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Help with Diagramming Lawgic

notwilliamwallacenotwilliamwallace Alum Member
edited April 2016 in Logic Games 1049 karma
Hey Guys,

Needed some help with diagramming a couple of sentences into lawgic and their contraceptives.

1. If Aliens or Ghosts are in the house, then Tim and John are not in the house...Is the info I have written below correct?

- The way I have diagrammed this is like: A or G --> /T and /J. Taken further, this becomes: A --> /T, A --> /J ; G --> /T, G --> J. If Aliens are in the house, then Tom is not there. If Aliens are in the house, then John is not there. If Ghosts are in the house, then Tom is not there. If Ghosts are in the house, then John is not there.
- Contrapositive: T or J --> /A and /G. Taken further, this becomes: T --> /A, T --> /G ; J --> /A, J --> /G. If Tom is in the house, then Aliens are not the house. If Tom is in the house, then Ghosts are not in the house. If John is in the house, then Aliens are not in the house. If John is in the house, then Ghosts are in the house.

2. Totally confused about diagramming "Tom will play really well for his soccer team if John or Bill, but not both, play on the same team as him."

I understand that John or Bill, but not both will be diagrammed as /J <--> B, where contrapositive is J <--> /B. But how do I diagram the info about Tom. Is this correct? /J <--> B --> T? If yes, what will be the contrapositive?

Thanks for your help, as always.
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Comments

  • quinnxzhangquinnxzhang Member
    611 karma
    Regarding (1), yes, your diagram and inferences are correct, assuming "Tim and John are not in the house" is to be interpreted as "neither Tim nor John is in the house".

    Regarding (2), I personally prefer to use the exclusive-or ("XOR") for these instances, but your biconditional diagram is equivalent. What you have is correct. I also recommend using parentheses to make it clear which connective takes the widest scope -- i.e. '(~J ↔ B) → T'.

    I think the parentheses will also help you see what the contrapositive is -- i.e. '~T → ~(~J ↔ B)'. If you want to distribute the negation into the biconditional, this becomes ~T → (J ↔ B)'.

  • notwilliamwallacenotwilliamwallace Alum Member
    1049 karma
    Thank you @quinnxzhang ... that was very helpful.
  • runiggyrunruniggyrun Alum Inactive Sage Inactive ⭐
    2481 karma
    @quinnxzhang that was a brilliant explanation. The use of parentheses is imo crucial to keep everything straight, especially if you have to draw contrapositives.
    I Just wanted to add that I've come across two games that make use of this kind of conditional (one was something about if yews are out in then either laurels or oaks but not both are in) from PTB, and there was another one in a more recent PT that I can't seem to find right now. So, they definitely come in handy!
  • notwilliamwallacenotwilliamwallace Alum Member
    1049 karma
    Hey @quinnxzhang (or anyone else reading), I get the lawgic diagram that you have included: (~J ↔ B) → T, however is the following wrong?

    "Tom will play really well for his soccer team if John or Bill, but not both, play on the same team as him." ... (/J and B) or (J and /B) --> T ; /T --> (J and B) or (/J and /B)
  • LSATislandLSATisland Free Trial Inactive Sage
    1878 karma
    Looks good @notwilliamwallace
  • quinnxzhangquinnxzhang Member
    edited April 2016 611 karma
    @notwilliamwallace, yes those inferences are correct. Note that '(/J and B) or (J and /B)' is just a statement of exclusive-or, which, as noted above, is equivalent to the '(~J ↔ B)' translation you originally proposed.

    However, again, I recommend being a bit more strict with parentheses so that you can see that the conditional takes wider scope over the disjunction, i.e. '((~J&B) ∨ (J&~B)) → T'.

    This will also let you see that the consequent of the contrapositive, '~T → ~((~J&B) ∨ (J&~B))', can be worked out via De Morgan's laws. '~((~J&B) ∨ (J&~B))' is equivalent to '~(~J&B) & ~(J&~B)' by one application of De Morgan's. Applying De Morgan's one level further gives you '(J∨~B) & (~J∨B)'. Converting this to DNF gives you '(J&B) ∨ (~J&~B)', which is what you have and which is equivalent to the '(J ↔ B)' translation above.
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