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Extension theorem (Existence)

  1. May 19, 2010 #1
    1. The problem statement, all variables and given/known data

    "Given a function H, assigning a value, 0 or 1, to each atomic sentence, define a sequence ℑ0, ℑ1, ℑ2, ℑ3,... of functions, as follows:
    0 is just H.

    Given a function ℑn, assigning a value, either 0 or 1, to the sentences of degree less than or equal to n, define the function ℑn+1, assigning a value, either 0 or 1, to the sentences of degree less than or equal to n+1, as follows: If φ has degree less than or equal to n, ℑn+1(φ) = ℑn(φ)."

    I confused with the english, (sorry i'm not good in english). please clarify this


    H({0,1}) = ℑ0, ℑ1, ℑ2, ℑ3,...
    (does this mean "Given a function H, assigning a value, 0 or 1, to each atomic sentence, define a sequence ℑ0, ℑ1, ℑ2, ℑ3,... of functions"???)

    and what "as follows: ℑ0 is just H." means? as follows? does is mean "such as"?

    and i'll post the next question after this answered, because i wan to clear this first.

    i'm sorry but the concise language is too concise for me i guess. help
     
  2. jcsd
  3. May 19, 2010 #2
    A thought experiment

    Let x - 2 = 0 from this it follows that x = 2.

    Understand the meaning now ?

    Let try to formulate this another way let [tex]f(x) = x^2[/tex] which is defined as follows

    [tex] - 1 \leq x \leq 1 [/tex] that is the general meaning of the term "defined as follows" :)
     
  4. May 19, 2010 #3
    so, anything after "as follows :" is the domain of the function?

    which means,

    H({0,1}) = {ℑ0, ℑ1, ℑ2, ℑ3,... }, and the domain is "ℑ0 is just H"

    T_T i don't think it is correct of what i was doing
     
  5. May 19, 2010 #4
    For which subject is this for? Discrete mathematics ?
     
  6. May 19, 2010 #5
    hmm, i'm doing self-study, and i guess the subject is "Mathematical Logic".
     
  7. May 19, 2010 #6
    Thats not my field but found some notes online in more plain english....

    http://www.math.psu.edu/simpson/courses/math557/logic.pdf [Broken]

    Maybe they can help you :)
     
    Last edited by a moderator: May 4, 2017
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