# Aristotelian logic Lesson 7 Plato on the left

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Aristotelian logic Lesson 7

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. Lesson 7 2

Aristotelian logic n n n The Greece philosopher and logician Aristotle examined before more than 2000 years Subject – Predicate propositions and arguments (syllogisms) formed from them: All S are P Sa. P affirmo No S is P Se. P nego Some S are P Si. P affirmo Some S are not P So. P nego 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) Přednáška Lesson 7 6 3

the square of opposition universal positive Sa. P negative Se. P existential Si. P Sa. P So. P, Sa. P |= Se. P, Si. P |= So. P, Sa. P |= Si. P, Si. P |= Sa. P, So. P Se. P Si. P Se. P |= Sa. P So. P |= Si. P Se. P |= Sop, So. P |= Se. P Lesson 7 contradictory contrary subalternative 4

Logical square – reversals Si. P Pi. S Se. P Pe. S Some students are married Some married are students No man is a tree No tree is a man Sa. P |= Pi. S Se. P |= Po. S All teachers are public agents. |= Some public agents are teachers. No toxic mushroom is eatable. |= Some of the eatable mushrooms are not toxic. 5

Proofs of entailments in the Square universal positive Sa. P negative Se. P existential Si. P So. P Sa. P So. P, Se. P Si. P contradictory (diagonal) All S are P It is not true that some S are not P Proof (de Morgan): 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)] Lesson 7 6

Proofs (continuing) Sa. P Se. P Si. P So. P Sa. P |= Se. P, Se. P |= Sa. P 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 SU PU then SU cannot be a subset of the complement of PU; hence SU is not a subset of PU 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 SU PU (of complement) then SU cannot be a subset of PU; hence SU is not a subset of PU 7

Proofs (continuing) Si. P |= So. P, So. P |= Si. P sub-contrary It is not true that Some S are P |= Some S are not P x [S(x) P(x)] If SU PU (a subset of the complement) and SU (is nonempty) then the intersection of SU and the complement of PU is nonempty as well: (SU PU) , 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) Sa. P |= Si. P, Se. P |= Sop, subalternative Si. P |= Sa. P, So. P |= Se. P 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 Lesson 7 8

The Square - reversals Si. P Pi. S Se. P Pe. S 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)] Sa. P |= Pi. S Se. P |= Po. S 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)]

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-P II. P-M III. M-P IV. P-M S-M M-S n 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) n We do not have to memorize valid modes because they are easily derivable by means of Venn’s diagram method. n 10

Aristotelian logic: syllogisms Venn’s diagrams a set-theoretical method of proving the validity of syllogism arguments. The method proceeds as follows: n We represent the truth domains of predicates S, P, M as mutually non-disjoint circles. n 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. n Finally, check whether the so-represented truth of the premises ensure that the conclusion is true as well. Lesson 7 11

John Venn 1834 - 1923 Cambridge Přednáška 6 12

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

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

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

Sylogisms – validness verification n Some politics are wise No wise man is a populist n Some politics are not populists x [Pl(x) Po(x)] n First: the universal premise 2. x [Pl(x) W(x)] x [W(x) Po(x)] W There are no wise people (W) who’d be populists (Po). So we hatch the intersection of W and Po Po 1 st premise: the intersection of W and Pl is nonempty put a cross Pl 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 16

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

Universal premises not |= existence All golden mountains are golden n All golden mountains are mountains n | Some mountains are golden n Example of Bertrand Russell (1872 -1970) Invalid Argument 18

Syllogisms – validness verification n 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) n Some vehicles have a wheel n n x [V(x) W(x)] V C 2 nd premise: area C is a subset of W: hatch 1 st premise: Area C must be a subset of V: hatch W 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

Wider use, not only syllogisms P 1: All statesmen are politicians P 2: Some statesmen are intelligent P 3: Some politicians are not statesmen Z 1: ? Some politicians are not intelligent Z 2: ? Some politicians are intelligent S x [S(x) P(x)] x [S(x) I(x)] x [P(x) S(x)] x [P(x) I(x)] ? P P 1: crosshatch S / P P 2: put the cross into the intersection of S and I 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 Lesson 7 20

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

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

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

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

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

Venn’s diagrams x [P(x) Q(x)] x [Q(x) R(x)] ----------- x [P(x) R(x)] p 1: All students learn to think logically. p 2: Who learns to think logically won’t loose. -------------------------z: Some students won’t loose. Remark: P Q 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! ! R Now we check if the argument is valid or not. Přednáška 6 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. 26

Venn’s diagrams p 1: All students learn to think logically. p 2: Who learns to think logically won’t loose. x [P(x) Q(x)] x [Q(x) R(x)] -------------------------z: Some students won’t loose. ----------- x [P(x) R(x)] x P(x) Students exist (implicit assumption) P 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. Q R Now we can put the cross to the intersection of P and R – it is nonempty. Přednáška 6 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 27

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

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

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