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Proposition Calculus

  • 08-09-2005 02:21PM
    #1
    Yook
    Closed Accounts Posts: 437 ✭✭


    Basically my problem is not being able to prove any of the axioms, even the simplest p = p (where = is equivalence).

    Does anyone know how to do these, is there method to their madness?
    Like could anyone solve:

    Axoim (3.32): p V q = p V ¬q = p (where = is equvalence, cant find the one with 3 lines :p )

    Thanks for any help in advance!


Comments

  • #2
    ecksor
    Moderators, Social & Fun Moderators Posts: 10,501 Mod ✭✭✭✭ecksor


    Axioms aren't supposed to be proved, they're supposed to be an apparently consistent set of assumptions from which theorems can be derived.

    http://mathworld.wolfram.com/Axiom.html


  • #3
    Yook
    Closed Accounts Posts: 437 ✭✭Yook


    ecksor wrote:
    Axioms aren't supposed to be proved, they're supposed to be an apparently consistent set of assumptions from which theorems can be derived.

    http://mathworld.wolfram.com/Axiom.html

    Sorry its theorem 3.32, using the others with numbers below.


  • #4
    Adblock
    Closed Accounts Posts: 110 ✭✭Adblock


    Propositional Calculus is rooted in with boolean algreba very closely..

    to solve a simple truth like this, you just replace the letters with binary digits
    like
    p=0 // p and q can be any varity of 0,0 1,1, 01 01 etc
    q=1

    p V q = p V ¬q = (p i think this should be q) other wise the "V" is the Logical And

    0 V 1 = 0 V not1 = 1

    so 0 V 1 reads (zero "Logical [OR]" one )
    0 V 1 =[look up the truth table at] http://educ.queensu.ca/~compsci/resources/BoolLogic/summary.html

    The only problem is some lecturers/people differ on the symbol used for and and or xor etc i do believe the teacher/person u have is using V to mean LogicalAND other wise the =p should be = q
    V is normall used to repersent the Logial OR operator , but it must be And


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