Hey TLS,
Reviewing my In/Out logic chain fundamentals and a question came to mind. Appreciate any help.
Let's say we have an In/Out game with 4 elements ABCD. And the logic chain we have is A > ~B > ~C > D. From the A/B "not both" rule and the C/D "or" rule, we know that at least one of C/D is IN, and at least one of A/B is OUT. Can we draw any additional inference(s) from the B/C rule, and if so, how could we represent it? I think the answer is No, just leave that rule alone and do not put it into your chart. But in theory, at least one of A/C is out as well, but then there is overlap between the inferences at least one of A/C out and at least one of A/B out, which can get confusing, especially if you allocate 2 slots for these inferences (since they could just take up one OUT slot).
So it it best to just use each variable once for inference sake (A/B and C/D inferences), and then just refer back to your chain for individual questions? Hope my question is clear.
Thank you!!
Quick In/Out logic chain question

 Posts: 1
 Joined: Thu Sep 14, 2017 10:30 am
Re: Quick In/Out logic chain question
Bigbear, I am quite confused with this question. This is one of my first posts and I am also studying for the LSAT.
So the players are ABCD and its an in an out set up
I think the Rules are:
1) C or D is in
2) A/B cannot both be in.
3) What does the B/C rule state?
Until you clarify the exact rules of the game, I think it will be hard to get any help from others. Also remember that just because A/B cant both be in.. doesnt mean they both cant be out.
So the players are ABCD and its an in an out set up
I think the Rules are:
1) C or D is in
2) A/B cannot both be in.
3) What does the B/C rule state?
Until you clarify the exact rules of the game, I think it will be hard to get any help from others. Also remember that just because A/B cant both be in.. doesnt mean they both cant be out.

 Posts: 107
 Joined: Tue Jun 06, 2017 10:34 pm
Re: Quick In/Out logic chain question
BigBear85 wrote:Hey TLS,
Reviewing my In/Out logic chain fundamentals and a question came to mind. Appreciate any help.
Let's say we have an In/Out game with 4 elements ABCD. And the logic chain we have is A > ~B > ~C > D. From the A/B "not both" rule and the C/D "or" rule, we know that at least one of C/D is IN, and at least one of A/B is OUT. Can we draw any additional inference(s) from the B/C rule, and if so, how could we represent it? I think the answer is No, just leave that rule alone and do not put it into your chart. But in theory, at least one of A/C is out as well, but then there is overlap between the inferences at least one of A/C out and at least one of A/B out, which can get confusing, especially if you allocate 2 slots for these inferences (since they could just take up one OUT slot).
So it it best to just use each variable once for inference sake (A/B and C/D inferences), and then just refer back to your chain for individual questions? Hope my question is clear.
Thank you!!
I think the easiest way to think about this particular example is by framing it (edit: that is, trying to work out all of the possible outcomes by following the logic chain as far as we can) and allowing the inferences to work themselves out. Everything's connected, so we're down to two (edit: three, derp. just woke up) possibilities:
Either:
In: A, D Out: B, C
Or:
In: C, B Out: A, D (which you get if you just take the contrapositive of each rule)
Or: [edit]
In: D, B Out: C, A (which you get by starting with B in)
(I'm assuming the rules are A>~B, ~B>~C, and ~C>D.)
edit number 596: so in short, I'm not totally sure what your question is.
 wmbuff
 Posts: 232
 Joined: Mon Sep 04, 2017 6:26 pm
Re: Quick In/Out logic chain question
sev wrote:BigBear85 wrote:Hey TLS,
Reviewing my In/Out logic chain fundamentals and a question came to mind. Appreciate any help.
Let's say we have an In/Out game with 4 elements ABCD. And the logic chain we have is A > ~B > ~C > D. From the A/B "not both" rule and the C/D "or" rule, we know that at least one of C/D is IN, and at least one of A/B is OUT. Can we draw any additional inference(s) from the B/C rule, and if so, how could we represent it? I think the answer is No, just leave that rule alone and do not put it into your chart. But in theory, at least one of A/C is out as well, but then there is overlap between the inferences at least one of A/C out and at least one of A/B out, which can get confusing, especially if you allocate 2 slots for these inferences (since they could just take up one OUT slot).
So it it best to just use each variable once for inference sake (A/B and C/D inferences), and then just refer back to your chain for individual questions? Hope my question is clear.
Thank you!!
I think the easiest way to think about this particular example is by framing it (edit: that is, trying to work out all of the possible outcomes by following the logic chain as far as we can) and allowing the inferences to work themselves out. Everything's connected, so we're down to two (edit: three, derp. just woke up) possibilities:
Either:
In: A, D Out: B, C
Or:
In: C, B Out: A, D (which you get if you just take the contrapositive of each rule)
Or: [edit]
In: D, B Out: C, A (which you get by starting with B in)
(I'm assuming the rules are A>~B, ~B>~C, and ~C>D.)
edit number 596: so in short, I'm not totally sure what your question is.
So, the rules reduce to:
~A v ~B
B v ~C
C v D
This set can be satisfied by:
A, D
B, C, D
B, C
B, D
D
Unless you place constraints on the numbers in and out, you have those five potential In groups.
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