quick formal logic question Forum

[JJDancer]

Every B --> A
Every B --> D

Therefore some As are Ds

Is this valid?

[NewtonLied]

Yes.

[macaulian]

No it is invalid. If I tell you if something is a snozzberry it tastes like steak, and then I tell you that is something is a snozzberry it tastes like applesauce, that does not imply something exists that tastes like steak and tastes like applesauce. That would require that a snozzberry exists.

b->a
b->d
does not imply for some X, a and d, since I did not give you b. This is predicate logic.

[NewtonLied]

wrong. snozzberries taste like steak AND applesauce. Therefore some things that taste like steak (snozzberries) also taste like applesauace.

That is as long as you assume snozzberries exist. It depends on if one of the premises of the question is that there is at least one B, and the way it's phrased (every B) makes it sound as though that IS one of the assumptions.

[chocho]

i think it would only work if there are B's. without the B to connect them, u cant say there are some with the two characteristics.


basically im trying to say what is above me, but he wrotes its betta.

[JJDancer]

ok so assuming that yes Bs do exist, this would be valid right?
I'm thinking Venn diagram proves this too...

All Bs are As
All Bs are Ds

So All Bs fit inside A (and inside D) and some Ds have to overlap..

[yoni45]

For the purposes of the LSAT, when given All X are Y, you're allowed to assume that and least 1 X exists.

[ might run counter to some things, but take it as-is for the LSAT, and you'll sleep better at night... =) ]

Given that, yes, it's valid.

[Shrimps]

All blueberries are blue
All blueberries are berries

Therefore, there are some blue things that are berries, and there are some berries that are blue.

[NewtonLied]

All blueberries are blue
All blueberries are berries

Therefore, there are some blue things that are berries, and there are some berries that are blue.

Ugh, your conclusion is correct, but that logic is atrocious.

[iamtaw]

Every B --> A
Every B --> D

Therefore some As are Ds

Is this valid?

No.

If the information you had was:
Every B -> A
Every B -> D
There is at least one B.

in this case you can say that some As are Ds.
You can't claim existence from 2 conditionals

[thinkbig]

All blueberries are blue
All blueberries are berries

Therefore, there are some blue things that are berries, and there are some berries that are blue.

Ugh, your conclusion is correct, but that logic is atrocious.

you need to assert that blueberries exist. what if there is not at least 1 blueberry?

[NewtonLied]

All blueberries are blue
All blueberries are berries

Therefore, there are some blue things that are berries, and there are some berries that are blue.

Ugh, your conclusion is correct, but that logic is atrocious.

you need to assert that blueberries exist. what if there is not at least 1 blueberry?

In most formal logic classes you can usually take an empty set to be the special case. If it isn't defined as empty, you can generally assume it isn't. That's been my experience anyway. I think as far as the LSAT goes it's certainly a safe bet.

[SOCRATiC]

OMFG.... TITCR:

All B are A
All B are D
Therefore some A are D
-----------

"All B are A" means that some A are B. If some A are B, and if all B are D, then it must be true that some A are in fact D.

Also:Don't pay attention to what anyone else says.

[macaulian]

For the purposes of the LSAT, when given All X are Y, you're allowed to assume that and least 1 X exists.

[ might run counter to some things, but take it as-is for the LSAT, and you'll sleep better at night... =) ]

Given that, yes, it's valid.

I disagree, they can have a LG where they have two conditionals for which the antecedent cannot ever happen. For instance, If ray goes first, then susan goes second, and if ray goes first, jill goes third. If they asked the question that required you to determine if it possible for susan to go second and jill to go third, this could be false. For instance, if I told you susan and jill cannot be placed consecutively (if I give you this, you can conclude that ray never goes first).

Whenever you simplify the real rules of logic, you are making a mistake. There is no such thing as "for LSAT purposes". If they ever asked that question, did not give you that b exists, and the answer was that some a's are d's, people would flip out and they would have to cancel the question, because that is false. Making assumptions that are not given is why alot of people do poorly on the test.

[thinkbig]

All blueberries are blue
All blueberries are berries

Therefore, there are some blue things that are berries, and there are some berries that are blue.

Ugh, your conclusion is correct, but that logic is atrocious.

you need to assert that blueberries exist. what if there is not at least 1 blueberry?

In most formal logic classes you can usually take an empty set to be the special case. If it isn't defined as empty, you can generally assume it isn't. That's been my experience anyway. I think as far as the LSAT goes it's certainly a safe bet.

The important thing to realize is that, assuming at least 1 blueberry exists, it is true that there are some blue things that are berries. There could be blue things that are not berries, and there could be berries that are not blue. But the conjunction of the two sets guaranties that if all A are B and all A are C, then some B are C.

[thinkbig]

OMFG.... TITCR:

All B are A
All B are D
Therefore some A are D
-----------

"All B are A" means that some A are B. If some A are B, and if all B are D, then it must be true that some A are in fact D.

Also:Don't pay attention to what anyone else says.

+1. Good explanation.

[SOCRATiC]

OMFG.... TITCR:

All B are A
All B are D
Therefore some A are D
-----------

"All B are A" means that some A are B. If some A are B, and if all B are D, then it must be true that some A are in fact D.

Also:Don't pay attention to what anyone else says.

All Boys are human. --> Some humans are boys.
All Boys are dumb.

Some humans are boys + All boys are dumb = Some humans are dumb.

[macaulian]

All blueberries are blue
All blueberries are berries

Therefore, there are some blue things that are berries, and there are some berries that are blue.

Ugh, your conclusion is correct, but that logic is atrocious.

you need to assert that blueberries exist. what if there is not at least 1 blueberry?

In most formal logic classes you can usually take an empty set to be the special case. If it isn't defined as empty, you can generally assume it isn't. That's been my experience anyway. I think as far as the LSAT goes it's certainly a safe bet.

Ok, this is very wrong. First of all, this is predicate logic, not (edit) basic symbolic logic, which is often called propositional logic (thanks for the correction). Second, as someone who has taken logic, written papers on the subject, and taught a logic class, I can tell you that anyone who says "empty set" in reference to PL is BSing you. Once again, you cannot assume something exists in logic unless it is given. For example, here is modus ponens:

a->b
a
therefore, b

Without providing a, you cannot prove b. Never assume something not given. Most of the time something may or may not be true, and you are saying that is something may or may not be true, assume that it is true, that is absolutely wrong.

[macaulian]

OMFG.... TITCR:

All B are A
All B are D
Therefore some A are D
-----------

"All B are A" means that some A are B. If some A are B, and if all B are D, then it must be true that some A are in fact D.

Also:Don't pay attention to what anyone else says.

+1. Good explanation.

To bad it is absolutely and utterly wrong. See above. If b->a does not mean there exists a that is b. It never has and never will.

[NewtonLied]

For the purposes of the LSAT, when given All X are Y, you're allowed to assume that and least 1 X exists.

[ might run counter to some things, but take it as-is for the LSAT, and you'll sleep better at night... =) ]

Given that, yes, it's valid.

I disagree, they can have a LG where they have two conditionals for which the antecedent cannot ever happen. For instance, If ray goes first, then susan goes second, and if ray goes first, jill goes third. If they asked the question that required you to determine if it possible for susan to go second and jill to go third, this could be false. For instance, if I told you susan and jill cannot be placed consecutively (if I give you this, you can conclude that ray never goes first).

Whenever you simplify the real rules of logic, you are making a mistake. There is no such thing as "for LSAT purposes". If they ever asked that question, did not give you that b exists, and the answer was that some a's are d's, people would flip out and they would have to cancel the question, because that is false. Making assumptions that are not given is why alot of people do poorly on the test.

If they say:
some lawyers are athletes
some lawyers are politicians

and the answer is some politicians are athletes, would you contest that based on the fact that there might not be lawyers? More importantly, would you really not just assume there are lawyers when you answer the question?

[thinkbig]

OMFG.... TITCR:

All B are A
All B are D
Therefore some A are D
-----------

"All B are A" means that some A are B. If some A are B, and if all B are D, then it must be true that some A are in fact D.

Also:Don't pay attention to what anyone else says.

+1. Good explanation.

To bad it is absolutely and utterly wrong. See above. If b->a does not mean there exists a that is b. It never has and never will.

I was assuming that B exists. Not safe to assume on the LSAT, but IF any B exists, then it is also A. Therefore, IF any B exists, then there is some A that is also B.

[SOCRATiC]

Ok, this is very wrong. First of all, this is predicate logic, not symbolic logic. Second, as someone who has taken logic, written papers on the subject, and taught a logic class, I can tell you that anyone who says "empty set" in reference to PL is BSing you. Once again, you cannot assume something exists in logic unless it is given. For example, here is modus ponens:

a->b
a
therefore, b

Without providing a, you cannot prove b. Never assume something not given. Most of the time something may or may not be true, and you are saying that is something may or may not be true, assume that it is true, that is absolutely wrong.

In Symbolic Logic, you have Propositional Calculus AND Predicate Logic. Predicate logic is a subdivision of Symbolic Logic, you dimwit. The demonstration of your ignorance of this fact completely undermines the credibility boost that you attempted to achieve by mentioning your credentials (taking a logic course, writing papers, and teaching the god damned subject).

[macaulian]

For the purposes of the LSAT, when given All X are Y, you're allowed to assume that and least 1 X exists.

[ might run counter to some things, but take it as-is for the LSAT, and you'll sleep better at night... =) ]

Given that, yes, it's valid.

I disagree, they can have a LG where they have two conditionals for which the antecedent cannot ever happen. For instance, If ray goes first, then susan goes second, and if ray goes first, jill goes third. If they asked the question that required you to determine if it possible for susan to go second and jill to go third, this could be false. For instance, if I told you susan and jill cannot be placed consecutively (if I give you this, you can conclude that ray never goes first).

Whenever you simplify the real rules of logic, you are making a mistake. There is no such thing as "for LSAT purposes". If they ever asked that question, did not give you that b exists, and the answer was that some a's are d's, people would flip out and they would have to cancel the question, because that is false. Making assumptions that are not given is why alot of people do poorly on the test.

If they say:
some lawyers are athletes
some lawyers are politicians

and the answer is some politicians are athletes, would you contest that based on the fact that there might not be lawyers? More importantly, would you really not just assume there are lawyers when you answer the question?

It may be reasonable to assume the existence of a lawyer, but is it reasonable to assume the existance of a snozzberry or a situation where ray goes first? Yes in the first, not in the two examples I have given, so the answer to the question is that for something a and d is true does not follow, as this is general. Just because a line of reasoning works in some specific cases does not mean it is valid. If I can provide one counter example (I have given two), then the reasoning is invalid. Your example proves that you must first determine whether the anticedent of the two conditionals is ever true. By reasoning that lawyers exist your are making that determination.

[thinkbig]

For the purposes of the LSAT, when given All X are Y, you're allowed to assume that and least 1 X exists.

[ might run counter to some things, but take it as-is for the LSAT, and you'll sleep better at night... =) ]

Given that, yes, it's valid.

I disagree, they can have a LG where they have two conditionals for which the antecedent cannot ever happen. For instance, If ray goes first, then susan goes second, and if ray goes first, jill goes third. If they asked the question that required you to determine if it possible for susan to go second and jill to go third, this could be false. For instance, if I told you susan and jill cannot be placed consecutively (if I give you this, you can conclude that ray never goes first).

Whenever you simplify the real rules of logic, you are making a mistake. There is no such thing as "for LSAT purposes". If they ever asked that question, did not give you that b exists, and the answer was that some a's are d's, people would flip out and they would have to cancel the question, because that is false. Making assumptions that are not given is why alot of people do poorly on the test.

If they say:
some lawyers are athletes
some lawyers are politicians

and the answer is some politicians are athletes, would you contest that based on the fact that there might not be lawyers? More importantly, would you really not just assume there are lawyers when you answer the question?

Good point. Part of the LSAT instructions are that you need to apply standard real world assumptions. You have to assume that lawyers exist. If it is a special specific case (If Jim is in line before Julia, then Jim is third) then you can;t assume that the antecedent is extant unless it is declared.

[iamtaw]

For the purposes of the LSAT, when given All X are Y, you're allowed to assume that and least 1 X exists.

[ might run counter to some things, but take it as-is for the LSAT, and you'll sleep better at night... =) ]

Given that, yes, it's valid.

I disagree, they can have a LG where they have two conditionals for which the antecedent cannot ever happen. For instance, If ray goes first, then susan goes second, and if ray goes first, jill goes third. If they asked the question that required you to determine if it possible for susan to go second and jill to go third, this could be false. For instance, if I told you susan and jill cannot be placed consecutively (if I give you this, you can conclude that ray never goes first).

Whenever you simplify the real rules of logic, you are making a mistake. There is no such thing as "for LSAT purposes". If they ever asked that question, did not give you that b exists, and the answer was that some a's are d's, people would flip out and they would have to cancel the question, because that is false. Making assumptions that are not given is why alot of people do poorly on the test.

If they say:
some lawyers are athletes
some lawyers are politicians

and the answer is some politicians are athletes, would you contest that based on the fact that there might not be lawyers? More importantly, would you really not just assume there are lawyers when you answer the question?

also just by purely using logical language,
if you just have
1. B -->A
2. B -->D
you cannot get the conclusion
A&D