display all the ideas for this combination of philosophers
5 ideas
| 10300 | Logical consequence can be defined in terms of the logical terminology [Shapiro] |
| Full Idea: Informally, logical consequence is sometimes defined in terms of the meanings of a certain collection of terms, the so-called 'logical terminology'. | |
| From: Stewart Shapiro (Higher-Order Logic [2001], 2.4) | |
| A reaction: This seems to be a compositional account, where we build a full account from an account of the atomic bits, perhaps presented as truth-tables. |
| 10259 | The two standard explanations of consequence are semantic (in models) and deductive [Shapiro] |
| Full Idea: The two best historical explanations of consequence are the semantic (model-theoretic), and the deductive versions. | |
| From: Stewart Shapiro (Philosophy of Mathematics [1997], 7.2) | |
| A reaction: Shapiro points out the fictionalists are in trouble here, because the first involves commitment to sets, and the second to the existence of deductions. |
| 13626 | Semantic consequence is ineffective in second-order logic [Shapiro] |
| Full Idea: It follows from Gödel's incompleteness theorem that the semantic consequence relation of second-order logic is not effective. For example, the set of logical truths of any second-order logic is not recursively enumerable. It is not even arithmetic. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], Pref) | |
| A reaction: I don't fully understand this, but it sounds rather major, and a good reason to avoid second-order logic (despite Shapiro's proselytising). See Peter Smith on 'effectively enumerable'. |
| 13637 | If a logic is incomplete, its semantic consequence relation is not effective [Shapiro] |
| Full Idea: Second-order logic is inherently incomplete, so its semantic consequence relation is not effective. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 1.2.1) |
| 10257 | Intuitionism only sanctions modus ponens if all three components are proved [Shapiro] |
| Full Idea: In some intuitionist semantics modus ponens is not sanctioned. At any given time there is likely to be a conditional such that it and its antecedent have been proved, but nobody has bothered to prove the consequent. | |
| From: Stewart Shapiro (Philosophy of Mathematics [1997], 6.7) | |
| A reaction: [He cites Heyting] This is a bit baffling. In what sense can 'it' (i.e. the conditional implication) have been 'proved' if the consequent doesn't immediately follow? Proving both propositions seems to make the conditional redundant. |