxa. AfCCgaCaACxAwxa. 10 A f ANANgaANaAANxAwxa | ------------------- 10 AfANANgaANaAANxAwxa  We can't use the rule of inference {AN\alpha$$\beta$, $\alpha$} $\vdash$ $\beta$ here in this form. Fortunately "f" only appears once in 10, so we can substitute f with Nf. Then we have

10 AN f ANANgaANaAANxAwxa
|
-------------------
10    AfANANgaANaAANxAwxa


Re-lettering ANANgaANaAANxAwxa we thus obtain

11 ANANxyANyAANzAuzy. CCxyCyACzAuzy.

3  AN ANAN x y A z  A  u     w   ANANuxAzAwx
|     |     |
--   ------ |
11    ANAN x y A Ny A ANzAuz y


Thus applying the rule we obtain

12 ANANANzAuzxANyAyx. CCCzAuzxCyAyx.

8  AN AN a         AN x A w x AbANANuxANxAwx
|            |   | |
---------    |   | |
12    AN ANANzAuzx AN y A y x


Here we can figure out that x, w of 8 and y, x of 12 will end up as the same variable. This holds even if we did change the "x" and "w" of 8 to say "c" and "d" respectively. Notice that "b" and "u" only appear on the right hand side of 8. Thus, we can infer

13 AbANANaxANxAxx. AbCCaxCxAxx.

Now we can use 13 as the major and minor. Notice that 13 has a property in common with 10. The property consists of the fact that the left side consists of a single letter which is NOT found anywhere on the right hand side. For any such formula where it gets used as the major premise, the most general formula we can deduce consists of the right hand formula. I'll call this MT 1. As you can see from the following diagrams:

A  a  b   A a  b
---       --
Acd       Ne


We simply substitute "a" with NAcd, or NNe depending on what the minor is. Since a does not appear anywhere in b, we can then infer b. So, with 13 as the major and 13 as the minor, and re-lettering, we infer

14 ANANxyANyAyy. CCxyCyAyy.

3   AN ANAN x y A z  A u w ANANuxAzAwx
| |   |    | |
| |   --   | |
14     ANAN x y A Ny A y y


Thus, by substitution and the "detachment" rule we can infer:

15 ANANyxANyAyx. CCyxCyAyx.

3  AN ANAN x y A z  A u w ANANuxAzAwx
| |   |    | |
| |   --   | |
15    ANAN y x A Ny A y x


So, we can infer

16 ANANyyANyAxy. CCyyCyAxy. (this contains the law of identity Cyy).

8  AN AN a    AN x A w x AbANANuxANxAwx
|       |   | |
----    |   | |
16    AN ANyy AN y A x y


Notice that "b" and "u" do not appear on the left hand side in 8. Doing some re-lettering we can thus infer

17 AxANANyzANzAuz. AxCCyzCzAuz.

Now taking 17 as the major and 17 as the minor and using MT 1, we know that the most general formula we can obtain (up to re-lettering) is

18 ANANxyANyAzy. CCxyCyAzy.

7   AN AN AN x y A w  x ANANuxAzAwx
| |   |  |
| |   -- ---
18     AN AN x y A Ny Azy


Notice that x in 7 getting substituted with Azy comes the choice that will work, and then x in the lower we need not worry about since it only appears once. "z" and "u" only appear on the right hand side of 7. Thus, from 7 and 18 we can infer

19 ANANuAzyAvANyAzy. CCuAzyAyCyAzy.

3 AN ANAN x y A z  A u w ANANuxAzAwx
| |   |    | |
| |   --   | |
18   ANAN x y A Ny A z y


The substitutions here thus allow us to infer from 3 and 18

20 ANANzxANyAyx. CCzxCyAyx.

3  AN ANAN x y A z  A u w ANANuxAzAwx
| |   |    | |
| |   --   | |
20    ANAN z x A Ny A y x


So, from 3 and 20 we can infer

21 ANANyzANyAxz. CCyzCyAxz.

21 AN A N y     z ANyAxz
|     |
---- ------
18    A N ANxy ANyAzy


Note that "x" only appears on the right hand side of 21. We can thus infer from 21 and 18

22 ANANxyAzANyAuy. CCxyAzCyAuy.

3  AN ANAN x y A z A u  w ANANuxAzAwx
| |   |   |  |
| |   |   -- ---
22    ANAN x y A z A Ny Auy


Thus, from 3 and 22 (and the rule of inference) we can infer

23 ANANNxyAzAAuxy. CCNxyAzAAuxy.

3  AN ANAN x  y A z A u   w ANANuxAzAwx
|  |   |   |   |
-- |   |   --- |
23    ANAN Nx y A z A Aux y


So, from 3 and 23 we can infer

24 ANANAxyNyAzAuNy. CCAxyNyAzAuNy.

19 AN AN u       A z y AvANyAzy
|         | |
-------   | ----
24    AN ANAxyNy A z AuNy


Notice that "v" only appears on the right hand side of 19. So, from 19 and 24 we can infer

25 AxANAyNzAuAyNz. AxCAyNzAuAyNz.

Now "x" only appears on the left hand side of 25. Thus, by MT 1, we know that the rule of inference at work here allows us to infer from 25 and 25

26 ANAxNyAzAxNy. CAxNyAzAxNy.

3 AN ANA Nx y  A z A u w ANANuxAzAwx
-- |    |   | |
|  --   |   | --
26   ANA x  Ny A z A x Ny


Here the lower "x" gets substituted by the upper "Nx". So, the upper "u" gets substituted by "Nx" as it appears in the upper formula 3. Thus, from 3 and 26 we may infer

27 ANANNxxAyANzx. CCNxxAyCzx.

19 AN AN u     A z y AvANyAzy
|       | |
-----   | ----
27    AN ANNxx A y ANzx


So, from 19 and 27 we may infer

28 AxANANyzAuANyz. AxCCyzAuCyz.

Since "x" only appears on the left hand side of 28, from 28 and 28 we can infer

29 ANANxyAzANxy. CCxyAzCxy.

3  AN ANAN x y A z A u  w ANANuxAzAwx
| |   |   |  |
| |   |   -- |
29    ANAN x y A z A Nx y


Thus, from 3 and 29 we may infer

30 ANANNxxAyAzx. CCNxxAyAzx.

19 ANAN u    A z y AvANyAzy
|      | |
-----   | ---
30  AN ANNxx A y Azx


"v" only appears on the right hand side of 19. So, from 19 and 30 we may infer

31 AxANAyzAuAyz. AxCAyzAuAyz.

3   AN ANAN x  y A z A u w ANANuxAzAwx
|  |   |   | |
-- |   |   | |
30     ANAN Nx x A y A z x


So, from 3 and 31 we may infer 32

32 ANANxNyAzAyNy. CCxNyAzAyNy.

Note that 31 has "x" only appearing on the left hand side of 31, and only "x" appears on the left side of 31. Thus, from 31 and 31 MT allows us to infer

33 ANAxyAzAxy. CAxyAzAxy.

21 AN AN y   z ANyAxz
|   |
--- -----
33    AN Axy AzAxy


Notice that "x"" only appears on the right hand side of 21. Thus, from 21 and 33 we may infer

34 ANAxyAzAuAxy. CAxyAzAuAxy.

3  AN ANA Nx y A z A u w ANANuxAzAwx
-- |   |   | |
|  |   |   | ---
34    ANA x  y A z A u Axy


In 34 if we substitute x with Nx, then Axy becomes ANxy. Thus, w in 3 will get substituted with ANxy. Doing this allows us to infer from 3 and 34

35 ANANuxAzAANxyx. CCuxAzACxyx.

3 AN ANAN x y A z A u    w ANANuxAzAwx
| |   |   |    |
| |   |   ---- |
35   ANAN u x A z A ANxy x


So, from 3 and 35 we can infer

36 ANANANxyzAuAxz. CCCxyzAuAxy.

  3   AN ANAN x    y A z A u w ANANuxAzAwx
|    |   |   | |
---- |   |   | |
36     ANAN ANxy z A u A x z


So, from 3 and 36 we may infer

37 ANANxANxyAzAuANxy. CCxCxyAzAuCxy.

  36 ANANAN x y  z AuAxz
| |  |
| -- ------
32   ANAN x Ny AzAyNy


"u" only appears on the right hand side of 36. Thus, we can infer from 36 and 32

38 AxAyAzAuNu.

 36 ANANAN x y  z AuAxz
| |  |
| | -----
20   ANAN z x ANyAyx


So, from 36 and 30 we may infer

39 AxAyANzAzu. AxAyCzAzu.

For the next steps I'll just write Dx.x..., D38.38 gives us

40 AxAyAzNz.

D40.40 gives us

41 AxAyNy.

D41.41 gives us

42 AxNx. The law of the excluded middle thus is the answer to "the life universe and everything"!!!!

Now, what do we have for D42.42? Well, x/Nx in 42 yields ANxNNx. Then using this we can see

  AN x NN x
|
----
42 AxNx


Thus, from 42 and 42 we may infer

43 NNAxNx.

D39.39 yields

44 AyANzAzu. AyCzAzu.

And D44.44 yields

45 ANxAxy. CxAxy (this corresponds to one of the disjunction introduction rules of natural deduction).

Note the relationship between 45 and the desired formula ANxAxy. If we can deduce ANAxAyzAxAzy, ANxAxy will follow. But, we'll end up getting ANxAyx a different way.

Suppose we have some formula !:AN a b, where a consists of a single variable, and that variable does not appear in b. In such a case, D!.x, where x indicates any formula yields b with it's instance of "a" substituted by the formula "x" (this has a relationship to what sometimes gets called "specialized assertion" in that from CpCCpqq and Cpp, "specialized assertion, CCCppqq follows). Notice that 45 has a single variable on the left hand side. Thus, from 45 and 43 we obtain

46 ANNAxNxy. CNAxNxy (the negation of the law of the excluded middle implies anything... thus if the law the excluded middle were false, anything is the answer to the question of the life, universe, and everything).

46 consists of the very first formula which only has a single letter appearing on the right hand side of the formula.

  18 AN AN x     y ANyAzy
|     |
----- |
46    AN NAxNx y


"z" only appears on the right hand side of 18, and thus we infer

47 ANxAyx. --- Axiom 2. CxAyx... this corresponds to another of the disjunction introduction rules of natural deduction.

  37 ANAN x     A Nx y AzAuANxy
|       -- |
-----   |  ---
30   AN ANNxx A y  Azx


Fortunately here "y" only appears once in 30. "z" and "u" only appear on the right hand side of 37. Consequently, we infer

49 AxAyANANNzzAuz. AxAyCCNzzAuz.

D49.49 yields

50 AxANANNyyAzy. AxCCNyyAzy.

D50.50 yields

51 ANANNxxAyx. CCNxxAyx.

  37 ANAN x     A Nx y AzAuANxy
|       -- |
-----   |  |
51   AN ANNxx A y  x


y only appears once in 51, and thus we don't need to worry about what the substitution of y with Nx does to any of the other substitutions needed here. "z" and "u" don't appear on the left hand side of 37. Thus, we may infer from 37 and 51

52 AxAyANANNzzz. AxAyCCNzzz.

D52.52 yields

53 AxANANNyyy. AxCCNyyy.

D53.53 yields

54 ANANNxxx. CCNxxx. This corresponds to the law of Clavius.

  36  ANANAN x  y z AuAxz
|  | |
-- | |
54    ANAN Nx x x


"u" only appears on the right hand side of 36. Thus, from 36 and 54 we may infer

55 AxANyy. AxCyy.

D55.55 yields

56 ANxx. Cxx. This corresponds to the weak law of identity.

  3 AN AN ANxy A z  Auw ANANuxAzAwx
---    |  ---
|      -- |
55   A  x    A Ny y


So we can see that y from 56 gets substituted with Auw, and thus z/NAuw. Consequently, from 3 and 56 we may infer

57 ANANuxANAuwAwx. CCuxCAuwAwx.

  57 AN AN u x ANAuwAwx
| |
| |
56    AN x x


Thus from 57 and 56 we may infer

58 ANAxyAyx. CAxyAyx. This corresponds to disjunction commuting.

  57 AN A Nu x ANAuwAwx
-- |
|  --
42    A x  Nx


Since x in 42 gets substituted with Nu, x in 57 (x in 57 is not the same as x in 57 because of their relative positions in their formulas) gets substituted with NNu. "w" only appears on the right hand side of 57. Thus from 57 and 42 follows

59 ANAxyAyNNx. CAxyAyNNx.

58 AN A x y Ayx | | ---- --- 58 A NAxy Ayx

So, from 58 and 58 follows

60 AAxyNAyx.

57 AN A N u x ANAuwAwx | | --- --- 58 A N Axy Ayx

Thus from 57 and 58 we may infer

61 ANAAxyzAzAyx. CAAxyzAzAyx.

Now D58.x the right hand side and the left hand side of any formula "x" which starts with "A". Thus, D58.59 yields

62 AAxNNyNAyx.

  57 AN AN u     x ANAuwAwx
|     |
----- -----
61    AN AAxyz AzAyx


Thus, from 57 and 61 we may infer

63 ANAAAxyzuAuAzAyx. CAAAxyzuAuAzAyx.

61 allows us to infer from AA$\alpha$$\beta$$\gamma$ to A$\gamma$A$\beta$$\alpha. Thus, 61 and 62 allow us to infer 64 ANAyxANNyx. CAyxCNyx. This has more to it than the rule of disjunctive elimination, but it also enables a rule of disjunctive elimination. 47 ANxAyx, allows to infer A \alphap where \vdash$$\alpha$, and p does not appear in $\alpha$. Thus, 47 and 58 allow us to infer

65 AxANAyzAzy. AxCAyzAxy.

   3 AN A NANxy A z    A u w ANANuxAzAwx
-----   |      | |
|       ----   | |
65   A x     A NAyz A z y


x only appears once in 65, and thus whatever we substitute it with does not affect any other substitution in 65. Thus, from 3 and 65 we may infer

66 ANANzxANAyzAyx. CCzxCAyzAyx.

66  AN AN z   x A NAyzAyx
|   |
--- ---
58     AN Axy Ayx


"y" only appears on the right hand side of 66. Thus, from 66 and 58 we may infer

67 ANAxAyzAxAzy. CAxAyzAxAzy.

66 AN AN z     x A NAyzAyx
|     |
----- |
54    AN ANNxx x


So, from 66 and 54 we may infer

68 ANAxANNyyAxy. CAxCNyyAxy.

Now 67 allows us to infer from A$\alpha$A$\beta$$\gamma to A\alphaA\gamma$$\beta$. Thus, from 67 and 61 we may infer

69 ANAAxyzAAyxz. CAAxyzAAyxz.

68  AN A x    ANN y y Axy
|        | |
----     | |
64     A NAyx ANN y x


Thus, the "x" and "y" in 64 need to match. Thus x/NAxx, y/x in 68, and y/x in 64 enables us to infer

70 ANAxxx. CAxxx. This is 1st axiom of the 3 axiom system.

64 allows us to infer from A$\alpha$$\beta to ANN\alpha$$\beta$. Thus, from 64 and 69 we may infer

72 ANNNAAxyzAAyxz CNNAAxyzAAyxz.

59 allows us to infer from A$\alpha$$\beta to A\betaNN\alpha. Thus, from 59 and 60 we may infer 73 ANAyxNNAxy. CAyxNNAxy. 73 allows us to infer from A\alpha$$\beta$ to NNA$\beta$$\alpha. Thus, from 73 and 66 we may infer ANANzxANAyzAyx 74 NNAANAyzAyxNANzx. NNAANAyzAyxNCzx. 72 allows us to infer from NNAA\alpha$$\beta$$\gamma to AA\beta$$\alpha$$\gamma. Thus, from 72 and 74 we may infer 75 AAAyxNAyzNANzx. AAAyxNAyzNCzx. 63 allows us to infer from AAA\alpha$$\beta$$\gamma$$\delta$ to A$\delta$A$\gamma$A$\beta$$\alpha$. Thus, from 63 and 75 we may infer

76 ANANxyANAzxAyz. CCxyCAzxAyz, and this is the 3rd axiom.

I've counted 18 times (19 if you count it as getting used twice in the first detachment) that the single axiom here got used. It does not seem uncommon that a single axiom for a known and complicated logical system can do this much work.

I found a proof that the three axiom set under condensed detachment yields the Meredith axiom with OTTER [2]!

----> UNIT CONFLICT at 1271.80 sec ----> 274965 [binary,274964.1,2.1] $ANS(Meredith1). Length of proof is 98. Level of proof is 41. ---------------- PROOF ---------------- 1 [] -P(A(N(x),y))| -P(x)|P(y). 2 [] -P(A(N(A(N(A(N(p),q)),A(r,A(s,t)))),A(N(A(N(s),p)),A(r,A(t,p)))))|$ANS(Meredith1).
3 [] P(A(N(A(x,x)),x)).
4 [] P(A(N(x),A(y,x))).
5 [] P(A(N(A(N(x),y)),A(N(A(z,x)),A(y,z)))).
11 [hyper,1,5,4] P(A(N(A(x,y)),A(A(z,y),x))).
12 [hyper,1,5,3] P(A(N(A(x,A(y,y))),A(y,x))).
19 [hyper,1,11,4] P(A(A(x,A(y,z)),N(z))).
20 [hyper,1,11,3] P(A(A(x,y),N(A(y,y)))).
23 [hyper,1,5,12] P(A(N(A(x,A(y,A(z,z)))),A(A(z,y),x))).
26 [hyper,1,12,4] P(A(x,N(x))).
38 [hyper,1,11,19] P(A(A(x,N(y)),A(z,A(u,y)))).
58 [hyper,1,23,38] P(A(A(x,y),A(z,N(x)))).
76 [hyper,1,12,58] P(A(N(x),A(x,y))).
85 [hyper,1,11,76] P(A(A(x,A(y,z)),N(y))).
86 [hyper,1,5,76] P(A(N(A(x,y)),A(A(y,z),x))).
112 [hyper,1,86,26] P(A(A(N(x),y),x)).
144 [hyper,1,86,112] P(A(A(x,y),A(N(x),z))).
149 [hyper,1,11,112] P(A(A(x,y),A(N(y),z))).
166 [hyper,1,3,149] P(A(N(x),x)).
175 [hyper,1,5,166] P(A(N(A(x,y)),A(y,x))).
198 [hyper,1,5,175] P(A(N(A(x,A(y,z))),A(A(z,y),x))).
200 [hyper,1,175,149] P(A(A(N(x),y),A(z,x))).
206 [hyper,1,175,85] P(A(N(x),A(y,A(x,z)))).
213 [hyper,1,175,20] P(A(N(A(x,x)),A(y,x))).
214 [hyper,1,175,19] P(A(N(x),A(y,A(z,x)))).
224 [hyper,1,198,198] P(A(A(x,A(y,z)),N(A(x,A(z,y))))).
239 [hyper,1,198,86] P(A(A(x,A(y,z)),N(A(x,y)))).
243 [hyper,1,198,12] P(A(A(x,y),N(A(x,A(y,y))))).
244 [hyper,1,198,11] P(A(A(x,A(y,z)),N(A(x,z)))).
265 [hyper,1,5,206] P(A(N(A(x,y)),A(A(z,A(y,u)),x))).
310 [hyper,1,11,213] P(A(A(x,A(y,z)),N(A(z,z)))).
311 [hyper,1,5,213] P(A(N(A(x,A(y,y))),A(A(z,y),x))).
325 [hyper,1,5,214] P(A(N(A(x,y)),A(A(z,A(u,y)),x))).
378 [hyper,1,175,224] P(A(N(A(x,A(y,z))),A(x,A(z,y)))).
386 [hyper,1,175,239] P(A(N(A(x,y)),A(x,A(y,z)))).
395 [hyper,1,175,243] P(A(N(A(x,A(y,y))),A(x,y))).
403 [hyper,1,175,244] P(A(N(A(x,y)),A(x,A(z,y)))).
490 [hyper,1,175,310] P(A(N(A(x,x)),A(y,A(z,x)))).
635 [hyper,1,378,311] P(A(N(A(x,A(y,y))),A(x,A(z,y)))).
659 [hyper,1,5,386] P(A(N(A(x,A(y,z))),A(A(y,A(z,u)),x))).
712 [hyper,1,395,325] P(A(N(A(A(x,A(y,z)),z)),A(x,A(y,z)))).
716 [hyper,1,395,265] P(A(N(A(A(x,A(y,z)),y)),A(x,A(y,z)))).
728 [hyper,1,5,403] P(A(N(A(x,A(y,z))),A(A(y,A(u,z)),x))).
815 [hyper,1,5,490] P(A(N(A(x,A(y,y))),A(A(z,A(u,y)),x))).
969 [hyper,1,5,635] P(A(N(A(x,A(y,A(z,z)))),A(A(y,A(u,z)),x))).
1112 [hyper,1,378,659] P(A(N(A(x,A(y,z))),A(x,A(y,A(z,u))))).
1355 [hyper,1,5,712] P(A(N(A(x,A(A(y,A(z,u)),u))),A(A(y,A(z,u)),x))).
1370 [hyper,1,5,716] P(A(N(A(x,A(A(y,A(z,u)),z))),A(A(y,A(z,u)),x))).
1429 [hyper,1,378,728] P(A(N(A(x,A(y,z))),A(x,A(y,A(u,z))))).
1673 [hyper,1,378,815] P(A(N(A(x,A(y,y))),A(x,A(z,A(u,y))))).
2037 [hyper,1,1112,175] P(A(N(A(x,y)),A(y,A(x,z)))).
2154 [hyper,1,1355,659] P(A(A(x,A(y,z)),N(A(z,A(x,y))))).
2174 [hyper,1,1370,728] P(A(A(x,A(y,z)),N(A(y,A(x,z))))).
2220 [hyper,1,1429,175] P(A(N(A(x,y)),A(y,A(z,x)))).
2509 [hyper,1,5,1673] P(A(N(A(x,A(y,A(z,z)))),A(A(y,A(u,A(v,z))),x))).
2785 [hyper,1,395,2037] P(A(N(A(x,A(x,y))),A(x,y))).
2799 [hyper,1,5,2037] P(A(N(A(x,A(y,z))),A(A(z,A(y,u)),x))).
2870 [hyper,1,175,2154] P(A(N(A(x,A(y,z))),A(y,A(z,x)))).
2922 [hyper,1,175,2174] P(A(N(A(x,A(y,z))),A(y,A(x,z)))).
3110 [hyper,1,5,2220] P(A(N(A(x,A(y,z))),A(A(z,A(u,y)),x))).
3763 [hyper,1,1355,2799] P(A(A(x,A(y,z)),N(A(z,A(y,x))))).
4303 [hyper,1,2922,5] P(A(N(A(x,y)),A(N(A(N(y),z)),A(z,x)))).
4634 [hyper,1,378,3110] P(A(N(A(x,A(y,z))),A(x,A(z,A(u,y))))).
5266 [hyper,1,175,3763] P(A(N(A(x,A(y,z))),A(z,A(y,x)))).
6123 [hyper,1,2509,4303] P(A(A(N(A(N(x),y)),A(z,A(u,y))),N(A(y,x)))).
6536 [hyper,1,5,4634] P(A(N(A(x,A(y,A(z,u)))),A(A(y,A(u,A(v,z))),x))).
7277 [hyper,1,395,5266] P(A(N(A(x,A(y,A(y,x)))),A(y,x))).
7278 [hyper,1,378,5266] P(A(N(A(x,A(y,z))),A(A(y,x),z))).
8091 [hyper,1,175,6123] P(A(N(A(x,y)),A(N(A(N(y),x)),A(z,A(u,x))))).
8607 [hyper,1,6536,4303] P(A(A(N(A(N(x),y)),A(z,A(u,y))),N(A(z,x)))).
9514 [hyper,1,378,7277] P(A(N(A(x,A(y,A(y,x)))),A(x,y))).
10244 [hyper,1,5,8091] P(A(N(A(x,A(y,z))),A(A(N(A(N(z),y)),A(u,A(v,y))),x))).
10976 [hyper,1,175,8607] P(A(N(A(x,y)),A(N(A(N(y),z)),A(x,A(u,z))))).
11405 [hyper,1,5,9514] P(A(N(A(x,A(y,A(z,A(z,y))))),A(A(y,z),x))).
11878 [hyper,1,10244,200] P(A(A(N(A(N(x),y)),A(z,A(u,y))),A(N(x),v))).
12780 [hyper,1,5,10976] P(A(N(A(x,A(y,z))),A(A(N(A(N(z),u)),A(y,A(v,u))),x))).
13760 [hyper,1,11405,11878] P(A(A(N(x),y),A(N(A(N(x),z)),A(u,A(v,z))))).
15901 [hyper,1,12780,175] P(A(A(N(A(N(x),y)),A(z,A(u,y))),N(A(x,z)))).
21742 [hyper,1,175,15901] P(A(N(A(x,y)),A(N(A(N(x),z)),A(y,A(u,z))))).
27283 [hyper,1,969,21742] P(A(A(N(A(N(x),y)),A(z,A(u,y))),N(A(x,A(u,y))))).
29852 [hyper,1,175,27283] P(A(N(A(x,A(y,z))),A(N(A(N(x),z)),A(u,A(y,z))))).
33412 [hyper,1,5,29852] P(A(N(A(x,A(y,A(z,u)))),A(A(N(A(N(y),u)),A(v,A(z,u))),x))).
33413 [hyper,1,29852,13760] P(A(N(A(N(A(N(x),y)),A(z,A(u,v)))),A(w,A(N(A(N(x),v)),A(z,A(u,v)))))).
35671 [hyper,1,33412,5266] P(A(A(N(A(N(x),y)),A(z,A(u,y))),N(A(y,A(u,x))))).
35681 [hyper,1,33412,378] P(A(A(N(A(N(x),y)),A(z,A(u,y))),N(A(x,A(y,u))))).
35741 [hyper,1,2785,33413] P(A(N(A(N(A(N(x),y)),A(z,A(u,v)))),A(N(A(N(x),v)),A(z,A(u,v))))).
35991 [hyper,1,175,35671] P(A(N(A(x,A(y,z))),A(N(A(N(z),x)),A(u,A(y,x))))).
36069 [hyper,1,175,35681] P(A(N(A(x,A(y,z))),A(N(A(N(x),y)),A(u,A(z,y))))).
36769 [hyper,1,5,35991] P(A(N(A(x,A(y,A(z,u)))),A(A(N(A(N(u),y)),A(v,A(z,y))),x))).
37300 [hyper,1,5,36069] P(A(N(A(x,A(y,A(z,u)))),A(A(N(A(N(y),z)),A(v,A(u,z))),x))).
42526 [hyper,1,7278,35741] P(A(A(N(A(N(x),y)),N(A(N(A(N(x),z)),A(u,A(v,y))))),A(u,A(v,y)))).
42579 [hyper,1,37300,42526] P(A(A(N(A(N(x),y)),A(z,A(u,y))),A(N(A(N(v),u)),N(A(N(A(N(v),w)),A(x,A(y,u))))))).
42982 [hyper,1,2870,42579] P(A(N(A(N(x),y)),A(N(A(N(A(N(x),z)),A(u,A(v,y)))),A(N(A(N(u),v)),A(w,A(y,v)))))).
49321 [hyper,1,5,42982] P(A(N(A(x,A(N(y),z))),A(A(N(A(N(A(N(y),u)),A(v,A(w,z)))),A(N(A(N(v),w)),A(v6,A(z,w)))),x))).
55086 [hyper,1,49321,144] P(A(A(N(A(N(A(N(x),y)),A(z,A(u,v)))),A(N(A(N(z),u)),A(w,A(v,u)))),A(x,v6))).
66272 [hyper,1,712,55086] P(A(N(A(N(A(N(x),y)),A(z,A(u,v)))),A(N(A(N(z),u)),A(x,A(v,u))))).
255003 [hyper,1,7278,66272] P(A(A(N(A(N(x),y)),N(A(N(A(N(z),u)),A(x,A(y,v))))),A(z,A(v,y)))).
257552 [hyper,1,36769,255003] P(A(A(N(A(N(x),y)),A(z,A(u,y))),A(N(A(N(v),x)),N(A(N(A(N(y),w)),A(v,A(x,u))))))).
259289 [hyper,1,2870,257552] P(A(N(A(N(x),y)),A(N(A(N(A(N(z),u)),A(x,A(y,v)))),A(N(A(N(y),z)),A(w,A(v,z)))))).
265726 [hyper,1,5,259289] P(A(N(A(x,A(N(y),z))),A(A(N(A(N(A(N(u),v)),A(y,A(z,w)))),A(N(A(N(z),u)),A(v6,A(w,u)))),x))).
270214 [hyper,1,265726,144] P(A(A(N(A(N(A(N(x),y)),A(z,A(u,v)))),A(N(A(N(u),x)),A(w,A(v,x)))),A(z,v6))).
274964 [hyper,1,712,270214] P(A(N(A(N(A(N(x),y)),A(z,A(u,v)))),A(N(A(N(u),x)),A(z,A(v,x))))).
274965 [binary,274964.1,2.1] \$ANS(Meredith1).


[1]- W. McCune, "Prover9 and Mace4", http://www.cs.unm.edu/~mccune/Prover9, 2005-2014.
[2]- W. McCune, "OTTER and Mace2", http://www.mcs.anl.gov/research/projects/AR/otter/, 1988-2014.

This sort of problem (not this problem though, at least not in the following link) appears on the metamath site.

Ted Ulrich types this: "Can a fully-automated proof of the sufficiency of Meredith's 21-character single axiom, CCCCCpqCNrNsrtCCtpCsp [this isn't the same axiom as the original post, Meredith found a few single axioms which work for two-valued propositional logic] / ((((p⊃q)⊃(~r⊃~s))⊃r)⊃t)⊃((t⊃p)⊃(s⊃p)), for classical logic in C and N be found that equals (or shortens) Meredith's own derivation of Syl = CCpqCCqrCpr / ((p⊃q)⊃((q⊃r)⊃(p⊃r)), Scotus = CpCNpq / p⊃(~p⊃q), and Clavius = CCNppp / (~p⊃p)⊃p using just 41 condensed detachments?"

Ted Ulrich continued " UPDATE: Larry Wos has (with OTTER) now discovered a proof using only 38 applications of condensed detachment. See Automated reasoning and the discovery of missing and elegant proofs, L. Wos and G. Peiper, Rinton, Paramus, 2003, sections 3.1 through 3.3."

There's plenty of other questions of this type that one can ask, since there exist several sets of axioms for classical propositional calculus. Arthur Prior's textbook Formal Logic has three appendixes. The first consists of postulate sets for logical calculi. The second on methods of proof. The preface to the second edition reads "There is also abundant material for exercises in simply verifying some of the relations asserted to hold between postulate-sets in this [the first] Appendix, using to this end, the techniques sketched in the one that follows it."

That these sorts of problems have gotten done have enabled us to know that there exist several different ways to approach propositional calculi axiomatically (even with only one axiom), even with axiom sets where every axiom is independent of every other axiom.