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Aristotelian logic

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**1. **Aristotelian logic
Lesson 7

**2. **Lesson 7 2 Platoon (the left one) Aristotle (the right one) 384 – 322
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.

**3. **Prednáška 6 3 Aristotelian logic The Greece philosopher and logician Aristotle examined before more than 2000 years Subject – Predicate propositions and arguments formed from them:
All S are P SaP affirmo
No S is P SeP nego
Some S are P SiP affirmo
Some S are not P SoP 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)

**4. **Lesson 7 4 Aristotelian logic: the square positive negative
universal SaP SeP
existential SiP SoP
SaP ? ?SoP, SeP ? ?SiP contradictory
SaP |= ?SeP, SeP |= ?SaP contrary
?SiP |= SoP, ?SoP |= SiP subcontrary
SaP |= SiP, SeP |= Sop, subalternative
?SiP |= ?SaP, ?SoP |= ?SeP

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

**6. **Lesson 7 6 Proofs of entailments in the logical square positive negative
universal SaP SeP
existential SiP SoP
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)]

**7. **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 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

**8. **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 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)
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

**9. **Logical square - reversals SiP ? PiS SeP ? 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 |= PiS SeP |= 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)]

**10. **10 Aristotelian sylogisms 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 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.

**11. **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:
Hatch the areas which correspond to empty sets (universal premises)
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.

**12. **Prednáška 6 12 John Venn 1834 - 1923 Cambridge

**13. **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)]

**14. **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)]

**15. **Prednáška 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)]

**16. **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)]

**17. **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)]

**18. **18 Universal premises not |= existence All golden mountains are golden
All golden mountains are mountains
|? Some mountains are golden
Example of Bertrand Russell
(1872-1970)

**19. **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)]

**20. **Lesson 7 20 Wider use, not only syllogisms P1: All statesmen are politicians ?x [S(x) ? P(x)]
P2: Some statesmen are intelligent ?x [S(x) ? I(x)]
P3: Some politicians are not statesmen ?x [P(x) ? ?S(x)]
Z1: ?? Some politicians are not intelligent ?x [P(x) ? ?I(x)] ?
Z2: ?? Some politicians are intelligent ?x [P(x) ? I(x)] ?

**21. **Lesson 7 21 Venn’s diagrams p1: All gardeners are skillful. ?x [P(x) ? Q(x)]
p2: All skillful are intelligent. ?x [Q(x) ? R(x)]
p3: There is at least one gardener. ?x P(x) )
--------------------------------------------------- ----------------------
z: Some gardeners are intelligent. ?x [P(x) ? R(x)]

**22. **Lesson 7 22 Venn’s diagrams p1: All gardeners are skillful. ?x [P(x) ? Q(x)]
p2: All skillful are intelligent. ?x [Q(x) ? R(x)]
p3: There is at least one gardener. ?x P(x) )
--------------------------------------------------- ----------------------
z: Some gardeners are intelligent. ?x [P(x) ? R(x)]

**23. **Lesson 7 23 Venn’s diagrams p1: All students can think logically. ?x [S(x) ? L(x)]
p2: Only intelligent people can think logically. ?x [L(x) ? I(x)]
--------------------------------------------------- ----------------------
z: All students are intelligent people. ?x [S(x) ? I(x)]

**24. **Lesson 7 24 Venn’s diagrams p1: All students can think logically. ?x [S(x) ? L(x)]
p2: Only intelligent people can think logically. ?x [L(x) ? I(x)]
--------------------------------------------------- ----------------------
z: All students are intelligent people. ?x [S(x) ? I(x)]

**25. **Lesson 7 25 Venn’s diagrams p1: All students learn to think logically. ?x [P(x) ? Q(x)]
p2: Who learns to think logically won’t loose. ?x [Q(x) ? R(x)]
--------------------------------------------------- ----------------------
z: Some students won’t loose. ?x [P(x) ? R(x)]

**26. **Prednáška 6 26 Venn’s diagrams p1: All students learn to think logically. ?x [P(x) ? Q(x)]
p2: 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!

**27. **Prednáška 6 27 Venn’s diagrams p1: All students learn to think logically. ?x [P(x) ? Q(x)]
p2: 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.

**28. **Lesson 7 28 Areas definition 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)

**29. **29 Venn’s diagrams and syllogisms p1: No bird is a mammal. ?x [P(x) ? ?S(x)]
p2: Some birds are runners ?x [P(x) ? B(x)]
----------------------------------- ----------------------
z: Some runners are not mammals. ?x [B(x) ? ?S(x)]

**30. **Lesson 7 30 Venn’s diagrams p1: Some rulers are cruel. ?x [V(x) ? K(x)]
p2: No good housekeeper is cruel. ?x [H(x) ? ?K(x)]
--------------------------------------------------- ----------------------
z: Some rulers are not good housekeepers. ?x [V(x) ? ?H(x)]