Formal Logic

Comments

• 1 Like

  Cant Get Right Yearly + Live Member Sage ⭐ 🍌 7Sage Tutor

  January 2017 27899 karma

  That's an interesting question. I'm leaning towards no, but not with as much confidence as I'd like!
  
  So with the (X ---> Y) as a sufficient condition, we aren't making any determinations about whether or not that relationship holds up. X may ---> Y or it may not. All we know is that IF it does, then Z. So, in Z & X ---> Y, it seems to me that we're having to satisfy this condition first. Z is a necessary condition, so we could theoretically deny the sufficient condition in which case Z is free to go either way. But by denying the sufficient, we're denying the relationship between X and Y. So they have nothing to do with each other. So in that scenario, Y is not strictly necessary from Z & X.
• 1 Like

  Sami Yearly + Live Member Sage ⭐ 7Sage Tutor

  January 2017 10806 karma

  
  (X ---> Y) ---> Z is lawgically equivalent to:
  (/X or Y) --->Z this is equivalent to:
  /(/X or Y) or Z this can be simplified to:
  (X and /Y) or Z
  I don't think this can be simplified any further.
  
  This above statement {(X and /Y) or Z} gives us two possibilities:
  /Z-----> X and /Y
  /(X and /Y)--> Z
  
  This means whenever Z doesn't happen we will have both X and /Y occur.
  And if we don't have (X and /Y), then we have to have Z occur.
  
  
  This was tough for me and I am not sure about this at all. @"Cant Get Right" what do you think?
  
• 3 Likes

  Jonathan Wang Yearly Sage ⭐

  January 2017 edited January 2017 6869 karma

  If all apples are bad, then candy is awesome. (A->B) -> C
  
  If candy is not awesome, then not all apples are bad. ~C -> ~(A->B)
  
  So now, your proposed inference:
  
  If candy is awesome and I have apples, then I have something bad. C and A -> B
  
  What do you think?