Difference between revisions of "Positive propositional calculus"
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− | A [[Propositional calculus|propositional calculus]] in the language | + | {{TEX|done}} |
+ | A [[Propositional calculus|propositional calculus]] in the language $\{\&,\lor,\supset\}$ specified by the following 8 axiom schemes: | ||
− | + | $$A\supset(B\supset A),\quad(A\supset(B\supset C))\supset((A\supset B)\supset(A\supset C)),$$ | |
− | + | $$A\&B\supset A,\quad A\&B\supset B,\quad A\supset(B\supset A\&B),$$ | |
− | + | $$A\supset A\lor B,\quad B\supset A\lor B,\quad(A\supset C)\supset((B\supset C)\supset(A\lor B)\supset C),$$ | |
− | and the [[Modus ponens|modus ponens]] derivation rule. This calculus contains the part of the intuitionistic propositional calculus I (see [[Intuitionism|Intuitionism]]) that is not dependent on negation: Any propositional formula not containing | + | and the [[Modus ponens|modus ponens]] derivation rule. This calculus contains the part of the intuitionistic propositional calculus I (see [[Intuitionism|Intuitionism]]) that is not dependent on negation: Any propositional formula not containing $\neg$ (negation) is derivable in the positive propositional calculus if and only if it is derivable in I. One obtains the calculus I if one adds two axiom schemes to the positive propositional calculus: |
− | 1) | + | 1) $\neg A\supset(A\supset B)$ (antecedent negation law), |
− | 2) | + | 2) $(A\supset B)\supset((A\supset\neg B)\supset\neg A)$ (reductio ad absurdum law). |
To derive I, instead of 2) one can take the weaker scheme: | To derive I, instead of 2) one can take the weaker scheme: | ||
− | 2') | + | 2') $(A\supset\neg A)\supset\neg A$ (law of partial reductio ad absurdum). |
See also [[Implicative propositional calculus|Implicative propositional calculus]]. | See also [[Implicative propositional calculus|Implicative propositional calculus]]. |
Latest revision as of 08:00, 12 August 2014
A propositional calculus in the language $\{\&,\lor,\supset\}$ specified by the following 8 axiom schemes:
$$A\supset(B\supset A),\quad(A\supset(B\supset C))\supset((A\supset B)\supset(A\supset C)),$$
$$A\&B\supset A,\quad A\&B\supset B,\quad A\supset(B\supset A\&B),$$
$$A\supset A\lor B,\quad B\supset A\lor B,\quad(A\supset C)\supset((B\supset C)\supset(A\lor B)\supset C),$$
and the modus ponens derivation rule. This calculus contains the part of the intuitionistic propositional calculus I (see Intuitionism) that is not dependent on negation: Any propositional formula not containing $\neg$ (negation) is derivable in the positive propositional calculus if and only if it is derivable in I. One obtains the calculus I if one adds two axiom schemes to the positive propositional calculus:
1) $\neg A\supset(A\supset B)$ (antecedent negation law),
2) $(A\supset B)\supset((A\supset\neg B)\supset\neg A)$ (reductio ad absurdum law).
To derive I, instead of 2) one can take the weaker scheme:
2') $(A\supset\neg A)\supset\neg A$ (law of partial reductio ad absurdum).
See also Implicative propositional calculus.
References
[1] | A. Church, "Introduction to mathematical logic" , 1 , Princeton Univ. Press (1956) |
[2] | D. Hilbert, P. Bernays, "Grundlagen der Mathematik" , 1–2 , Springer (1968–1970) |
Positive propositional calculus. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Positive_propositional_calculus&oldid=13329