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Can Hilbert (style) axioms prove the following tautology?

$$A\wedge(C\rightarrow B)\oplus(A\wedge B\leftrightarrow C)\rightarrow(C\rightarrow(A\rightarrow B))\qquad\text{algebraic style} $$$$ (((A\wedge(C\rightarrow B))\oplus((A\wedge B)\leftrightarrow C)))\rightarrow(C\rightarrow(A\rightarrow B))\qquad\text{formal style} $$

Here $\wedge$ means logical and, $\oplus$ means exclusive-or, and $\leftrightarrow$ means logical equivalence.

Due to Wikipedia/Hilbert style the axioms below describe classical propositional logic:

P1. $\phi \to \phi$

P2. $\phi \to \left( \psi \to \phi \right)$

P3. $\left( \phi \to \left( \psi \rightarrow \xi \right) \right) \to \left( \left( \phi \to \psi \right) \to \left( \phi \to \xi \right) \right)$

P4. $\left ( \lnot \phi \to \lnot \psi \right) \to \left( \psi \to \phi \right)$

How to go? Is it necessary to add definitions of connectives?

hardmath
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Lehs
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    Axioms that do not use all the connectives you do will not prove any tautology that depends on your connectives. – Colin McLarty Sep 26 '14 at 17:09
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    @Colin McLarty: but if I add a list of transformations? – Lehs Sep 26 '14 at 17:12
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    The title isn't suposed to replace the first line of your question. The answer is yes anyway. – Git Gud Sep 26 '14 at 17:15
  • @Git Gud: what do you mean? Should the first line be in the title? – Lehs Sep 26 '14 at 17:31
  • I mean that in the previous version of the question, there was no question at all, unless one reads the title. Are you asking for a formal proof of the statement? – Git Gud Sep 26 '14 at 17:34
  • @Git Gud: I understand, there should be an explicit question in the text. Well, I sure would like to see some sketch of a proof. – Lehs Sep 26 '14 at 17:38
  • Personally I think no one would attempt this, it's ridiculously long. Note that the statement is written with a lot of abbreviations (you're only supposed to use $\to$ and $\neg$). Rewriting the statement in the given language alone makes it awfully long, let alone a proof of it. – Git Gud Sep 26 '14 at 17:42
  • @Lehs Hilbert's axioms are complete so yes if you either regard your connectives as abbreviations for expressions in Hilbert's calculus, or add axioms completely characterizing you connectives in terms of his, then his axioms will derive this tautology. – Colin McLarty Sep 26 '14 at 18:56
  • @Colin McLarty: thanks for the answer and the question! "How to prove that the Hilbert axioms can prove any tautology?" – Lehs Sep 26 '14 at 19:02
  • @ColinMcLarty There is no axiom set for all connectives. It might comes as possible to come up with a sufficient axiom set for all 4 unary connectives and all 16 binary connectives in 2-valued logic, but even then more work would come as needed to set axioms for all ternary, 4-ary, etc. connectives. – Doug Spoonwood Oct 01 '14 at 22:05
  • What is $\oplus$ ? –  Oct 01 '14 at 23:46
  • @Christian Remling: $\oplus$ is XOR. – Lehs Oct 01 '14 at 23:47
  • The semantical approach that has been suggested a few times already seems best I think (check that your statement is valid for arbitrary truth values of $A,B,C$ and then refer to completeness). –  Oct 01 '14 at 23:59
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    @Doug Spoonwood: surely the set of all true statements of propositional logic, in the propositional language with all possible connectives, would be a complete and computable set of axioms for that language. – Carl Mummert Oct 02 '14 at 10:58
  • It's unclear to me what precedence "and" versus "xor" will have in the antecedent of the proposition. Shouldn't we have an extra pair of parenthesis (or more) to clear things up in $[A\wedge(C\rightarrow B)\oplus(A\wedge B\leftrightarrow C)]$ ? – hardmath Oct 02 '14 at 13:20
  • @hardmath: I guess you're right, but I myself think of $\wedge$ as multiplication and $\oplus$ as addition in an algebraic style. I'll edit. – Lehs Oct 02 '14 at 15:59
  • @Lehs: Thanks, that clears things up enough for me. I'll nominate it for reopening. – hardmath Oct 02 '14 at 17:00
  • @CarlMummert I find your statement interesting. In such a system you would only list out some of the axioms/axiom schema before writing a proof. All proofs would show a path from a proper and finite subset of the axioms/axiom schema. – Doug Spoonwood Oct 02 '14 at 17:40
  • @hardmath: thanks for that! – Lehs Oct 02 '14 at 19:13
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    I'm trying to get OTTER to prove this using the subformula strategy right now. As a longer C-N tautology to prove, try to prove that EEpEqrEEpqr where Epq abbreviates NCCpqNCqp. Doing the transformations here we have EEpNCCqrNCrqENCCpqNCqpr, then ENCCpNCCqrNCrqNCNCCqrNCrqpNCCNCCpqNCqprNCrNCCpqNCqp and finally NCCNCCpNCCqrNCrqNCNCCqrNCrqpNCCNCCpqNCqprNCrNCCpqNCqpNCNCCNCCpqNCqprNCrNCCpqNCqpNCCpNCCqrNCrqNCNCCqrNCrqp. I counted 58 symbols in the CJKaCcbEKabcCcCab tautology. I count 106 symbols in the "E" tautology EEpEqrEEpqr. – Doug Spoonwood Oct 23 '14 at 18:56
  • @Dough, I really appreciate LIFO-stacks and the parenthesis free notation (most the reversed form) of Lukasiewicz, but doesn't it seems contra-productive to minimize the number of axioms and start all proofs from that ascetic position? If the axioms for Boolean rings was deduced, once for all, the calculations would be simplified and the needed effort minimized. – Lehs Oct 23 '14 at 19:47
  • The axioms for Boolean rings are equational. There is no notion of equality in propositional calculus. The primitive rules of inference usable differ from an equational theory (where you can use replacement as a primitive rule of inference) to a Hilbert-Frege propositional calculus with uniform substitution and detachment. Replacement can work out as a derivable rule of inference in plenty of propositional calculi, but that is not always the case. Replacement is always a rule of inference in an equational theory such as that of Boolean rings (so long as you fully parenthesize everything). – Doug Spoonwood Oct 23 '14 at 20:39
  • Formal first-order theories presuppose a propositional logic. A propositional calculus basically only presupposes rules of inference and certain axioms. Minimizing the number of axioms and rules of inference means that you've assumed less when producing a proof. A propositional calculus proof assumes less than a proof that something equals the special element "1", where (a+-a)=1, in a Boolean ring. – Doug Spoonwood Oct 23 '14 at 20:51
  • @Dough: I am sure you are right but I'm equally sure that you are wrong. Equations constructed with a mid equivalence do works with replacement of equivalent expressions and I'm sure that all axioms of this form for Boolean rings can be proved by minimized systems. And the "special element" can be replaced by $a\rightarrow a$. And any propositional tautology can be proved by calculations in Boolean rings (just as with proof-tables). – Lehs Oct 23 '14 at 22:09
  • "I'm sure that all axioms of this form for Boolean rings can be proved by minimized systems" I don't know what you mean by "minimized system" here. I know that I would mean some system with just one rule of inference and one axiom. Or at least one axiom and very few rules of inference. There do exist propositional calculi minimized in that sense that they have one rule of inference and one axiom. There also exist single axioms for Boolean Algebra (I don't know about the rules part there though)... http://www.cs.unm.edu/~mccune/papers/basax/v12.pdf – Doug Spoonwood Oct 23 '14 at 22:38
  • While working through converting a "natural deduction" proof to an axiomatic proof using condensed detachment, it seems that [(((A∧(C→B))⊕((A∧B)↔C)))→(C→(A→B))] is a special case of the more general tautology [(((A∧(C→B))⊕((A∧B)↔C)))→(D→(A→B))] [note the "D"]. I've confirmed this via truth table. Also, suppose it false. Then (D→(A→B)) is false, and thus D is true and (A→B) is false. So, A is true, and B false. The antecedent simplifies to [(C→0)⊕(0↔C)], where "0" indicates falsity. [(0→0)⊕(0↔0)]=[1⊕1]=0. [(1→0)⊕(0↔1)]=[0⊕0]=0. Therefore, the more general tautology holds. – Doug Spoonwood Oct 24 '14 at 16:40
  • You're right about the generalization @Dough, I transformed your expression via Zhegalkin polynomials into a Boolean ring and the resulting polynomial became 1. – Lehs Oct 24 '14 at 17:33
  • @Dough: Also $((A\wedge(C\Rightarrow B))\oplus((A\wedge B)\Leftrightarrow C))\Rightarrow(A \Rightarrow B)$ is a tautology. – Lehs Oct 24 '14 at 23:31
  • CCCNCxNCyzuNCNCvNCyNCwNzdCyCez is also more general. – Doug Spoonwood Oct 26 '14 at 06:34

3 Answers3

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You can see Elliott Mendelson, Introduction to mathematical logic (4ed - 1997) : PROPOSITION 1.14 (COMPLETENESS THEOREM).

The proof (due to Kalmar,1935) allows us to build a proof of a tautology $\mathcal B$ whatever.

Assume $\mathcal B$ is a tautology, and let $B_1, \ldots, B_k$ be the statement letetrs in $\mathcal B$. For any truth value assignment to $B_1, \ldots, B_k$ we have :

$B_1', \ldots, B_k' \vdash \mathcal B$

where let $B_i'$ be $B_i$ if $B_i$ takes the value $T$, and let $B_i'$ be $\lnot B_i$ if $B_i$ takes the value $F$.

Thus, due to the fact that your formula $\mathcal F$ has only three statement letters : $A,B,C$, it is enough to rewrite it in "primitive" notation, i.e. using only $\rightarrow$ and $\lnot$ (call it : $\mathcal F'$) and then apply the procedure described in the above Proposition.

Starting from example from the truth value assignment $v(A)=v(B)=v(C)=T$, we have :

$A,B,C \vdash \mathcal F'$.

Mauro ALLEGRANZA
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I use condensed detachment notated by "D" in the proof analysis. Dx.y indicates that condensed detachment gets applied to formula x with form Cx y, and y has form x' such that x and x' have a most general unifier.

So we want to prove CCCNCaNCcbNCCNCaNbcNCcNCaNbNCNCCNCaNbcNCcNCaNbNCaNCcbCcCab. I'll note that as I worked through this proof I had to several times find the antecedent of a formula, or where particular formulas would "match". The simplicity of the algorithm for formulas in Lukasiewicz notation helped a lot! Here's a proof:

axiom      1 CxCyx. 
axiom      2 CCxCyzCCxyCxz. 
axiom      3 CCNxNyCyx. 
D1.1       4 CxCyCzy.
D2.2       5 CCCxCyzCxyCCxCyzCxz.
D1.5       6 CxCCCyCzuCyzCCyCzuCyu.
D2.1       7 CCxyCxx.
D7.1       8 Cxx.
D2.3       9 CCCNxNyyCCNxNyx.
D1.3      10 CxCCNyNzCzy.
D1.4      11 CxCyCzCuz.
D2.8      12 CCCxyxCCxyy.
D2.10     13 CCxCNyNzCxCzy.
D13.1     14 CNxCxy.
D13.14    15 CNNxCyx.
D2.14     16 CCNxxCNxy.
D1.14     17 CxCNyCyz.
D12.17    18 CCCNxCxyzz. 
D5.11     19 CCxCCyCzyuCxu.
D5.4      20 CCxCCyxzCxz.
D20.15    21 CNNxx.
D1.2      22 CxCCyCzuCCyzCyu.
D2.22     23 CCxCyCzuCxCCyzCyu.
D23.1     24 CCxyCCzxCzy.
D3.21     25 CxNNx.
D24.25    26 CCxyCxNNy.
D18.26    27 CNxNNCxy.
D13.26    28 CCNxyCNyx.
D3.27     29 CNCxyx.
D28.15    30 CNCxyNy.

break

D19.24    31 CCCxyzCyz.
D31.9     32 CxCCNyNxy.
D2.32     33 CCxCNyNxCxy.
D33.16    34 CCNxxx.
D1.23     35 CxCCyCzCuvCyCCzuCzv.
D2.35     36 CCxCyCzCuvCxCyCCzuCzv.
D1.36     37 CxCCyCzCuCvwCyCzCCuvCuw.
D2.37     38 CCxCyCzCuCvwCxCyCzCCuvCuw.
D1.38     39 CxCCyCzCuCvCwdCyCzCuCCvwCvd.
D2.39     40 CCxCyCzCuCvCwdCxCyCzCuCCvwCvd.
D1.21     41 CxCNNyy.
D1.41     42 CxCyCNNzz.
D1.42     43 CxCyCzCNNuu.
D1.43     44 CxCyCzCuCNNvv.
D1.44     45 CxCyCzCuCvCNNww. 
D40.45    46 CxCyCzCuCCvNNwCvw. 
D38.46    47 CxCyCzCCuCvNNwCuCvw. 
D1.30     48 CxCNCyzNz.
D1.48     49 CxCyCNCzuNu.
D1.49     50 CxCyCzCNCuvNv.
D1.50     51 CxCyCzCuCNCvwNw. 
D36.47    52 CxCyCCzCuCvNNwCzCuCvw. 
D23.52    53 CxCCyCzCuCvNNwCyCzCuCvw. 
D2.53     54 CCxCyCzCuCvNNwCxCyCzCuCvw. 
D54.51    55 CxCyCzCuCNCvNww. 
D24.1     56 CCxyCxCzy. 
D56.1     57 CxCyCzx. 
D1.57     58 CxCyCzCuy. 
D1.56     59 CxCCyzCyCuz.
D2.59     60 CCxCyzCxCyCuz.

break

D60.58    61 CxCyCzCuCvy. 
D40.55    62 CxCyCzCuCCNCvNCwdwCNCvNCwdd. 
D38.62    63 CxCyCzCCuCNCvNCwdwCuCNCvNCwdd. 
D36.63    64 CxCyCCzCuCNCvNCwdwCzCuCNCvNCwdd. 
D23.64    65 CxCCyCzCuCNCvNCwdwCyCzCuCNCvNCwdd. 
D2.65     66 CCxCyCzCuCNCvNCwdwCxCyCzCuCNCvNCwdd. 
D66.61    67 CxCyCzCuCNCvNCyww. 
D1.17     68 CxCyCNzCzu.
D1.68     69 CxCyCzCNuCuv.
D1.69     70 CxCyCzCuCNvCvw.
D1.70     71 CxCyCzCuCvCNwCwd. 
D40.71    72 CxCyCzCuCCvNwCvCwd. 
D38.72    73 CxCyCzCCuCvNwCuCvCwd. 
D36.73    74 CxCyCCzCuCvNwCzCuCvCwd. 
D23.74    75 CxCCyCzCuCvNwCyCzCuCvCwd. 
D1.11     76 CxCyCzCuCvu. 
D2.75     77 CCxCyCzCuCvNwCxCyCzCuCvCwd.
D77.76    78 CxCyCzCNuCvCuw. 
D40.78    79 CxCyCzCNuCCvuCvw. 
D38.79    80 CxCyCzCCNuCvuCNuCvw. 
D36.80    81 CxCyCCzCNuCvuCzCNuCvw. 
D34.81    82 CxCCyCzCNuCvuCyCzCNuCvw. 
D2.82     83 CCxCyCzCNuCvuCxCyCzCNuCvw. 
D83.67    84 CxCyCzCNuCNCvNCyuw. 
D56.57    85 CxCyCzCux. 
D38.85    86 CCxyCzCuCCvxCvy. 
D36.86    87 CCxyCzCCuCvxCuCvy. 
D23.87    88 CCxyCCzCuCvxCzCuCvy. 
D2.88     89 CCCxyCzCuCvxCCxyCzCuCvy.
D89.84    90 CCCNCxNCyzuvCyCwCNzv.

break 

D1.29     91 CxCNCyzy.
D1.91     92 CxCyCNCzuz.
D1.92     93 CxCyCzCNCuvu.
D1.93     94 CxCyCzCuCNCvwv. 
D38.94    95 CxCyCzCCuNCvwCuv. 
D36.95    96 CxCyCCzCuNCvwCzCuv. 
D23.96    97 CxCCyCzCuNCvwCyCzCuv. 
D2.97     98 CCxCyCzCuNCvwCxCyCzCuv.
D98.90    99 CCCNCxNCyzuNCvwCyCdCNzv. 
D38.51   100 CxCyCzCCuNCvwCuNw. 
D36.100  101 CxCyCCzCuNCvwCzCuNw. 
D34.101  102 CxCCyCzCuNCvwCyCzCuNw. 
D2.102   103 CCxCyCzCuNCvwCxCyCzCuNw.
D103.99  104 CCCNCxNCyzuNCNCvwdCyCeCNzNw. 
D38.44   105 CxCyCzCCuNNvCuv. 
D36.105  106 CxCyCCzCuNNvCzCuv. 
D23.106  107 CxCCyCzCuNNvCyCzCuv. 
D2.107   108 CCxCyCzCuNNvCxCyCzCuv.
D108.104 109 CCCNCxNCyzuNCNCvNwdCyCeCNzw. 
D38.109  110 CCCNCxNCyzuNCNCvNCwdeCyCfCCNzwCNzd. 
D36.110  111 CCCNCxNCyzuNCNCvNCwdeCyCCfCNzwCfCNzd. 
D23.111  112 CCCNCxNCyzuNCNCvNCwdeCCyCfCNzwCyCfCNzd.   
D2.112   113 CCCCNCxNCyzuNCNCvNCwdeCyCfCNzwCCCNCxNCyzuNCNCvNCwdeCyCfCNzd.
D113.11  114 CCCNCxNCyzuNCNCvNCwdeCyCwCNzd.
D23.114  115 CCCNCxNCyzuNCNCvNCwdeCCywCyCNzd. 
D2.115   116 CCCCNCxNCyzuNCNCvNCwdeCywCCCNCxNCyzuNCNCvNCwdeCyCNzd.
D1.8     117 CxCyy
D116.117 118 CCCNCxNCyzuNCNCvNCywdCyCNzw. 
D1.56    119 CxCCyzCyCuz.            
D2.119   120 CCxCyzCxCyCuz.

break

D120.118 121 CCCNCxNCyzuNCNCvNCywdCyCeCNzw.  
D23.101  122 CxCCyCzCuNCvwCyCzCuNw.  
D2.122   123 CCxCyCzCuNCvwCxCyCzCuNw. 
D123.121 124 CCCNCxNCyzuNCNCvNCyNCwdeCyCfCNzNd.
D108.124 125 CCCNCxNCyzuNCNCvNCyNCwNdeCyCfCNzd. 
D1.34    126 CxCCNyyy.
D1.126   127 CxCyCCNzzz.
D1.127   128 CxCyCzCCNuuu. 
D36.128  129 CxCyCCzCNuuCzu. 
D23.129  130 CxCCyCzCNuuCyCzu. 
D2.130   131 CCxCyCzCNuuCxCyCzu.
D131.125 132 CCCNCxNCyzuNCNCvNCyNCwNzdCyCez.

You might want to observe that 132 has eight variables. $\alpha$/$\beta$ indicates that formula $\alpha$ gets substituted with $\beta$.

132 x/a  133 CCCNCaNCyzuNCNCvNCyNCwNzdCyCez.
133 y/c  134 CCCNCaNCczuNCNCvNCcNCwNzdCcCez.
134 z/b  135 CCCNCaNCcbuNCNCvNCcNCwNbdCcCez.
135 u/NCCNCaNbcNCcNCaNb   136 CCCNCaNCcbNCCNCaNbcNCcNCaNbNCNCvNCcNCwNbdCcCeb.
136 v/CNCaNbc 137 CCCNCaNCcbNCCNCaNbcNCcNCaNbNCNCCNCaNbcNCcNCwNbdCcCeb.
137 w/a  138 CCCNCaNCcbNCCNCaNbcNCcNCaNbNCNCCNCaNbcNCcNCaNbdCcCeb.
138 d/NCaNCcb 139 CCCNCaNCcbNCCNCaNbcNCcNCaNbNCNCCNCaNbcNCcNCaNbNCaNCcbCcCeb.

So, CCCNCxNCyzuNCNCvNCyNCwNzdCyCez is a more general formula than 139. How did I prove something like CCCNCaNCcbNCCNCaNbcNCcNCaNbNCNCCNCaNbcNCcNCaNbNCaNCcbCcCab? First I set up a "natural deduction" style proof where all of the variables behave like constants. Then I assumed the antecedents "CCNCaNCcbNCCNCaNbcNCcNCaNbNCNCCNCaNbcNCcNCaNbNCaNCcb", "c", and "b". I thought that if I assumed the negation of the conclusion "Nb" I could then derive a contradiction. Then I would use the formula CCNxNyCCNxyx to get to "b". But how to do that? I've found that sometimes when you have a formula like CCNCaNCcbNCCNCaNbcNCcNCaNbNCNCCNCaNbcNCcNCaNbNCaNCcb, it can help to try and prove that its antecedent holds true... in this case "CNCaNCcbNCCNCaNbcNCcNCaNb". Then we can detach its consequent "CNCaNCcbNCCNCaNbcNCcNCaNb". So, if we assume NCaNCcb, can we get to NCCNCaNbcNCcNCaNb? Well, if we assume any negation of a conditional NCxy, we can infer that "x" holds and "Ny" holds. Now I did find another path than I originally got on, but trying to prove the antecedent of an assumed conditional did help, and inferring from a formula of the form NCxy to "x" and "Ny" did help here. Here's what I did for the set-up:

I used some lemmas:

L1. CNNxx.
L2. CNxCxy.
L3. CNCxyNy.
L4. CNCxyx.
L5. CCNxxx.

and set up the proof with this deduction:

hypothesis 1 | CCNCaNCcbNCCNCaNbcNCcNCaNbNCNCCNCaNbcNCcNCaNbNCaNCcb. (ante)
hypothesis 2 || c.
hypothesis 3 ||| a.
hypothesis 4 |||| Nb.
hypothesis 5 ||||| NCaNCcb.
D5.L3      6 ||||| NNCcb.
D(L1).6    7 ||||| Ccb.
D7.2       8 ||||| b.
D(L2).4    9 ||||| Cbx.
D9.8      10 ||||| x.
Ci 5-10   11 |||| CNCaNCcbx.
D1.11     12 |||| NCNCCNCaNbcNCcNCaNbNCaNCcb.
D(L4).12  13 |||| NCCNCaNbcNCcNCaNb
D(L3).13  14 |||| NNCcNCaNb.
D(L1).14  15 |||| CcNCaNb.
D15.2     16 |||| NCaNb.
D(L3).16  17 |||| NNb.
D(L1).17  18 |||| b.
Ci 4-18   19 ||| CNbb.
D(L5).19  20 ||| b.
Ci 3-20   21 || Cab.
Ci 2-21   22 | CcCab.
Ci 1-22   23 CCCNCaNCcbNCCNCaNbcNCcNCaNbNCNCCNCaNbcNCcNCaNbNCaNCcbCcCab.

Then I used the conversion procedure outlined by standard proofs of the Deduction Meta-Theorem along with condensed detachment.

Community
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Doug Spoonwood
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-1

To do this you would want to first define the other connectives. I will use Lukasiewicz/Polish notation. For your "∧" I write "K". For your "→" I write "C". For your "⊕", following A. N. Prior, I will write "J". For "$\lnot$", "N". For "↔" I will write E.

The axioms become

P2 CpCqp.

P3 CCpCqrCCpqCpr.

P4 CCNpNqCqp.

I will note that P1 is NOT an axiom given that a requirement for axioms is that they are independent, and I simply fail to understand this new standard where axioms don't have to be independent.

Now we'll need definitions for the other connectives, which in effect will end up behaving like axioms. The symbol "$\delta$" indicated a variable function of one argument. This means we can make substitutions for $\delta$.

PK: C$\delta$Kpq$\delta$NCpNq.

PE: C$\delta$Epq$\delta$NCCpqNCqp.

PJ: C$\delta$Jpq$\delta$CCpqNCqp.

Your formula is CJKaCcbEKabcCcCab.

Using definition PK we can tell that CJKaCcbEKabcCcCab is equivalent to

CJNCaNCcbENCaNbcCcCab.

Using definition PE we can tell that CJNCaNCcbENCaNbcCcCab is equivalent to

CJNCaNCcbNCCNCaNbcNCcNCaNbCcCab.

Using definiton PJ we can tell that CJNCaNCcbNCCNCaNbcNCcNCaNbCcCab is equivalent to

CCCNCaNCcbNCCNCaNbcNCcNCaNbNCNCCNCaNbcNCcNCaNbNCaNCcbCcCab.

To try to prove it using Prover9, or OTTER you'll only need to parenthesize all of the axioms and use a unary predicate (which you may think of as meaning "provable" or "is formally provable"), and rewrite the variables as "x", "y" and "z" respectively for "a", "c", and "b" or "p", "q" and "r"... something like this:

P2: P(C(x,C(y,x))).

P3: P(C(C(x,C(y,z)),C(C(x,y),C(x,z)))).

P4: P(C(C(N(x),N(y)),C(y,x))).

Parenthesize P(CCCNCxNCzyNCCNCxNyzNCzNCxNyNCNCCNCxNyzNCzNCxNyNCxNCzyCzCxy), and add this as the goal. This formula is:

 P(C(C(C(N(C(x,N(C(z,y)))),N(C(C(N(C(x,N(y))),z),N(C(z,N(C(x,N(y)))))))),N(C(N(C(C(N(C(x,N(y))),z),N(C(z,N(C(x,N(y))))))),N(C(x,N(C(z,y))))))),C(z,C(x,y)))).

You'll also want an assumption which enables Prover9 to prove things like it were using condensed detachment:

-P(C(x,y)) | -P(x) | P(y). (it is not the case that either it is provable that x implies y, or it is not the case that x is provable, or y is provable). Then you might find a proof equivalent to a proof using condensed detachment. You'll want to prove that since you have

PK: C$\delta$Kpq$\delta$NCpNq.

PE: C$\delta$Epq$\delta$NCCpqNCqp.

PJ: C$\delta$Jpq$\delta$CCpqNCqp.

you can prove

PK reversed: C $\delta$NCpNq $\delta$Kpq.

PE reversed: C $\delta$NCCpqNCqp $\delta$Epq.

PJ reversed: C $\delta$CCpqNCqp $\delta$Jpq.

And it does come as possible to prove each of those, since it has gotten proven that from

C $\delta$ P $\delta$ R

we can derive

C $\delta$ R $\delta$ P

where "R" and "P" are constants, and we only have substitution and detachment as the rules of inference.

Another approach could involve proving CCNpqCCNpNqp (or CCNpNqCCNpqp). Then one could set up a natural-deduction like demonstration of the C-N fomrula you've referred to, and then use the proof of the deduction meta-theorem to produce a formal proof.

Doug Spoonwood
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