display all the ideas for this combination of texts
10 ideas
| 13643 | Aristotelian logic is complete [Shapiro] |
| Full Idea: Aristotelian logic is complete. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 2.5) | |
| A reaction: [He cites Corcoran 1972] |
| 8077 | Stoic propositional logic is like chemistry - how atoms make molecules, not the innards of atoms [Chrysippus, by Devlin] |
| Full Idea: In Stoic logic propositions are treated the way atoms are treated in present-day chemistry, where the focus is on the way atoms fit together to form molecules, rather than on the internal structure of the atoms. | |
| From: report of Chrysippus (fragments/reports [c.240 BCE]) by Keith Devlin - Goodbye Descartes Ch.2 | |
| A reaction: A nice analogy to explain the nature of Propositional Logic, which was invented by the Stoics (N.B. after Aristotle had invented predicate logic). |
| 20791 | Chrysippus has five obvious 'indemonstrables' of reasoning [Chrysippus, by Diog. Laertius] |
| Full Idea: Chrysippus has five indemonstrables that do not need demonstration:1) If 1st the 2nd, but 1st, so 2nd; 2) If 1st the 2nd, but not 2nd, so not 1st; 3) Not 1st and 2nd, the 1st, so not 2nd; 4) 1st or 2nd, the 1st, so not 2nd; 5) 1st or 2nd, not 2nd, so 1st. | |
| From: report of Chrysippus (fragments/reports [c.240 BCE]) by Diogenes Laertius - Lives of Eminent Philosophers 07.80-81 | |
| A reaction: [from his lost text 'Dialectics'; squashed to fit into one quote] 1) is Modus Ponens, 2) is Modus Tollens. 4) and 5) are Disjunctive Syllogisms. 3) seems a bit complex to be an indemonstrable. |
| 13651 | A set is 'transitive' if contains every member of each of its members [Shapiro] |
| Full Idea: If, for every b∈d, a∈b entails that a∈d, the d is said to be 'transitive'. In other words, d is transitive if it contains every member of each of its members. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 4.2) | |
| A reaction: The alternative would be that the members of the set are subsets, but the members of those subsets are not themselves members of the higher-level set. |
| 13647 | Choice is essential for proving downward Löwenheim-Skolem [Shapiro] |
| Full Idea: The axiom of choice is essential for proving the downward Löwenheim-Skolem Theorem. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 4.1) |
| 13631 | Are sets part of logic, or part of mathematics? [Shapiro] |
| Full Idea: Is there a notion of set in the jurisdiction of logic, or does it belong to mathematics proper? | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], Pref) | |
| A reaction: It immediately strikes me that they might be neither. I don't see that relations between well-defined groups of things must involve number, and I don't see that mapping the relations must intrinsically involve logical consequence or inference. |
| 13654 | It is central to the iterative conception that membership is well-founded, with no infinite descending chains [Shapiro] |
| Full Idea: In set theory it is central to the iterative conception that the membership relation is well-founded, ...which means there are no infinite descending chains from any relation. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 5.1.4) |
| 13640 | Russell's paradox shows that there are classes which are not iterative sets [Shapiro] |
| Full Idea: The argument behind Russell's paradox shows that in set theory there are logical sets (i.e. classes) that are not iterative sets. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 1.3) | |
| A reaction: In his preface, Shapiro expresses doubts about the idea of a 'logical set'. Hence the theorists like the iterative hierarchy because it is well-founded and under control, not because it is comprehensive in scope. See all of pp.19-20. |
| 13666 | Iterative sets are not Boolean; the complement of an iterative set is not an iterative sets [Shapiro] |
| Full Idea: Iterative sets do not exhibit a Boolean structure, because the complement of an iterative set is not itself an iterative set. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 7.1) |
| 13653 | 'Well-ordering' of a set is an irreflexive, transitive, and binary relation with a least element [Shapiro] |
| Full Idea: A 'well-ordering' of a set X is an irreflexive, transitive, and binary relation on X in which every non-empty subset of X has a least element. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 5.1.3) | |
| A reaction: So there is a beginning, an ongoing sequence, and no retracing of steps. |