Timeline for How can we formalize the claim that these two languages are equally expressive?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Aug 3 at 20:26 | comment | added | Julius Hamilton | I haven’t read this thoroughly, but there are standard definitions of logical equivalence in model theory, in case that would help. | |
| Aug 3 at 18:38 | answer | added | emesupap | timeline score: 1 | |
| Aug 3 at 18:34 | comment | added | WillG | @mudskipper (and Mauro) So, if we are using truth tables to define meaning, then perhaps you have the following definition in mind: Two formulas ɸ and ψ "mean the same thing" iff they contain the same variables and their truth tables are identical. Makes sense to me. Is that what you're claiming? | |
| Aug 3 at 18:34 | answer | added | Bumble | timeline score: 2 | |
| Aug 3 at 18:15 | comment | added | mudskipper | In every first order prop. calculus (with all the required logical connectives, so with at least 'not' and 'and' for instance) two arbitrary true propositions will be equivalent. In other words, that fact is not relevant to establish that T and T' have the same expressivity. The new axiom schema (plus associated truth table; e.g. using induction on the application of the schema) fully establishes that T and T' have the same expressivity - so, there is no further question of justification possible. | |
| Aug 3 at 13:34 | comment | added | Mauro ALLEGRANZA | Regarding "meaning" the meaning of propositional connectives is the respective truth tables. | |
| Aug 3 at 13:02 | comment | added | WillG | Perhaps we must simply declare certain axioms to be “definitional”, and then take as given that definitional axioms preserve meaning (and then expand on this a bit to cover multiple replacements). But to my mind, this is an additional sort of convention or assumption that ought to be stated explicitly—nothing about the axiom introducing ∨ automatically “says” it’s a definition, and the function of the axiom is the same as that of the original axioms. | |
| Aug 3 at 12:57 | comment | added | WillG | @MauroALLEGRANZA I see that, but my question is how we can justify the claim that the definitional axioms preserve meaning, and how we can translate such a claim into claims that more complex propositions (after many replacements/axiom applications) have identical meanings. | |
| Aug 3 at 12:31 | comment | added | Mauro ALLEGRANZA | You have the original logical axioms of T with let say Modus Ponens as only rule of inference. In them the disjunction symbol never occur, so you cannot produce theorem with it. The only way is using the "definitional axiom" that allows you to replace a suitable formula with the conditional with the corresponding version with the disjunction. This is the only way you can have formulas of T'. | |
| Aug 3 at 6:06 | answer | added | J D | timeline score: 1 | |
| Aug 2 at 3:22 | comment | added | WillG | Basically I'm Bob, and I want to know Alice's response at this point. | |
| Aug 2 at 3:19 | comment | added | WillG | Bob: "But if that's the criterion for ψ and ψ* to 'mean the same thing', then I could just as well let ψ* be the formula 'C ➝ C'. Then T’ still proves ψ ➝ ψ* and ψ* ➝ ψ. In other words, according to this criterion, ψ 'means the same thing' as every true formula. So what grounds do we really have for saying 'A ➝ (¬A ➝ B)' means the same thing as 'A ➝ A ∨ B'? | |
| Aug 2 at 3:18 | comment | added | WillG | I'm imagining a dialogue something like this. Alice: "T and T’ are identical theories." Bob: "No, because T’ can prove ψ, where ψ is 'A ➝ A ∨ B', whereas T does not even have ψ in its language." Alice: "Ok, but ψ means the same thing as ψ*, where ψ* is 'A ➝ (¬A ➝ B)'. And ψ* is a theorem of T." Bob: "What makes you say that ψ and ψ* 'mean the same thing'?" Alice: "Well, T’ can prove ψ ➝ ψ* and ψ* ➝ ψ. So by that criterion, we should say that ψ* means the same thing as ψ." | |
| Aug 2 at 3:10 | comment | added | WillG | @Miss_Understands I'm just pointing out that the arbitrary mapping is possible, which means we should have some justification for why the original mapping was the "correct" one. | |
| Aug 2 at 0:39 | comment | added | Miss Understands | I don't understand. You proved that T and T' are the same theory because of the syntactic substitution. But then for some reason, you want to confuse this by mapping elements of T and T' arbitrarily. Why would you want to do that? What am I missing? | |
| Aug 1 at 22:23 | history | asked | WillG | CC BY-SA 4.0 |