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Aristotelian logic Lesson 7

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2 Plato (on the left) Aristotle (on the right) 384 – 322 (B.C.) Mine is the first step and therefore a small one, though worked out with much thought and hard labor. You, my readers or hearers of my lectures, if you think I have done as much as can fairly be expected of an initial start... will acknowledge what I have achieved and will pardon what I have left for others to accomplish.

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Přednáška 6Lesson 7 3 Aristotelian logic The Greece philosopher and logician Aristotle examined before more than 2000 years Subject – Predicate propositions and arguments (syllogisms) formed from them: All S are PSaPaffirmo No S is PSePnego Some S are PSiPaffirmo Some S are not PSoPnego All concepts S, P are nonempty. From the contemporary point of view, Aristotle developed a fragment of first-order predicate logic. Propositional logic was examined by the stoics at that time, who opposed Aristotle and laid down the fundamentals of the predicate logic apart from other things. (Viz František Gahér: Stoická sémantika a logika)

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Lesson 7 4 the square of opposition positivenegative universalSaPSeP existentialSiPSoP SaP SoP, SeP SiP contradictory SaP |= SeP,SeP |= SaP contrary SiP |= SoP, SoP |= SiP subcontrary SaP |= SiP, SeP |= Sop,subalternative SiP |= SaP, SoP |= SeP

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5 Logical square – reversals SiP PiSSeP PeS Some students are married Some married are students No man is a tree No tree is a man SaP |= PiSSeP |= PoS All teachers are public agents. |= Some public agents are teachers. No toxic mushroom is eatable. |= Some of the eatable mushrooms are not toxic.

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Lesson 7 6 Proofs of entailments in the Square positivenegative universalSaPSeP existentialSiPSoP SaP SoP, SeP SiP contradictory (diagonal) All S are P It is not true that some S are not P Proof (de Morgan): x [S(x) P(x)] x [S(x) P(x)] No S is P It is not true that some S are P Proof (de Morgan): x [S(x) P(x)] x [S(x) P(x)]

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7 Proofs (continuing) SaP SeP SiP SoP SaP |= SeP,SeP |= SaP contrary All S are P |= It is not true that no S is P x [S(x) P(x)] |= x [S(x) P(x)] Proof (semantic): If S U P U then S U cannot be a subset of the complement of P U ; hence S U is not a subset of P U No S is P |= It is not true that All S are P x [S(x) P(x)] |= x [S(x) P(x)] Proof (semantic): If S U P U (of complement) then S U cannot be a subset of P U ; hence S U is not a subset of P U

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Lesson 7 8 Proofs (continuing) SiP |= SoP, SoP |= SiP sub-contrary It is not true that Some S are P |= Some S are not P x [S(x) P(x)] x [S(x) P(x)] If S U P U (a subset of the complement) and S U (is nonempty) then the intersection of S U and the complement of P U is nonempty as well: (S U P U ) , hence x [S(x) P(x)] Similarly: It is not true that Some S are not P |= Some S are P (with the assumption of nonemptiness) SaP |= SiP, SeP |= Sop,subalternative SiP |= SaP, SoP |= SeP The proofs of the remaining entailments are analogical: All S are P; hence Some S are P, and so on… All of them are valid on the assumption of nonemptiness

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The Square - reversals SiP PiSSeP PeS Some S are P Some P are S No S is P No P is S x [S(x) P(x)] x [P(x) S(x)] x [S(x) P(x)] x [P(x) S(x)] SaP |= PiSSeP |= PoS All S are P |= Some P are S No S is P |= Some P is not S x [S(x) P(x)] x S(x) |= x [P(x) S(x)] x [S(x) P(x)] x P(x) |= x [P(x) S(x)]

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10 Aristotelian syllogisms Simple arguments formed by combination of three predicates S, P, M, where M is a mediate predicate. M does not occur in the conclusion; the conclusion is thus of a form S-P I.M-PII. P-MIII. M-PIV. P-M S-M S-M M-S M-S Valid modes are: I.aaa, eae, aii, eio (barbara, celarent, darii, ferio) II.aoo, aee, eae, eio (baroco, camestres, cesare, festino) III.oao, aai, aii, iai, eao, eio (bocardo, darapti, datisi, disamis, felapton, ferison) IV.aai, aee, iai, eao, eio (bamalip, calemes, dimatis, fesapo, fresison) We do not have to memorize valid modes because they are easily derivable by means of Venn’s diagram method.

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Lesson 7 11 Aristotelian logic: syllogisms Venn’s diagrams a set-theoretical method of proving the validity of syllogism arguments. The method proceeds as follows: We represent the truth domains of predicates S, P, M as mutually non-disjoint circles. First, illustrate the situation when the premises are true as follows: 1.Hatch the areas which correspond to empty sets (universal premises) 2.Mark with a cross the areas which are nonempty (existential premises); put a cross to an area only if there is no other area into which it might be possible to put the cross as well. Finally, check whether the so-represented truth of the premises ensure that the conclusion is true as well.

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Přednáška 6 12 John Venn Cambridge

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13 Syllogisms and Venn’s diagrams All family houses are privates. x [F(x) P(x)] Some realties are family houses. x [R(x) F(x)] Some realties are privates. x [R(x) P(x)] FP R The 1 st premise: F / P is empty The 2 nd premise: the intersection of R and F is nonempty put a cross Important: Deal first with universal premises, and only afterwards with existential ones.

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14 Syllogisms and Venn’s diagrams All of family houses are privates x [F(x) P(x)] Some realties are family houses. x [R(x) F(x)] Some realties are privates x [R(x) P(x)] FP R The 1 st premise: F / P is empty The 2 nd premise: the intersection of R and F is nonempty a cross Check whether the conclusion is valid: the intersection of R and P is nonempty; the argument is valid

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Přednáška 6 15 Syllogisms and Venn’s diagrams All badgers are art collectors x [B(x) A(x)] Some art collectors live in caves x [A(x) C(x)] Some badgers live in caves x [B(x) C(x)] BA C According to the 1 st premise there is no badger that would not be an art collector hatch According to the 2 nd premise the inter- section of A and C is nonempty; but we don’t know where to put the cross. Thus we cannot put it there. Argument is invalid ? ?

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16 Sylogisms – validness verification Some politics are wise x [Pl(x) W(x)] No wise man is a populist x [W(x) Po(x)] Some politics are not populists x [Pl(x) Po(x)] W PoPo Pl First: the universal premise 2. There are no wise people (W) who’d be populists (Po). So we hatch the intersection of W and Po 1 st premise: the intersection of W and Pl is non- empty put a cross Check the truth of the conclusion: the intersection of Pl and the complement of Po must be nonempty, which is so: the truth is guaranteed… the argument is valid

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17 Syllogisms – validness verification All cars are vehicles x [C(x) V(x)] All cars have a wheel x [C(x) W(x)] Some vehicles have a wheel x [V(x) W(x)] C V W 1 st premise: the area C must be a subset of V: hatch 2 nd premise: area C is a subset of W: hatch Validness is not guarantied; there is no cross in the intersection of V and W! The Argument is invalid

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18 Universal premises not |= existence All golden mountains are golden All golden mountains are mountains | Some mountains are golden Example of Bertrand Russell ( ) Invalid Argument

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Syllogisms – validness verification All cars are vehicles x [C(x) V(x)] All cars have a wheel x [C(x) W(x)] The cars exists (implicit assumption) x C(x) Some vehicles have a wheel x [V(x) W(x)] C V W 1 st premise: Area C must be a subset of V: hatch 2 nd premise: area C is a subset of W: hatch The validness is guarantied … there is a cross in the intersection of V and W … the Argument is valid 3 rd premise: we put a cross into area C

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Lesson 7 20 Wider use, not only syllogisms P 1 : All statesmen are politicians x [S(x) P(x)] P 2 : Some statesmen are intelligent x [S(x) I(x)] P 3 : Some politicians are not statesmen x [P(x) S(x)] Z 1 : ? Some politicians are not intelligent x [P(x) I(x)] ? Z 2 : ? Some politicians are intelligent x [P(x) I(x)]? SP I P 1 : crosshatch S / P P 2 : put the cross into the intersection of S and I P 3 : cannot put a cross … we don’t know where exactly Z 1 : doesn’t follow from premises… no cross Z 2 : follows … cross

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Lesson 7 21 Venn’s diagrams p 1 :All gardeners are skillful. x [P(x) Q(x)] p 2 :All skillful are intelligent. x [Q(x) R(x)] p 3 : There is at least one gardener. x P(x) ) z:Some gardeners are intelligent. x [P(x) R(x)] P Q R 1 st premise: there is no P that wouldn’t be a Q (de Morgan) hatch 2 nd premise: there is no Q that wouldn’t be an R (de Morgan) hatch 3 rd premise: P is not empty cross

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Lesson 7 22 Venn’s diagrams p 1 : All gardeners are skillful. x [P(x) Q(x)] p 2 : All skillful are intelligent. x [Q(x) R(x)] p 3 : There is at least one gardener. x P(x) ) z: Some gardeners are intelligent. x [P(x) R(x)] PQ R Now we check the conclusion. There is a cross in the intersection of P and R. Hence the conclusion is true. The argument is Valid

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Lesson 7 23 Venn’s diagrams p 1 :All students can think logically. x [S(x) L(x)] p 2 :Only intelligent people can think logically. x [L(x) I(x)] z:All students are intelligent people. x [S(x) I(x)] SL I 1 st premise: there is no entity which is in S and not in L. (De Morgan) hatch 2 nd premise: there is no entity which is in L and not in I. (De Morgan) hatch

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Lesson 7 24 Venn’s diagrams p 1 : All students can think logically. x [S(x) L(x)] p 2 : Only intelligent people can think logically. x [L(x) I(x)] z: All students are intelligent people. x [S(x) I(x)] SL I Now we check if the diagram represents our conclusion. The conclusion says that all entities which are in S are also in I. It’s true… So the Argument is valid.

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Lesson 7 25 Venn’s diagrams p 1 : All students learn to think logically. x [P(x) Q(x)] p 2 :Who learns to think logically won’t loose. x [Q(x) R(x)] z:Some students won’t loose. x [P(x) R(x)] PQ R 1 st premise: There is no entity that is in P and not in Q. (De Morgan) hatch 2 nd premise: There is no entity that is in Q and not in R. (De Morgan) hatch

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Přednáška 6 26 Venn’s diagrams p 1 : All students learn to think logically. x [P(x) Q(x)] p 2 : Who learns to think logically won’t loose. x [Q(x) R(x)] z: Some students won’t loose. x [P(x) R(x)] Remark: In Aristotelian logic this argument is assumed to be valid due to the assumption of nonempty concepts. But from universal premises we cannot deduce the existence! PQ R Now we check if the argument is valid or not. The conclusion says that there is an entity that is in P and in R. Diagram doesn’t prove it – no cross. Argument is invalid. !

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Přednáška 6 27 Venn’s diagrams p 1 : All students learn to think logically. x [P(x) Q(x)] p 2 : Who learns to think logically won’t loose. x [Q(x) R(x)] Students exist (implicit assumption) x P(x) z: Some students won’t loose. x [P(x) R(x)] Remark: In Aristotelian logic all the concepts are assumed to be nonempty. If we add the implicit assumption that the students exist, then the argument is valid. PQ R Now we can put the cross to the intersection of P and R – it is nonempty. The conclusion says that there is an entity in the intersection of the set P and the set R. There is at least one entity (cross). Argument is valid

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28 Venn’s diagrams and syllogisms p 1 : No bird is a mammal. x [P(x) S(x)] p 2 : Some birds are runners x [P(x) B(x)] z:Some runners are not mammals. x [B(x) S(x)] PS B 1 st premise: there is no entity in P and S. hatch 2 nd premise: there is al least one entity in the intersection of P and B cross Check the conclusion: the intersection of P and complement of S is nonempty, Argument is valid

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Lesson 7 29 Venn’s diagrams p 1 : Some rulers are cruel. x [V(x) K(x)] p 2 : No good housekeeper is cruel. x [H(x) K(x)] z:Some rulers are not good housekeepers. x [V(x) H(x)] HK V Now check the conclusion: V and the complement of H nonempty; Argument is valid. Then the 1 st premise: put the cross to the intersection of V and K First 2 nd premise: crosshatch the intersection of H and K

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Lesson 7 30 Definition of sets: formulae with free variables A:S(x) P(x) M(x) B: S(x) P(x) M(x) C: S(x) P(x) M(x) D: S(x) P(x) M(x) E: S(x) P(x) M(x) F: S(x) P(x) M(x) G: S(x) P(x) M(x) H: S(x) P(x) M(x) S P M A B C D E F G H

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