Namespaces
Variants
Actions

Difference between revisions of "Implicative propositional calculus"

From Encyclopedia of Mathematics
Jump to: navigation, search
(Importing text file)
 
m (tex encoded by computer)
 
Line 1: Line 1:
A [[Propositional calculus|propositional calculus]] using only the primitive connective <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050300/i0503001.png" /> (implication). Examples of an implicative propositional calculus are the complete (or classical) implicative propositional calculus given by the axioms
+
<!--
 +
i0503001.png
 +
$#A+1 = 7 n = 0
 +
$#C+1 = 7 : ~/encyclopedia/old_files/data/I050/I.0500300 Implicative propositional calculus
 +
Automatically converted into TeX, above some diagnostics.
 +
Please remove this comment and the {{TEX|auto}} line below,
 +
if TeX found to be correct.
 +
-->
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050300/i0503002.png" /></td> </tr></table>
+
{{TEX|auto}}
 +
{{TEX|done}}
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050300/i0503003.png" /></td> </tr></table>
+
A [[Propositional calculus|propositional calculus]] using only the primitive connective  $  \supset $(
 +
implication). Examples of an implicative propositional calculus are the complete (or classical) implicative propositional calculus given by the axioms
 +
 
 +
$$
 +
p  \supset  ( q \supset p ) ,\ \
 +
( ( p \supset q )  \supset \
 +
( ( q \supset r )  \supset \
 +
( p \supset r ) ) ) ,
 +
$$
 +
 
 +
$$
 +
( ( p \supset q )  \supset  p )  \supset  p
 +
$$
  
 
and the rules of inference: [[Modus ponens|modus ponens]] and substitution; another example is the positive implicative propositional calculus given by the axioms
 
and the rules of inference: [[Modus ponens|modus ponens]] and substitution; another example is the positive implicative propositional calculus given by the axioms
  
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050300/i0503004.png" /></td> </tr></table>
+
$$
 +
p  \supset  ( q \supset p ) ,\ \
 +
( p  \supset  ( q \supset r ) )  \supset \
 +
( ( p \supset q )  \supset \
 +
( p \supset r ) )
 +
$$
  
and the same rules of inference. Every implicative formula, that is, a formula only containing the connective <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050300/i0503005.png" />, is deducible in complete (or positive) implicative propositional calculus if and only if it is deducible in classical (respectively, intuitionistic) propositional calculus. For any finite set <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050300/i0503006.png" /> of variables, among the formulas with variables in <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/i/i050/i050300/i0503007.png" /> there is only a finite number of pairwise inequivalent ones in the positive implicative propositional calculus (see [[#References|[3]]]). There exist undecidable finitely-axiomatizable implicative propositional calculi (see [[#References|[4]]]).
+
and the same rules of inference. Every implicative formula, that is, a formula only containing the connective $  \supset $,  
 +
is deducible in complete (or positive) implicative propositional calculus if and only if it is deducible in classical (respectively, intuitionistic) propositional calculus. For any finite set $  V $
 +
of variables, among the formulas with variables in $  V $
 +
there is only a finite number of pairwise inequivalent ones in the positive implicative propositional calculus (see [[#References|[3]]]). There exist undecidable finitely-axiomatizable implicative propositional calculi (see [[#References|[4]]]).
  
 
====References====
 
====References====
 
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  A. Church,  "Introduction to mathematical logic" , '''1''' , Princeton Univ. Press  (1956)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  J. Łukasiewicz,  A. Tarski,  "Untersuchungen über den Aussagenkalkül"  ''C.R. Soc. Sci. Letters Varsovie, Cl. III'' , '''23'''  (1930)  pp. 30–50</TD></TR><TR><TD valign="top">[3]</TD> <TD valign="top">  A. Diego,  "Sur les algèbres de Hilbert" , Gauthier-Villars  (1966)  ((translated from the Spanish))</TD></TR><TR><TD valign="top">[4]</TD> <TD valign="top">  M.D. Gladstone,  "Some ways of constructing a propositional calculus of any required degree of unsolvability"  ''Trans. Amer. Math. Soc.'' , '''118'''  (1965)  pp. 192–210</TD></TR></table>
 
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  A. Church,  "Introduction to mathematical logic" , '''1''' , Princeton Univ. Press  (1956)</TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  J. Łukasiewicz,  A. Tarski,  "Untersuchungen über den Aussagenkalkül"  ''C.R. Soc. Sci. Letters Varsovie, Cl. III'' , '''23'''  (1930)  pp. 30–50</TD></TR><TR><TD valign="top">[3]</TD> <TD valign="top">  A. Diego,  "Sur les algèbres de Hilbert" , Gauthier-Villars  (1966)  ((translated from the Spanish))</TD></TR><TR><TD valign="top">[4]</TD> <TD valign="top">  M.D. Gladstone,  "Some ways of constructing a propositional calculus of any required degree of unsolvability"  ''Trans. Amer. Math. Soc.'' , '''118'''  (1965)  pp. 192–210</TD></TR></table>

Latest revision as of 22:11, 5 June 2020


A propositional calculus using only the primitive connective $ \supset $( implication). Examples of an implicative propositional calculus are the complete (or classical) implicative propositional calculus given by the axioms

$$ p \supset ( q \supset p ) ,\ \ ( ( p \supset q ) \supset \ ( ( q \supset r ) \supset \ ( p \supset r ) ) ) , $$

$$ ( ( p \supset q ) \supset p ) \supset p $$

and the rules of inference: modus ponens and substitution; another example is the positive implicative propositional calculus given by the axioms

$$ p \supset ( q \supset p ) ,\ \ ( p \supset ( q \supset r ) ) \supset \ ( ( p \supset q ) \supset \ ( p \supset r ) ) $$

and the same rules of inference. Every implicative formula, that is, a formula only containing the connective $ \supset $, is deducible in complete (or positive) implicative propositional calculus if and only if it is deducible in classical (respectively, intuitionistic) propositional calculus. For any finite set $ V $ of variables, among the formulas with variables in $ V $ there is only a finite number of pairwise inequivalent ones in the positive implicative propositional calculus (see [3]). There exist undecidable finitely-axiomatizable implicative propositional calculi (see [4]).

References

[1] A. Church, "Introduction to mathematical logic" , 1 , Princeton Univ. Press (1956)
[2] J. Łukasiewicz, A. Tarski, "Untersuchungen über den Aussagenkalkül" C.R. Soc. Sci. Letters Varsovie, Cl. III , 23 (1930) pp. 30–50
[3] A. Diego, "Sur les algèbres de Hilbert" , Gauthier-Villars (1966) ((translated from the Spanish))
[4] M.D. Gladstone, "Some ways of constructing a propositional calculus of any required degree of unsolvability" Trans. Amer. Math. Soc. , 118 (1965) pp. 192–210
How to Cite This Entry:
Implicative propositional calculus. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Implicative_propositional_calculus&oldid=14066
This article was adapted from an original article by S.K. Sobolev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article