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In/out games bi-conditional confusion

Hans ZimmerHans Zimmer Member
in Logic Games 541 karma

Hi everyone,

So I had a discussion with a fellow 7sager about the conditional relationship v ----> /z and how it pertains to grouping/in/out games.

My understanding from the CC is that this relationship would not constitute a bi-conditional, because we can fail the sufficient condition /v, rendering the rule irrelevant. If it were a bi-conditional, /z would have to also be a sufficient condition for v, but this doesn't seem to be the case because as I just mentioned we can have v (v IN) and /v (v OUT). For this to be a bi-conditional would it not have to further be specified that we must have "either v or z, but not both" ?

The other 7sager mentioned that it would be a bi-conditional in a two group, non In/Out game. I was under the impression however that all two group grouping games can be conceived of as In/Out games -we seem to just arbitrarily assign one group as the in group and one group as the out group.

I would really appreciate any help because this has been running through my mind all day.

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  • jugolo96jugolo96 Alum Member
    edited July 2020 103 karma

    Sorry! I think I made this way more complex than it really is. You are correct in the way you are thinking through it so don't worry about it.

    In a normal situation v -->/z would be exactly as you put it. If V is in, then Z is in the other group. If Z is not in, we cannot say what follows, V and Z could both be out (basic rule that you got right, satisfying a necessary condition produces no inferences).

    Now imagine the following situation:

    We have two groups, Group A and Group B.
    We have the following elements: V, Z, X, Y
    All the elements can be in only one group and must all be utilized

    Rule: If V is in a group, then Z cannot be in that group (V-->/Z)

    This is a "camouflaged" biconditional. Essentially, since both letters must be used (in one of the groups, does not matter which). When V is in one of them, Z is in the other. And when Z is in one of the groups, then V must be in the other. What this rule really is:

    Va ->/Za and since we only have two groups, Va -> Zb (Za ->Vb)

    Vb ->/Zb and since we only have two groups, Vb ->Za (Zb ->Va)

    making it a bi-conditional:

    Va <-> Zb
    Vb <-> Za

    Since we only have two groups where we can place the items (binary setup), were both groups are treated the same, they must always be in the opposite group to one another, making it a biconditional.

    The rule itself is NOT strictly written as a biconditional, but since it just specifies a "group" and both A and B are groups (the only groups), it is camouflaging a biconditional relationship. Having to use all elements and having only two possible groups (both of which are treated the same), puts you in a situation where V and Z are always going to send the other to the opposite group.

    Now if I said:

    Rule: If V is in group A, then Z cannot be in that group (Va-->/Za)

    Now I do NOT have a biconditional

    Hope this helps!

  • Hans ZimmerHans Zimmer Member
    541 karma

    @jugolo96

    This clears things up!
    Thanks a lot for taking the time to write this out :)

  • jugolo96jugolo96 Alum Member
    103 karma

    No worries! :smiley:

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