Timeline for Dissolution of Tensors
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Apr 1, 2015 at 3:01 | review | Close votes | |||
| Apr 4, 2015 at 8:15 | |||||
| Mar 28, 2015 at 21:43 | comment | added | Todd Trimble | @DamianoMazza Okay, thanks for clarifying. I think what Zerkezhi was asking was essentially answered at the other thread, and is related more to what Emil wrote. But of course the question above was ambiguous, and your interpretation was a reasonable one. | |
| Mar 28, 2015 at 21:30 | comment | added | Damiano Mazza | @Todd actually I meant irreversibility in the proof-theoretic sense: given a proof of $\Gamma\vdash A\otimes B$, if you know that the last rule introduces $A\otimes B$, then you know there are proofs of $\Gamma_1\vdash A$ and $\Gamma_2\vdash B$ with $\Gamma_1+\Gamma_2=\Gamma$ but in general you still don't know how to determine $\Gamma_1$ and $\Gamma_2$. This is unrelated to the derivability of $A\otimes B\vdash A,B$ (the so-called "mix" rule), but I think it is still not what Zerkezhi was asking :-) | |
| Mar 26, 2015 at 19:53 | comment | added | Todd Trimble | Zerkezhi, I suggest you focus on Emil's comment instead. The answer I gave to your other question suggests that his interpretation is the relevant one. I think what Damiano is saying is that $A \otimes B \vdash A, B$ is not valid in linear logic, and that's true, but that's because a list of formulas to the right of the entailment symbol is to be interpreted as a linear disjunction, not conjunction. But I suspect that observation is not relevant here. | |
| Mar 26, 2015 at 19:05 | comment | added | Zerkezhi | @DamianoMazza Thank you, that's exactly what I wanted to know. Sadly not what I had hoped, but still, thank you. I have opened another question, of which I hoped you might be so kind as to respond to: [mathoverflow.net/q/201157/69644](Link) | |
| Mar 25, 2015 at 19:32 | comment | added | Damiano Mazza | I know linear logic quite well but unfortunately I don't understand the question. What do you mean by "opposite"? Are you asking about reversibility of the tensor rule? (If that's the case, the answer is no, the tensor rule is irreversible). | |
| Mar 24, 2015 at 21:43 | comment | added | Dima Pasechnik | When I voted to close this, it looked to me as a very vague question in terribly broken English... I've retracted my vote. | |
| Mar 24, 2015 at 18:22 | comment | added | Paul Taylor | May I suggest that some linear logician (not me, notwithstanding this book) give an explanation of $\otimes$ for the general enlightenment of MO readers. This requires that the question not be closed. I am not aware that Stefan Kohl and Alex Degtyarev are experts in this subject. | |
| Mar 24, 2015 at 16:56 | comment | added | Emil Jeřábek | It’s not quite clear to me what the question is, but anyway, a sequent $\Gamma,A,B\Longrightarrow\Delta$ is interderivable with $\Gamma,A\otimes B\Longrightarrow\Delta$. | |
| Mar 24, 2015 at 16:42 | review | Close votes | |||
| Mar 24, 2015 at 19:02 | |||||
| S Mar 24, 2015 at 16:30 | history | suggested | Tobias Fritz | CC BY-SA 3.0 |
fixed tag, typo and latex
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| Mar 24, 2015 at 16:25 | review | Suggested edits | |||
| S Mar 24, 2015 at 16:30 | |||||
| Mar 24, 2015 at 16:23 | review | First posts | |||
| Mar 24, 2015 at 16:39 | |||||
| Mar 24, 2015 at 16:20 | history | asked | Zerkezhi | CC BY-SA 3.0 |