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Implicational propositional calculus is a system of propositional calculus in which implication is the only logical connective, and all other connectives are defined with respect implication and a single false statement. Consider the system of implicational propositional calculus with the following two rules of inference: the Deduction Theorem, which states that if by assuming P you can conclude Q then P implies Q, and Modus Ponens, which states that if P and P implies Q then Q.

I'm guessing that this is not a complete system for the implicational propositional calculus, so my question is, what else do we need to add to make it complete? Is Peirce's Law all we need?

Any help would be greatly appreciated.

Thank You in Advance.

Keshav Srinivasan
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  • I think that some axioms (at least one) are needed. See Implicational propositional calculus. – Mauro ALLEGRANZA Feb 09 '14 at 08:25
  • You need something to separate this system from intuitionistic logic. I think excluded middle suffices? – Qiaochu Yuan Feb 09 '14 at 08:41
  • In the sub-entry List of logic systems you have that : "The positive implicational calculus is the implicational fragment of intuitionistic logic. The calculi below use modus ponens as an inference rule.", where "Intuitionistic logic [...] is commonly formulated with as the set of (functionally complete) basic connectives. It is not syntactically complete since it lacks excluded middle or Peirce's law ((A→B)→A)→A which can be added without making the logic inconsistent." – Mauro ALLEGRANZA Feb 09 '14 at 09:31
  • @QiaochuYuan Pierce's law implies the law of excluded middle, so the question is whether the Deuction Theorem + Modus Ponens + Pierce's law is complete. – Keshav Srinivasan Feb 09 '14 at 15:23
  • @MauroALLEGRANZA My system is different than the positive implicational calculus, because I have the deduction theorem rather than those other axioms. – Keshav Srinivasan Feb 09 '14 at 15:27
  • @QiaochuYuan The law of the excluded middle is a disjunction. The implicational propositional calculus only has implications. – Doug Spoonwood Feb 13 '14 at 02:36
  • @KeshavSrinivasan Pierce's law does NOT imply the law of the excluded middle here. There exist only implications in the implicational propositional calculus, and the law of the excluded middle is a disjunction. – Doug Spoonwood Feb 13 '14 at 02:37
  • @DougSpoonwood Yes, implication is the only logical connective in the implicational propositional calculus. But we can still define the other connectives in terms of implication, as long as we have a single false statement. See the Wikipedia article. – Keshav Srinivasan Feb 13 '14 at 09:44

4 Answers4

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In order to verifiy if Peirce's law is sufficient, when added to Deduction Theorem and modus ponens, we can try to verify if the (complete) axiom system for propositional logic of Elliott Mendelson, Introduction to Mathematical Logic (4th ed - 1997) [page 35] can be derived under these assumptions.

(A1) $\mathcal{B} \rightarrow ( \mathcal{C} \rightarrow \mathcal{B})$

We have that :

$B$ 1 --- assumption

$C$ 2 --- assumption

$B$ 3 --- assumption

$C \rightarrow B$ 4 --- deduction theorem from 2 and 3

$B \rightarrow (C \rightarrow B)$ 5 --- deduction theorem from 1 and 4.

(A2) $(\mathcal{B} \rightarrow ( \mathcal{C} \rightarrow \mathcal{D})) \rightarrow ((\mathcal{B} \rightarrow \mathcal{C}) \rightarrow (\mathcal{B} \rightarrow \mathcal{D}))$

We have that :

$B \rightarrow (C \rightarrow D)$ 1 --- assumption

$B \rightarrow C$ 2 --- assumption

$B$ 3 --- assumption

$C$ 4 --- modus ponens from 2 and 3

$C \rightarrow D$ 5 --- modus ponens from 1 and 3

$D$ 6 --- modus ponens from 4 and 5

$B \rightarrow D$ 7 --- deduction theorem from 3 and 6

$(B \rightarrow C) \rightarrow (B \rightarrow D)$ 8 --- deduction theorem from 2 and 7

$(B \rightarrow (C \rightarrow D)) \rightarrow ((B \rightarrow C) \rightarrow (B \rightarrow D))$ 9 --- deduction theorem from 1 and 8.

We still have to derive

(A3) $(\lnot \mathcal{C} \rightarrow \lnot \mathcal{B}) \rightarrow ((\lnot \mathcal{C} \rightarrow \mathcal{B}) \rightarrow \mathcal{C})$.

But, according to Implicational propositional calculus

the axiom system formed by (A1), (A2) and Peirce's law with the rule of inference modus ponens is semantically complete with respect to the usual two-valued semantics of classical propositional logic.

Note. Peirce's law is necessary, due to the fact that $\nvdash_{(A1)(A2)}(Peirce)$. The addition of Peirce’s law is sufficient, due to some results of Tarski and Bernays.

In order to have "full" propositional calculus, we have to add the falsum symbol ($\bot$) and an additional axiom : Ex Falso Quodlibet ($\bot \rightarrow \mathcal{A}$).

Mauro ALLEGRANZA
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  • "We still have to derive

    (A3) (¬C→¬B)→((¬C→B)→C)."

    No, you don't have to derive that. In fact, you shouldn't try to do so, because the implicational propositional calculus only has implication connectives in well-formed formulas.

     – Doug Spoonwood Feb 13 '14 at 02:44
  • @DougSpoonwood - yes, you are right. This is why I skipped the said derivation and jumped to the conclusion about Implicational propositional calculus. The issue was "reopened" in a new post about Propositional calculus. – Mauro ALLEGRANZA Feb 13 '14 at 08:44
  • @MauroALLEGRANZA Can you prove that if we have a false statement $\bot$ but we do not have ex falso quadlibet, then we won't get a complete system? – Keshav Srinivasan Feb 13 '14 at 18:20
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As Mauro's answer indicates you could join

  1. (⊥→A) [or C0a]

and Peirce's law

  1. (((A → B) → A) → A) [CCCabaa]

However, as Wajsberg's second paper on Metalogic of 1939 indicates (which you can find in the volume called "Polish Logic") you could just add the following formula as an axiom:

(((A → ⊥) → ⊥) → A) [CCCp00p]

Doug Spoonwood
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Not sure what exactly you mean , which book are you studying?

Positive implicational logic can mean many different things:

Classcal implicational logic (classical logic where implication is the only connective)

axioms (one of many, many possibilities)

  • $ P \to ( Q \to P) $
  • $ (P \to ( Q \to R)) \to (( P \to Q) \to ( P \to R )) $
  • $ ((P \to Q) \to P) \to P $

The above together with axioms for other connectives:

And:

  • $ P \to ( Q \to (P \land Q )) $ (and introduction)
  • $ (P \land Q ) \to P $ (and elimination)
  • $ (P \land Q ) \to Q $ (and elimination)

Or:

  • $ P \to (P \lor Q ) $ (or introduction)
  • $ P \to (Q \lor P ) $ (or introduction)
  • $ (P \to R) \to (( Q \to R) \to ((P \lor Q ) \to R)) $ (or elimination)

Or maybe something else alltogether, different authors have so there own names for the different systems.

Willemien
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  • This is the system I'm referring to: en.wikipedia.org/wiki/Implicational_propositional_calculus#Axiom_system The other connectives are just defined in terms of implication and a single false statement, as discussed in the article. So my question is, does the deduction theorem + modus ponens prove as much as that system, and if not what else do we need to add? – Keshav Srinivasan Feb 09 '14 at 20:21
  • @KeshavSrinivasan - in the Wikipedia article you are using : "Implication alone is not functionally complete as a logical operator because one cannot form all other two-valued truth functions from it. However, if one has a propositional formula which is known to be false and uses that as if it were a nullary connective for falsity, then one can define all other truth functions. So implication is virtually complete as an operator." The axioms are three, included Peirce's law, and modus ponens is needed. – Mauro ALLEGRANZA Feb 09 '14 at 21:51
  • @MauroALLEGRANZA Yes, that quote is exactly what I said - the other logical connectives are definable using implication and a single false statement. And yes, I'm aware that in the axiomatization given in the Wikipedia article, we have three axioms and then modus ponens. But my question is about what would happen if you replace those three axioms with the deduction theorem. And if the deduction theorem and modus ponens is not enough to get completeness, I want to know what else we need to add. – Keshav Srinivasan Feb 09 '14 at 22:21
  • @KeshavSrinivasan - in "standard" textbooks (like Mendelson) the two first axioms are used usually to pove "basic" theorems (like : $\vdash A \rightarrow A$) and then prove the DT. Are you able to derive the above only with Peirce's law ? – Mauro ALLEGRANZA Feb 09 '14 at 22:27
  • @MauroALLEGRANZA Are you asking whether the Deduction Theorem is provable using Pierce's law? I don't see what relevance that has. The question is what the deduction theorem can prove, not what is required to prove the deduction theorem. – Keshav Srinivasan Feb 09 '14 at 22:28
  • @KeshavSrinivasan - No; my question is : are enough Peirece'law and DT to deduce, for example $\vdash A \rightarrow A$ ? – Mauro ALLEGRANZA Feb 09 '14 at 22:35
  • @MauroALLEGRANZA Isn't that a trivial consequence of the deduction theorem? From A you can conclude that A, therefore A implies A. – Keshav Srinivasan Feb 09 '14 at 23:23
  • @KeshavSrinivasan Not sure what you mean: modus ponens (as inference rule ) together with the implication axioms give (axiomatic) classical logic, you can add the other connectives either by axioms or definitions. the deduction theorem , is a (metalogical) inference rule but is a bit deductable from modus ponens and the given axioms. $ \vdash P \to p $ is provable from the first two implication axioms (but the proof is rather complex, all axiomatic proofs are) does this answer your questions? – Willemien Feb 10 '14 at 00:14
  • @Willemien Yes, if you start with those axioms and Modus Ponens then you can prove the deduction theorem. But I want to go the other way: start with the deduction theorem and Modus Ponens and then prove those axioms, and if they don't suffice for this purpose, then what axioms do we need to add two those two inference rules to prove the standard axioms for implicational propositional calculus? – Keshav Srinivasan Feb 10 '14 at 01:31
  • @KeshavSrinivasan - perfect. Now, if you go on with the three other axioms of Mendelson, due to the fact that those axioms (with modus ponens) are enough for a complete proof system for propositional logic (Completeness Theorem), it's done. – Mauro ALLEGRANZA Feb 10 '14 at 07:18
  • @MauroALLEGRANZA That was the whole point of my question - to find out whether the Deduction Theorem plus Modus Ponens suffices to prove the standard axioms of implicational propositional calculus, or whether we need to add something else. – Keshav Srinivasan Feb 10 '14 at 09:38
  • I guess (but it is just a guess) that you can remove the axioms $ P \to ( Q \to P) $ and $ (P \to ( Q \to R)) \to (( P \to Q) \to ( P \to R )) $ , because I think you can proof them from modus ponens and the deduction theorem, but you still need the other axioms. (otherwise how do you use the other connectives) but this is all just a guess. – Willemien Feb 10 '14 at 16:22
  • @Willemien - of course, $\rightarrow$ is not enough for express all the "classical" propositional calculus. If you want to extend the implicational fragment to the full calculus, you have to use $\bot$ and define as usual $\lnot p$ as $p \rightarrow \bot$. In this way, you can derive (A3) of Mendelson, and it's done. – Mauro ALLEGRANZA Feb 10 '14 at 17:26
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You don't need to add more axioms actually here. Instead of adding Peirce's law CCCpqpp or some other formula as an axiom, you could just add the rule

CC$\alpha$$\beta$$\alpha$ $\vdash$ $\alpha$.

Doug Spoonwood
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  • I don't care about whether Pierce's law is added as an axiom or a rule of inference. All I'm interested is whether the addition of Pierce's law is necessary and sufficient to get a complete system. – Keshav Srinivasan May 26 '14 at 17:35
  • @KeshavSrinivasan The axiom and the rule of inference aren't the same thing. The axiom happens in the object language, the rule in the metalanguage. Adding either does allow you get a complete system for all formulas with just implications, because since you have the deduction meta-theorem you can move from the rule to the formula. – Doug Spoonwood May 26 '14 at 19:13