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Sybolic logic, help please

  1. Oct 31, 2007 #1
    I have a couple of homework problems that I can't get, hopefully one of you enlightened ones can help me.
    I have to give proofs of these valid arguments using only the 8 rules of inference (M.P. etc.), and 10 replacement rules(D.N. etc.) :

    (1)
    1.(A>E)>C
    2.C>~C
    /A

    (2)
    1.(A&G)>H
    2.A
    3.(I>~H)&(A>G)
    /G (Triple Bar, Biconditional) H

    (3)
    1.(A&K)>R
    2.K
    /A>R

    ">" is, "if then", "&" is the dot that resembles multiplication, but stands for the word "and". The numbers. "1., 2., 3.", are the premises. "/" is the conclusion
    I need help....too math like for me
    *symbolic Logic
     
    Last edited: Oct 31, 2007
  2. jcsd
  3. Oct 31, 2007 #2

    CompuChip

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    Welcome to PF.
    Can you please explain your notation a little more?
    What are the premises? Or is 1. the premise and 2. the conclusion for each part?
    What are the letters (are they all formulas? or do some have special meanings?) What do the @ and / stand for?
     
  4. Oct 31, 2007 #3
    I edited my first post to be more accurate
     
  5. Nov 1, 2007 #4

    CompuChip

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    Maybe it helps if you first write out the proofs for yourself. For example, in number three:
    1.(A&K)>R
    2.K
    /A>R

    Need to prove A -> R. So you must suppose A and show R or suppose not R and show not A (and then use some rule to flip them around). The latter is more complicated, so let's try the first one. You want to prove R and you in fact have a premise that has R as its conclusion, #1. The condition of this premise is (A&R). By assumption you already have A, so you still need to show that K. But this is just assumption 2.

    Now try to do the same for #1 and #2, and post it here so we can check if the problem is there or just in formally writing it down.

    Now, let me inverse the argument to a logical order:
    Suppose that A holds. By #2, K holds. So A and K hold, therefore R holds by #1. So assuming A, we proved R; therefore we can prove (without assumptions) that A implies R.

    Now again, try to convert the reasoning to such a format for 1) and 2) yourself.

    Finally, write this down in the correct formal way. You've basically done all the work already, just have to think which inference rules you need from one step to the next:
    1. A (assumption)
    2. (A&K) > R (premise 1)
    3. K (premise 2)
    4. A & K (&I 1, 3)
    5. R (>E 2, 4)
    6. A > R (>I 1, 5)
    where the last line invalidates (marks, daggers, deactivates, whatever you want to call it) the assumption on line 1.

    Finally, try to do this for 1) and 2) with the results you got above.

    Please post as much as you can do, so we can try to pinpoint where the problem is in your case.
     
    Last edited: Nov 1, 2007
  6. Nov 21, 2007 #5
    To solve these problems, I would recommend you to look at the conclusion first and work backwards from there.

    (1)
    1. (A > E) > C ____________Premise / Conclusion: A
    2. C > ~C ____________Premise
    3. ~C v ~C ____________CE 2
    4. ~C ____________DUP 2
    5. ~(A > E) ____________MT 1
    6. ~(~A v E) ____________CE 5
    7. A & ~E ____________DeM 6
    8. A ____________Simp 7

    (2)
    1. (A & G) > H ____________Premise / Conclusion: G=H
    2. A ____________Premise
    3. (I > ~H) & (A > G) ____________Premise
    4. A > (G > H) ____________Exp 1
    5. G > H ____________MP 2,4
    6. A > G ____________ Simp 3
    7. G ____________MP 2,6
    8. ~H v G ____________Add 7
    9. H > G ____________CE 8
    10. (G > H) & (H > G) ____________Conj 5,9
    11. G=H ____________BE 10

    (3)
    1. (A & K) > R ____________Premise / Conclusion: A > R
    2. K ____________Premise
    3. (K & A) > R ____________Comm 1
    4. K > (A > R) ____________Exp 3
    5. A > R ____________MP 2,4
     
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