Howdy, Stranger!

It looks like you're new here. If you want to get involved, click one of these buttons!

Sign In Apply for Membership

Categories

Diagramming problem

babybennybabybenny Free Trial Member
edited April 2014 in Logical Reasoning 156 karma
No A is B until C are both D and E.

Since there're two conditional expressions ("No~is~" and "~until~") in one sentence, I'm confused how it makes as a diagram (eg. A->B sth like that).
Which one is the sufficient condition and the necessary condition?

Please explain me.
Thanks!
Show Related Discussions

Comments

  • michellemoon0708michellemoon0708 Alum Member
    79 karma
    I'll give this a try

    No a is b = a-->\b
    C are both d and e = c-->d & e

    So, a-->\b "until" c--> d & e = [a-->b] --> [c-->d & e]
  • babybennybabybenny Free Trial Member
    156 karma
    Thank you so much!
    Sorry but I have one more question.

    You said [a-->b] --> [c-->d & e], so what is contrapositive?
  • michellemoon0708michellemoon0708 Alum Member
    79 karma
    Yeah no problem!

    I think the contrapositive is /[/e or /d-->/c]-->/[/b-->a]! However, the LSAT will not ask you to diagram such a complex conditional, so don't sweat this!
  • Jonathan WangJonathan Wang Yearly Sage ⭐
    edited April 2014 6866 karma
    Good try, michellemoon, but that's not quite right.

    "Until" is the conditional indicator here. Group 3. The two concepts are "No A is B" and "C is both D and E". In other words, the conditional statements (A -> /B) and (C -> D and E) are the two ideas we're relating here.

    When we use group 3, we pick a condition, negate it, and make it sufficient. Let's go with (A -> /B) as the condition you pick. Do you remember how to negate a conditional relationship? You have to negate the entire thing: /(A->/B). For clarity: that first slash should be slashing out the entire statement in parentheses. Then, drop the other half in the necessary.

    It ends up looking like this: /(A->/B) -> (C -> D and E)

    In English: IF A does not always imply /B, THEN C always implies both D and E. Or in other words, if there exists even one A that is also a B, then Cs are always Ds and Es.

    Contrapositive: /(C -> D and E) -> (A -> /B)

    In English: IF C does not always imply D and E, THEN A always implies /B. Or in other words, if there's even one C that isn't also both a D and an E, then As are always /Bs.

    Review the lesson on "Negating All Statements" if you don't remember the mechanics of this.
Sign In or Register to comment.