Ack. I don't think I ever posted this. It's probably too late to help anyone in the class, but maybe it'll help posterity.
The PDF:
https://docs.google.com/open?id=0Bywlg79R0_RDN1JtVjRjaVhFNG8
TeX:
https://docs.google.com/open?id=0Bywlg79R0_RDY1NKOFJLOWVVRzQ
Saturday, June 16, 2012
Thursday, May 31, 2012
Tuesday, May 29, 2012
Saturday, May 19, 2012
Wednesday, May 16, 2012
Monday, May 14, 2012
Sunday, May 13, 2012
Saturday, May 12, 2012
Scribe Notes for Friday, May 11
Wednesday, May 9, 2012
Wednesday, May 2, 2012
Scribe Notes for April 30
Wednesday, April 25, 2012
Scribe notes for April 20
Sunday, April 22, 2012
Scribe Notes: Wednesday, April 18th
Here is the PDF.
If I made any errors let me know and I will update the document.
If I made any errors let me know and I will update the document.
Friday, April 20, 2012
Thursday, April 19, 2012
Monday, April 16, 2012
Thursday, April 12, 2012
Halting Problem in Iambic Tetrameter
Since we just saw a proof of the undecidability of the halting problem, I thought I'd share with everyone my favorite version of the proof: Scooping the Loop Snooper
Wednesday, April 11, 2012
Monday, April 9, 2012
Wednesday, April 4, 2012
Monday, April 2, 2012
Turing's paper "On computable numbers, with an application to the Entscheidungsproblem" is freely available here, although in somewhat hard-to-use form.
Thursday, March 22, 2012
Friday, March 16, 2012
Thursday, March 15, 2012
Homework Question
So I'm not sure how often people look at the blog, but this seems a good a place as any.
For question 6, part a, is there something we are supposed to prove? It looks to me like a definition required for b-e, but I don't see a question for part a.
Also, while I'm here, is there still a take home final? Or is this last homework sorta the final or something.
For question 6, part a, is there something we are supposed to prove? It looks to me like a definition required for b-e, but I don't see a question for part a.
Also, while I'm here, is there still a take home final? Or is this last homework sorta the final or something.
Wednesday, March 14, 2012
Monday, March 12, 2012
Friday, March 9, 2012
The Notes Thus Far...
I just combined all the pdf files for any currently posted scribe notes, so that I wouldn't have to keep flipping through all the different files one by one. If anyone is interested, I've posted the file here.
Tuesday, March 6, 2012
Thursday, March 1, 2012
Notes for Feb 27
Wednesday, February 29, 2012
Notes for Wed. February 29
Due to an aberration in the spacetime metric brought on by the fact that today is February 29, the notes from today are being posted before the notes from Monday. Here are the notes and here is the TeX.
Includes:
Includes:
- An application of the Łoś-Vaught test
- A taste of computability
- Skolem's Paradox
- An introduction to the completeness theorem for uncountable languages.
Tuesday, February 28, 2012
Notes for Feb 24
Wednesday, February 22, 2012
Saturday, February 18, 2012
Scribe notes from February 17, 2012
Hi all,
Here are the scribe notes from Feb 17, 2012. (TeX code is available here.) In class, two notations for M were given: one without tilde and the other with tilde. The tilde notation works well for simple symbols such as ˜b, ˜c, and ˜t. But, I think, it is not for long terms such as ~f(t1,…,tk) (\widetilde{f(t_{1},\ldots,t_{k})}) .
Have a great weekend!
Best regards,
Tomoya Sato
Tomoya Sato
Friday, February 17, 2012
Thursday, February 16, 2012
Monday, February 13, 2012
Scribe notes from Feb 10, 2012
Here are the scribe notes from Feb 10, 2012. (TeX code is available here.) They introduce Frege-style systems for first order logic (aka Hilbert-style systems), as well as sequent calculus systems for first order logic (both LK and LKe systems).
Wednesday, February 8, 2012
Bob Chen emailed to ask "What is a resolution derivation? We couldn't
find the term in the 'Handbook of proof theory' (there it only mentions
resolution refutations)."
My answer: A resolution derivation is the same as a resolution refutation, except not requiring the final clause to be the empty clause. In particular, if there is a resolution refutation of a clause C from a set of clauses Γ, then Γ⊨C holds. (We proved this fact in class as part of proving the soundness of resolution.)
Sam
My answer: A resolution derivation is the same as a resolution refutation, except not requiring the final clause to be the empty clause. In particular, if there is a resolution refutation of a clause C from a set of clauses Γ, then Γ⊨C holds. (We proved this fact in class as part of proving the soundness of resolution.)
Sam
Monday, February 6, 2012
For Math 260 students (UCSD, Winter-Spring 2012).
This blog is intended for discussions accompanying the Introduction to Mathematical Logic course, during Winter and Spring 2012.
It will be set up so students can create new posts, and anyone can view or answer them. For this, if you are attending the course, please send me your preferred email for blog posting. It is probably best if you have a gmail address, but I expect it will work also if you do not have a gmail address.
Any discussions related to the course is fine, including discussions about, and hints for, homework problems.
It is a bit counter-intuitive how to use Blogger: To view a particular post including all of its comments, you must either click on the post title in the sidebar on the right side of the screen, or on the "Comments" link at the bottom of the past.
It is a bit counter-intuitive how to use Blogger: To view a particular post including all of its comments, you must either click on the post title in the sidebar on the right side of the screen, or on the "Comments" link at the bottom of the past.
You can use LaTeX commands in either posts or comments, which are implemented with MathJax and should work in almost all up-to-date browsers. Note that enclosing math in single $'s does not work. However, you may use "∖(" and "∖)" for inline math and double dollar signs or "∖[" and "∖]" for displayed math. Examples: $(y+\sqrt z)^{-1}$ and sin2x2. And, a displayed equation is: 23
Another displayed equation is here:
∀x∃y(x≤y∧y≤x↔x=y).
Another displayed equation is here:
∀x∃y(x≤y∧y≤x↔x=y).
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