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

[ampm]

Say we have a sentence such as the following:

If this is, as it seems to be, intended as an explanation of what it is for an argument to be deductively valid, it is apt to turn out either false, if 'A' or else trivial, if 'B'.

This can be compared to the more general form of such sentences: if A, then if B, then C.

How would this be diagrammed?

A -> (B -> C)?

How about the contrapositive: /(B->C) -> /A? Would this be interpreted like so: If B doesn't lead to C, then not A?

I'm assuming that in this case, to show that: (if B, then C) can ONLY happen by showing that (if B, then NOT C). Can someone confirm?

[Shmoopy]

This can be compared to the more general form of such sentences: if A, then if B, then C.
How would this be diagrammed?

A -> (B -> C)?

I agree with your diagramming.

How about the contrapositive: /(B->C) -> /A? Would this be interpreted like so: If B doesn't lead to C, then not A?

I'm assuming that in this case, to show that: (if B, then C) can ONLY happen by showing that (if B, then NOT C). Can someone confirm?

I think you formed the contrapositive correctly. The negation of "B -> C" is "B -> /C," which means that "/( B -> C) -> /A" is equivalent to "( B -> /C) -> /A."

I'm a little lost by your last statement. "B -> C" can be shown to be true by demonstrating that its contrapositive " /C -> /B" is true. If you showed that " B -> /C" is true, then you've shown that " B -> C" is false.

[TripTrip]

I'm assuming that in this case, to show that: (if B, then C) can ONLY happen by showing that (if B, then NOT C). Can someone confirm?

If you want to invalidate or prove that /(B->C), then yes. You could also prove (/C->B).

[ampm]

This can be compared to the more general form of such sentences: if A, then if B, then C.
How would this be diagrammed?

A -> (B -> C)?

I agree with your diagramming.

How about the contrapositive: /(B->C) -> /A? Would this be interpreted like so: If B doesn't lead to C, then not A?

I'm assuming that in this case, to show that: (if B, then C) can ONLY happen by showing that (if B, then NOT C). Can someone confirm?

I think you formed the contrapositive correctly. The negation of "B -> C" is "B -> /C," which means that "/( B -> C) -> /A" is equivalent to "( B -> /C) -> /A."

I'm a little lost by your last statement. "B -> C" can be shown to be true by demonstrating that its contrapositive " /C -> /B" is true. If you showed that " B -> /C" is true, then you've shown that " B -> C" is false.

Sorry I mistyped here. What I meant to say is what you typed: The negation of "B -> C" is "B -> /C," which means that "/( B -> C) -> /A" is equivalent to "( B -> /C) -> /A."

I was trying to say that furthermore: /( B -> C) can only equal ( B -> /C) and not (/B -> C).

I think we're on the same page here.

[ampm]

I'm assuming that in this case, to show that: (if B, then C) can ONLY happen by showing that (if B, then NOT C). Can someone confirm?

If you want to invalidate or prove that /(B->C), then yes. You could also prove the contrapositive: (/C->B).

So I don't know if to invalidate /(B->C), we can show (/C->B). To show that (B -> C) is false I feel like we can ONLY show (B -> /C).

So in a LR question which has the statement: (A -> (B -> C)) and it gives us (/B -> C) or (/C -> B), we can't conclusively state anything but if we're given (B -> /C), then we can state (/A).

[TripTrip]

(B -> C) and (/C -> B) cannot both be true together, because the resulting statement would be (/C -> B -> C), which is clearly illogical. Thus, if you know (/C -> B) then you also know /(B -> C).

However, even with that, you still can't say anything definitive about (A -> (B -> C)) because we've only discussed the necessary end of the arrow.

Just kidding.
(A -> (B -> C))
/(B -> C)
Then, /A

[foggynotion]

(B -> C) and (/C -> B) cannot both be true together

I wasn't sure if the original question was about how to symbolize and deal with sentences like what the OP originally asked about in a practically way while taking the test, or if it was moreso about symbolic logic. But if it's about symbolic logic, then those two statements can both be true at the same time. It sounds like you're looking at it like this: /C -> /B (the contrapostive of the first statement, quoted above) and /C -> B can't both be true at the same time because then you'd have /C leading to two contradictory ideas. But all that means is that /C can't be true (or rather, C must be true). If instead I use the contrapositive of the second statement above-- /B -> C --together with the first statement-- B -> C, then what you'd have is a pair of statements that collectively say, "if B is true, then C is true; and if B is not true, then C is true". So either way, C is true.

I know that on a practical level, it's probably easiest to think of the negation of a conditional in the way that's been described: /(A -> B) is being rewritten as A -> /B because you want to say: "If it's not true that A leads to B, then A doesn't lead to B." So it's being written as A -> /B. Which is probably fine, as long as you know what you're trying to express. But if the original question was about how to deal with a negated conditional using symbolic logic, then it's been my experience that /(A -> B) gets expressed as A & /B, because the only way a conditional can be false (at least in modern symbolic logic as I know it) is for the antecedent ("A" in my example) to be true and the consequent ("B" in my example) to be false. Sentences like:

A -> B
A -> /B

are not mutually exclusive, because both sentences would be true as long as A is false.

Sorry, I hope I'm not being a jerk, but like I said, if the interest here is how to formally/symbolically negate a conditional, then that's the way to do it.

[TripTrip]

(B -> C) and (/C -> B) cannot both be true together

I wasn't sure if the original question was about how to symbolize and deal with sentences like what the OP originally asked about in a practically way while taking the test, or if it was moreso about symbolic logic. But if it's about symbolic logic, then those two statements can both be true at the same time.

Mmmm no.
Watch:

Assume (/C -> B)
Assume (B -> C)
Combined: (/C -> B -> C)
Remove the unnecessary link: (/C -> C) (Hypothetical syllogism)
Reductio ad absurdum
Ergo: /((/C -> B) & (B -> C)) (It is not true that both are true)
QED

It sounds like you're looking at it like this: /C -> /B (the contrapostive of the first statement, quoted above) and /C -> B can't both be true at the same time because then you'd have /C leading to two contradictory ideas. But all that means is that /C can't be true (or rather, C must be true). If instead I use the contrapositive of the second statement above-- /B -> C --together with the first statement-- B -> C, then what you'd have is a pair of statements that collectively say, "if B is true, then C is true; and if B is not true, then C is true". So either way, C is true.

We're symbolizing the statements differently. You're right based on the symbolism you listed.

I know that on a practical level, it's probably easiest to think of the negation of a conditional in the way that's been described: /(A -> B) is being rewritten as A -> /B because you want to say: "If it's not true that A leads to B, then A doesn't lead to B." So it's being written as A -> /B. Which is probably fine, as long as you know what you're trying to express. But if the original question was about how to deal with a negated conditional using symbolic logic, then it's been my experience that /(A -> B) gets expressed as A & /B, because the only way a conditional can be false (at least in modern symbolic logic as I know it) is for the antecedent ("A" in my example) to be true and the consequent ("B" in my example) to be false. Sentences like:

A -> B
A -> /B

are not mutually exclusive, because both sentences would be true as long as A is false.

Sorry, I hope I'm not being a jerk, but like I said, if the interest here is how to formally/symbolically negate a conditional, then that's the way to do it.

No disagreement here.

[ScottRiqui]

Say we have a sentence such as the following:

If this is, as it seems to be, intended as an explanation of what it is for an argument to be deductively valid, it is apt to turn out either false, if 'A' or else trivial, if 'B'.

This can be compared to the more general form of such sentences: if A, then if B, then C.

How would this be diagrammed?

A -> (B -> C)?

How about the contrapositive: /(B->C) -> /A? Would this be interpreted like so: If B doesn't lead to C, then not A?

I'm assuming that in this case, to show that: (if B, then C) can ONLY happen by showing that (if B, then NOT C). Can someone confirm?

"if A, then if B, then C" is logically equivalent to (A AND B) -> C

Likewise, that would make the contrapositive ~C -> ~A OR ~B

There's no need to re-invent the wheel here, folks.

[Shmoopy]

"if A, then if B, then C" is logically equivalent to (A AND B) -> C

Likewise, that would make the contrapositive ~C -> ~A OR ~B

There's no need to re-invent the wheel here, folks.

No. Compare truth tables for A AND B and A-> B. For example, if both A and B are false then the conditional is true but the conjunction is false. Conditional and conjunction are completely different.

[ScottRiqui]

.

[homestyle28]

"if A, then if B, then C" is logically equivalent to (A AND B) -> C

Likewise, that would make the contrapositive ~C -> ~A OR ~B

There's no need to re-invent the wheel here, folks.

No. Compare truth tables for A AND B and A-> B. For example, if both A and B are false then the conditional is true but the conjunction is false. Conditional and conjunction are completely different.

The sentence is saying "if you have A, then if you have B you have C" The truth table for that is:

A B C
___|_
0 0 0
0 1 0
1 0 0
1 1 1

That's the same truth table as "A AND B".

Truth tables for 3 variable sentences have 8 lines.

[ScottRiqui]

"if A, then if B, then C" is logically equivalent to (A AND B) -> C

Likewise, that would make the contrapositive ~C -> ~A OR ~B

There's no need to re-invent the wheel here, folks.

No. Compare truth tables for A AND B and A-> B. For example, if both A and B are false then the conditional is true but the conjunction is false. Conditional and conjunction are completely different.

The sentence is saying "if you have A, then if you have B you have C" The truth table for that is:

A B C
___|_
0 0 0
0 1 0
1 0 0
1 1 1

That's the same truth table as "A AND B".

Truth tables for 3 variable sentences have 8 lines.

C is the result, not one of the inputs. "If you have A, then if you have B you have C". That means the only independent variables are A and B, hence the four-line truth table.

But looking at it further, I guess the English sentence isn't exactly equivalent to (A AND B) -> C, because the English sentence doesn't say anything about C one way or the other in the cases where you *don't* have A.

[Shmoopy]

C is the result, not one of the inputs. "If you have A, then if you have B you have C". That means the only independent variables are A and B, hence the four-line truth table.

C is an independent variable! The result column is the truth value of the entire 3-variable sentence. You wrote a truth table for A AND B and called the result column C.

Writing out the two truth tables in question is too much work for me right now, so consider one case where they are different:

If A = F, B = F, C = F

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

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

[ScottRiqui]

C is the result, not one of the inputs. "If you have A, then if you have B you have C". That means the only independent variables are A and B, hence the four-line truth table.

C is an independent variable! The result column is the truth value of the entire 3-variable sentence. You wrote a truth table for A OR B (not even A AND B, unless I'm confused by the 0s and 1s) and called the result column C.

Writing out the two truth tables in question is too much work for me right now, so consider one case where they are different:

If A = F, B = F, C = F

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

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

I still maintain that C is a dependent variable. In the sentence originally given ("if A, then if B, then C"), in the case where both A and B are true, the sentence is telling you that C is true as well. The reason why I was incorrect in saying that it's the same as C = (A AND B) is that unlike C = (A AND B), the English sentence doesn't tell you ANYTHING about C in the cases where A and/or B are false.

Here's a similar example: If I were to say "If it's Tuesday, and it's raining, then I'll pick you up from work", there's nothing in that rule that would prevent me from picking you up from work on a sunny Friday. My C = (A AND B) equivalence would be incorrect, because it's saying that unless it's both Tuesday AND raining, that I will NOT be picking you up from work.

If you're claiming that C is an independent variable, then what would be the result in your truth table when A is true, B is true, and C is false? Because that directly conflicts with the English-language rule given in the first post ("if A, then if B, then C").

[questcertainty]

(B -> C) and (/C -> B) cannot both be true together

I wasn't sure if the original question was about how to symbolize and deal with sentences like what the OP originally asked about in a practically way while taking the test, or if it was moreso about symbolic logic. But if it's about symbolic logic, then those two statements can both be true at the same time.

Mmmm no.
Watch:

Assume (/C -> B)
Assume (B -> C)
Combined: (/C -> B -> C)
Remove the unnecessary link: (/C -> C) (Hypothetical syllogism)
Reductio ad absurdum
Ergo: /((/C -> B) & (B -> C)) (It is not true that both are true)
QED

I don’t follow this reasoning: a proper reductio reduces some set of sentences to a contradiction by applying rules of inference. But the sentence (~C --> C) is not a contradiction; it’s truth value is contingent on the assignment of truth values to the sentence letter, C. When C is true, the conditional sentence is true; when false, it’s false. This follows from the truth-conditions of the material conditional: (F --> T) is true. You could also convince yourself of this by transforming the original conditional into the equivalent disjunction, which is (C v C) – obviously contingent, being true when C is true, false when C is false.

Also, foggynotion is right, the two sentences, (~C --> B) and (B --> C), can both be true, under the assignment (C=T, B=T) and (C=T, B=F).

[ScottRiqui]

But the sentence (~C --> C) is not a contradiction.

But in the context of diagramming the rules to an LSAT problem, wouldn't that *have* to be a contradiction? It would be like having a rule in an "in/out" game that says "If Charles is not chosen, then Charles is chosen". I guess you could have the rule and not run afoul of it, so long as Charles is always chosen (basically, if there were never a case that would 'trigger' the sufficient condition).

[Shmoopy]

If you're claiming that C is an independent variable, then what would be the result in your truth table when A is true, B is true, and C is false? Because that directly conflicts with the English-language rule given in the first post ("if A, then if B, then C").

I don't understand how you can say that C is not an independent variable when C is the necessary condition of the conditional. This is axiomatic. C can be true or false, and the values of A and B don't tell you which it is. The fact that you give me this scenario with A true, B true, and C false supports the idea that C is independent. Otherwise, you could just tell me A true and B true without specifying C. You do specify C, because C could be either true or false when A and B are true.

Anyway,

If A = T, B = T, C = F

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

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

A, B, and C determine the truth values of these two expressions. A and B do not determine the truth value of C because I can have A true, B true, C true, or A true, B true, C false. One of these two scenarios would be impossible if C were not independent.

[questcertainty]

But the sentence (~C --> C) is not a contradiction.

But in the context of diagramming the rules to an LSAT problem, wouldn't that *have* to be a contradiction? It would be like having a rule in an "in/out" game that says "If Charles is not chosen, then Charles is chosen". I guess you could have the rule and not run afoul of it, so long as Charles is always chosen (basically, if there were never a case that would 'trigger' the sufficient condition).

In the case of an LSAT game like this, yeah, it would be contradictory, but that’s due to collateral premises in conjunction with the conditional, not the material conditional alone. The problem with what was said before is that the reasoning is appealing to the simple material conditional (or so I assumed from the context of a syllogism and arrows), which isn’t enough to engender a contradiction in that case.

[Shmoopy]

the English sentence doesn't tell you ANYTHING about C in the cases where A and/or B are false.

Is this not the exact meaning of C being independent? A and B don't tell you anything about C. Therefore, C does not depend on A and B.

[ScottRiqui]

If you're claiming that C is an independent variable, then what would be the result in your truth table when A is true, B is true, and C is false? Because that directly conflicts with the English-language rule given in the first post ("if A, then if B, then C").

I don't understand how you can say that C is not an independent variable when C is the necessary condition of the conditional. This is axiomatic. C can be true or false, and the values of A and B don't tell you which it is. The fact that you give me this scenario with A true, B true, and C false supports the idea that C is independent. Otherwise, you could just tell me A true and B true without specifying C. You do specify C, because C could be either true or false when A and B are true.

Anyway,

If A = T, B = T, C = F

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

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

A, B, and C determine the truth values of these two expressions. A and B do not determine the truth value of C because I can have A true, B true, C true, or A true, B true, C false. One of these two scenarios would be impossible if C were not independent.

I guess I'm just confused, because I'm interpreting the sentence from the first post ("if A, then if B, then C") like a game rule. Going back to the example of an in/out game, a rule of that type would be something like "If A is chosen, then if B is chosen, C is chosen as well". You're using the status of A and B to determine the status of C, making it dependent on A and B.

In this case, the only time you would know anything for certain about C would be in the case where both A and B are chosen. In that case, C is chosen as well. In any other case where A, B or both of them are not chosen, you can't say anything definitive about C.

And thinking more about it, this *is* equivalent to a game rule that says "(A AND B) implies C". If both A and B are chosen, then C is chosen as well. If either A or B (or both) are not chosen, then you can't say anything about C one way or the other.

[ScottRiqui]

the English sentence doesn't tell you ANYTHING about C in the cases where A and/or B are false.

Is this not the exact meaning of C being independent? A and B don't tell you anything about C. Therefore, C does not depend on A and B.

A and B don't *necessarily* determine C, but they can. In the case where both A and B are true/chosen/whatever, the rule says that C is true/chosen/whatever as well.


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".

[ScottRiqui]

(double post)

[Shmoopy]

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.

[Shmoopy]

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).