Mathematical Logic 2022/2023 (official course code 569060)
THIS COURSE IS NO LONGER TAUGHT BY JOOST J. JOOSTEN.
Organisation
Here is an academic year calendar of the UB.
The lectures will take place
Wednesdays: 9:30 -- 11:30;
Fridays: 11:30 -- 12:30.
Calle Montalegre 6 in Aula ???.
The start date is ????. Official schedule information can be found here
The final grade is determined by
(A) Homework questions (this may include a mid-term exam); (60 %)
(B) Final Exam; (40 %).
All materials and assignments will also be placed on this page.
Joost J. Joosten is the lecturer of this course.
The literature will consist of among others a reader to be distributed among the participants.
The mathematical logic course constitutes for 6 European credits You can find the complete schedule here.
Likewise there is an official page with the course description.
(Feb 8 -- 12)
(Feb 15 -- 19)
(Feb 22 -- Mar26)
(Mar 1 -- 5)
(Mar 8 -- 12)
(Mar 15 -- 19)
(Mar 22 -- 26 )
(Mar 29 -- Apr 2)
(Apr 5 -- 9)
(Apr 12 -- 16)
(Apr 19 -- 23)
(Apr 26 -- 30)
(May 3 -- 7)
(May 10 -- May 14)
(May 17 -- May 21)
Question and answer
Question
Q. I have been trying to solve Exercise 4.5.10 My idea is to define formulas \phi_n such that if \phi_n is true in a world m, then if forces the structure of a binary tree of height n with root m.
I do not know if this approach is promising or not. If It is, I am facing three problems
1) How can I express the ‘preservation property’, i.e., if a propositional variable (except from p0) is true in a world n of the tree, then it is also true everywhere else above n?
2) While it is quite simple to force the split is the first step (the formula \neg p0 \wedge\Diamond(p0\wedge p1) \wedge\Diamond(p0\wedge \neg p1) works) it is no so simple to produce new branches in successive words.
3) I have made some progress in 1) and 2), but in general, when defining the formula \phi_{n+1}, I tend to use the formula \phi_{n} at least twice as part of the definition. This ends up in a exponential growth of the length of the formulas.
Answer
Your idea of forcing a binary splitting at each level is promising indeed. I like the way you isolate the technical problems you encounter.
This is a very good way of starting to organize your ideas and a first step of finding a good strategy to your solution.
At (1) I can indeed say: since the exercise does not consider transitivity you will have to repeat all your requirements for each level of successors.
So, if you want Property A. at all n future levels you will have to say Box A /\ Box Box A /\ ... Box^n A. That is not too elegant but grows linearly in the depth.
At (2) you may need to consider fresh variables. Also, you should be sure that your model cannot `'reuse'` worlds. So, that it cannot look back to
some earlier world. At (3) I can only say that you should carefully count and keep track of the number of copies. Enjoy working on it.
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