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Results tagged with lo.logic
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user 102684
first-order and higher-order logic, model theory, set theory, proof theory, computability theory, formal languages, definability, interplay of syntax and semantics, constructive logic, intuitionism, philosophical logic, modal logic, completeness, Gödel incompleteness, decidability, undecidability, theories of truth, truth revision, consistency.
23
votes
Accepted
Is there a form of choice that can elude Kunen's inconsistency theorem?
Work of Usuba combined with work of Woodin shows that if there is a Reinhardt cardinal $\kappa$ that is a limit of Lowenheim-Skolem cardinals, then there is a forcing extension in which $\kappa$ remai …
- 8,099
22
votes
A better way to explain forcing?
This answer is quite similar to Rodrigo's but maybe slightly closer to what you want.
Suppose $M$ is a countable transitive model of ZFC and $P\in M$. We want to find a process for adding a subset $G$ …
- 8,099
21
votes
Accepted
Are Berkeley cardinals easier to refute in ZFC than Reinhardt cardinals?
Yes, it is easier to refute Berkeleys than Reinhardts. There is a very simple refutation of Berkeleys in ZFC that is due to Woodin. It is part of the motivation for his contention that Berkeley cardin …
- 8,099
18
votes
Accepted
Is there a minimal inner model for determinacy?
Assuming AD, Woodin showed that no inner model that is missing a real correctly computes $\omega_1$. (This almost follows from Theorem 9 of Velickovic-Woodin's "Complexity of the set of reals of inner …
- 8,099
17
votes
Accepted
Can $Ord$ have nontrivial second-order elementary self-embeddings?
Answering this question would either require refuting choiceless large cardinals or getting close to refuting Woodin's HOD Conjecture.
First, if choiceless cardinals are consistent, one cannot rule ou …
- 8,099
17
votes
Accepted
Must there be a proper class of Reinhardt cardinals if there is a Reinhardt cardinal?
No, if the existence of a Reinhardt is consistent, then it is consistent with a Reinhardt cardinal that the class of inaccessible cardinals is bounded in the ordinals. Indeed, if $j : V\to V$ is a non …
- 8,099
16
votes
Forcings predicted by core model theory $+$$ZFC$ results proved by the method of core model ...
Question 1: A good example is Woodin's extender algebra. One reference describing the discovery of the extender algebra is the introduction to Neeman's book The Determinacy of Long Games. I am basical …
- 8,099
15
votes
Accepted
What is the error in this disproof of the $\Omega$-conjecture?
Woodin's theorem says that assuming the $\Omega$ Conjecture and the existence of a proper class of Woodin cardinals, the set $\mathcal V_\Omega$ of $\Pi_2$ sentences that hold in every universe of the …
- 8,099
15
votes
Accepted
Consistency strength of strongly compact cardinal
Steel [1] showed that if $\square_\kappa$ fails for some singular strong limit $\kappa$, then $\text{AD}$ holds in ${L(\mathbb R)}$. Since Solovay showed a strongly compact cardinal implies the failur …
- 8,099
13
votes
A “paradox” about the inner model problem
The issue is Thesis 2. The notion of a canonical inner model is vague, but I don't think that when inner model theorists use the term they mean to suggest that the model is obtained by iterating a sha …
- 8,099
13
votes
Accepted
Natural set-theoretic principles implying the Ground Axiom
The existence of a proper class of supercompact cardinals $\kappa$ that are indestructible by $\kappa$-directed closed forcing implies the Ground Axiom. This is because it implies the Continuum Coding …
- 8,099
12
votes
Accepted
Does $Π^1_{2n+1} = (Σ^1_{2n+1})^{M_{2n-1}}$?
Your conjecture is true. We assume PD throughout. The proof I see requires citing a number of facts from inner model theory and descriptive set theory. First, it uses Woodin's theorem characterizing t …
- 8,099
12
votes
Accepted
When do infinitary compactness numbers exist?
The compactness number for $\mathcal L_{\kappa,\kappa}$ is equal to the least $(\kappa,\infty)$-strongly compact cardinal. A cardinal is $(\kappa,\infty)$-strongly compact if for every set $X$, there …
- 8,099
11
votes
Accepted
If $U,D$ are $\kappa$-complete nonprincipal ultrafilters on $\kappa$ and $j_U(U) = j_D(D)$, ...
It is consistent with ZFC that the answer is no, but under the Ultrapower Axiom, the answer is yes, not only for $\kappa$-complete ultrafilters on $\kappa$, but also for arbitrary countably complete u …
- 8,099
10
votes
Accepted
What are the difficulties involved in proving that the Kunen inconsistency holds in $NGB$
Here is the strategy behind McCallum's attempted proof that there is no nontrivial elementary embedding $j :V\to V$.
Step 1 is to cite the theorem that if a $j :V\to V$ is consistent, then a $j : V\t …
- 8,099