Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with an SD derivation

  1. Jul 26, 2006 #1
    Ok, so I'm doing some logic review on my own. It's been awhile since I've done derivations, so I'm a little rusty. I'm just trying to use SD rules and not SD+. I'd appreciate any help you all can offer. Thanks in advance.

    Derive: (G&H)->J

    1. E->(F->G) assum.
    2. H->(G->I) assump
    3. (F->I)->(H->J) assump
    ______________________________________________
    4. |E assumption
    5. | |G&H assumption
    6. | |H 5 &E
    7. | |G->I 6,2 ->E
    8. | |F->G 4,1 ->E
    9. | | |F asummp
    10. | | |G 8,9 ->E
    11. | | |I 7,10 ->E
    12 | | F->I 9-11 ->I
    13. | | H->J 3, 12 ->E
    14. | | J 6, 13 ->E
    15. | (G&H)->J 5-14 ->I



    This is where I get stuck. I do get #15 out of the subderivation of E? Or am I going about this all wrong? Can you even derive this thing?

    Thanks
     
  2. jcsd
  3. Jul 26, 2006 #2

    honestrosewater

    User Avatar
    Gold Member

    Haha, that is a good question to ask. By my check, the tableaux closes without even needing (1). Your proof looks fine, but all you can get out of (4) is ~E or a formula of the form E -> p, e.g., E -> ((G & H) -> J). The key seems to lay with I. You can get I from (2) by assuming (G & H). Can you see how to get a formula from (3) that, effectively, makes you choose between ~I or J?

    On second thought, let me put it this way: The only way that F -> I can be false is for I to be ... ??
     
    Last edited: Jul 26, 2006
  4. Jul 26, 2006 #3
    Rule of thumb: whenever you're trying to prove a conditional claim, the first assumption you make should always be the antecedent of the conditional. You didn't follow this rule; that's why you're having trouble discharging your assumptions at the end.

    And the argument is valid: I did a truth-table that shows no row where all premises true and conclusion false.

    Derive: (G&H)->J

    1. E->(F->G) assum.
    2. H->(G->I) assump
    3. (F->I)->(H->J) assump
    ______________________________________________
    4. |G&H (assumption)
    5. | H 4, &E
    6. | G 4, &E
    7. | G -> I 2, 5, ->E
    8. || F assumption
    9. || I 6, 7, ->E
    10.| F -> I 8-9, ->I
    11.| H -> J 3, 10 ->E
    12.| J 11, 5 ->E
    13.(G&H) -> J 4-12, ->I QED

    Note that the first premise is irrelevant -- I never use it.
     
  5. Jul 26, 2006 #4

    honestrosewater

    User Avatar
    Gold Member

    Hello again, NickJ. :smile: FYI: We aren't allowed to give full solutions here unless the person is having serious problems.
     
  6. Jul 26, 2006 #5
    Thanks everyone. I appreciate it. You're hint hones helped out greatly. Thanks for taking the time NickJ to write out a sollution, even though you weren't supposed to. :biggrin:

    This is a great site, and I'm looking forward to reading and posting here.
     
  7. Aug 8, 2006 #6
    Oops! Now I know...and knowing is half the battle.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help with an SD derivation
  1. Need Some Help with SD (Replies: 6)

  2. SD Logics Help (Replies: 1)

Loading...