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!

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4 comments

  • Wednesday, Apr 16 2014

    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.

    3
  • Monday, Apr 14 2014

    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!

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  • Monday, Apr 14 2014

    Thank you so much!

    Sorry but I have one more question.

    You said [a-->b] --> [c-->d & e], so what is contrapositive?

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  • Monday, Apr 14 2014

    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]

    2

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