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Yes.


Apr
29
comment Formal second-order statements of Archimedean and completeness properties
Right, of course. Thank you. :^)
Apr
29
accepted Formal second-order statements of Archimedean and completeness properties
Apr
29
asked Formal second-order statements of Archimedean and completeness properties
Apr
10
comment Are these two statements equivalent?
There's several pages about this construction and translations in general here: math.stackexchange.com/questions/121388/…
Apr
3
revised Why there's difference between $\forall x \in T\ \exists y \in S\ F(x,y)$ and $\exists y \in S\ \forall x \in T\ F(x,y)$
typo: changed "true" to "false".
Apr
3
suggested suggested edit on Why there's difference between $\forall x \in T\ \exists y \in S\ F(x,y)$ and $\exists y \in S\ \forall x \in T\ F(x,y)$
Apr
2
comment Automorphisms of elementary extensions
You can answer your own question and accept the answer.
Mar
28
awarded  Nice Answer
Mar
28
answered Sheffer stroke the most important advance in logic?
Mar
27
answered What does Tarski mean by a “tautological operation” on a Boolean algebra?
Mar
27
comment What is the “Boolean algebra fragment of RA”?
@AndrásSalamon, also, to address the other connection with logics (the algebraic logic connection), note that (the 2 element) BA is equivalent to propositional logic, which is also well-known to be decidable.
Mar
27
comment What is the “Boolean algebra fragment of RA”?
@AndrásSalamon, I would take "boolean algebra is decidable" to mean that the (presumably first-order) theory of Boolean algebras is decidable, i.e., FOL + the three axioms B1-B3 listed in the article. The FO theory of BA is indeed decidable, as established by Tarski in 1949. See en.wikipedia.org/wiki/… or math.psu.edu/simpson/notes/master.pdf (page 30 4.2.2) or google for plenty of info. It's not some particular Boolean algebra that is decidable but rather the set of sentences about this class of structures.
Mar
25
revised Is the 'variable' in 'let $y=f(x)$' free, bound, or neither?
deleted 109 characters in body
Mar
25
comment Infinite Disjunctions and Conjunctions
@Hayden, I'm not familiar with Wolfe. If you don't already have Mendelson and can handle graduate texts, check out Shoenfield before getting Mendelson. And you might also still like Machover.
Mar
25
comment Why $\bar{A}A+\bar{A}B\Rightarrow B$?
@Voldemort, Consider the (true) sentence "if you have a child, then your mother has a grandchild". This doesn't assert that you have a child; it only asserts the truth of the implication as a whole, i.e., a certain relationship between you having a child and your mother having a grandchild. The author must be using "$\Rightarrow$" in a similar way. As you note, it isn't technically possible for $\bar{A}A$ to be true, so you have to think in slightly different terms -- either syntactically (in terms of deductions) or in terms of sets of models.
Mar
25
revised Is the 'variable' in 'let $y=f(x)$' free, bound, or neither?
added 28 characters in body
Mar
25
revised Is the 'variable' in 'let $y=f(x)$' free, bound, or neither?
added 20 characters in body
Mar
25
revised Is the 'variable' in 'let $y=f(x)$' free, bound, or neither?
added 20 characters in body
Mar
25
revised Is the 'variable' in 'let $y=f(x)$' free, bound, or neither?
added 20 characters in body
Mar
25
revised Is the 'variable' in 'let $y=f(x)$' free, bound, or neither?
added 560 characters in body