Diagramming: if A, then if B, then C.

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ScottRiqui
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Re: Diagramming: if A, then if B, then C.

Postby ScottRiqui » Tue Jan 15, 2013 11:11 pm

Shmoopy wrote:
ScottRiqui wrote:EDIT - ITT, are we not using "->" to mean "implies"? Because that's how I'm reading it. If I see a rule that says "(A AND B) -> C", then I'm using A and B to determine C. Specifically, "If A is true AND B is true, then this implies that C is true as well".


OK, I think I have got to the bottom of this disagreement. When I'm talking about an expression like (A AND B) -> C, I'm considering A, B, and C with separate truth values, and then I'm considering the truth value of the entire statement, which is determined by A, B, and C.

For example, if A = T, B = T, and C = F, I can plug all these truth values into the statement:

(A AND B) -> C = (T AND T) -> F = T -> F = F

This shows that A = T, B = T, and C = F yields ((A AND B) -> C) = F, so the entire expression has a truth value as well.

What you are doing is assuming that the entire expression is true. So if (A AND B) = T, and ((A AND B) -> C ) = T, then (T -> C) = T. This last conditional is only true when C is true, because (T -> T) = T and (T -> F) = F.Therefore, I don't think it's completely accurate to say that A AND B is true implies C is true, because this is only the case when the entire expression (A AND B) -> C is known to be true.

In the context of a logic game, we would probably know that the entire expression is true. IDK what the OP was even asking about, I'm just talking about logic in general at this point, but in other contexts, such as LR or just plain logic in general beyond the LSAT (which any LSAT prep should be grounded in IMO), the entire conditional could be either true or false. When we create truth tables, for example, we pick and choose combinations of A, B, and C being true or false (input) and see whether the entire expression is true (output).


That makes sense - I figured we were talking about two different things. In the context of a logic game rule, though, I still think C is a dependent variable. Imagine that you're given this rule:

"If a man's name is 'John' and he has red hair, then he's a criminal".

In this case, 'A' is whether he's named 'John', 'B' is whether he has red hair, and 'C' is whether he's a criminal. But they're not all independent variables - you use the state of A and B, along with the rule, to tell you about C. In this case, if both A and B are true (he's named 'John' and has red hair), then the rule tells you that C is true (he's a criminal).

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Shmoopy
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Re: Diagramming: if A, then if B, then C.

Postby Shmoopy » Tue Jan 15, 2013 11:31 pm

Shmoopy wrote:I just realized that I mixed up A -> (B-> C) with ( A -> B ) -> C. Derp. I will have to revise my original posts when I have time tonight.


Now I'm going to make truth tables comparing A -> (B -> C) and (A AND B) -> C. I am not assuming that either of these expressions is known to be true.

A B C A -> (B -> C)
T T T T
T T F F
T F T T
T F F T
F T T T
F T F T
F F T T
F F F T

A B C (A AND B) -> C
T T T T
T T F F
T F T T
T F F T
F T T T
F T F T
F F T T
F F F T


Hmm... they are the same. I guess I am wrong. In any case, I can show they are the same another way:

A -> (B -> C) = A -> (/B OR C) = /A OR (/B OR C) = /A OR /B OR C
(A AND B) -> C = /(A AND B) OR C = (/A OR /B) or C = /A OR /B OR C

That was fun I guess.




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