Here's my attempt at an invalid and valid argument. Someone correct me if I'm going about it wrong!
Invalid: All peaches have pits. Some pits are rotten. Therefore, some peaches are rotten.
The reason why this is invalid is because we don't know the peach is rotten. It could be because the air had weird chemicals in it, or maybe a bug was hiding in the peach. Simply knowing that some peach pits are rotten isn't enough to know that the peach itself is rotten.
Valid: Some peaches have pits. All pits are rotten. Therefore, some peaches are rotten.
This is valid because we know ALL of B is true. If all pits are rotten and we know some peaches have pits, we at least know some of the peaches are rotten.
What I have found that trips me up is thinking 'all A are B'. To me, it makes it sound like they are interchangeable ideas, when really it just means all of A is in the B group. It does not guarantee that if some (or most) of B is part of the C group, that that part of B has any A in it at all.
My understanding of this is that all A have B, but just because they have B doesn't mean that A will take on C. Because it can't be confirmed that just because you have B you also have C. The key indicator is "some". A is sufficient for B but does not guarantee that B brings about C.
All Options at Marys dessert shop is frozen yogurt, some frozen yogurt is mint flavored; therefore some of the Options at Marys shop is mint.
this in incorrect because the statement about mint is about all frozen yogurt, it doesn't tell us anything about if frozen yogurt at Marys dessert shop. It is just as likely she only serves vanilla and chocolate. we cannot disprove or prove she has mint, just that it is possible.
Remember: a valid conclusion about intersecting sets can only be made if the some arrow appears before the all arrow.
Ex.
Artists in the master class are painters. Some people who call themselves painters suck at art. Therefore, some artists in the master class suck at art.
Art's -> Pnt's <-s-> Suck@it
____
Art's <-s-> Suck@it
Thus is NOT valid
Again, just because the first set shares a characteristic with the second (painters), does not mean you can equate the two.
Ex.
Some artists in the master class are painters. All people who paint suck at art. Therefore, some artists in the master class suck at art.
Art's <-s-> Pnt's -> Suck@it
____
Art's <-s-> Suck@it
Here, we know 100% of the subset (painters) ALL share the quality we are discussing (sucking at it). So, we can easily conclude that the portion of artists in the class that are painters definitely suck at it.
im confused on the quantifiers. If we say all A's are B's, and we equate themselves logically, can they be interchangeable? A B and All B's are C's (or A = B), we cans say A B/C (because theyre interchangeable .
Why cant it work with some? All A's are B's (or A = B) and B C, therefore A/B C.
Is it all just contextual basing it on what sets youre talking about?
I think it helps to think about it like order of operations.
If you dump some of you A's into your B's
then dump all of your B bucket into your C's, then you will definitely have some A's in your C's.
But if you dump all of your A's into your B's
and then dump some of your B's into your C's, then you won't necessarily have any A's in your C's, you could, but its not guaranteed. If it is not guaranteed then it is not valid.
I think this reasoning also work for the most rule as well.
I have the same issue with this lesson as the previous. In V1 we were taught that EVERY which is the same as ALL before SOME is a valid argument structure as if ALL A are B and SOME A are C then Some B are C is a VALID ARGUMENT with the use of ALL before SOME. Can someone please clarify why the following is incorrect?
I think it would really be helpful to use the scoops example from previous lessons across these as well. It started to click for me when I read someones comment about scoops!
"Given that the previous argument is invalid and we can transform this argument into the previous one, then this one must be invalid as well."
I thought inferring only went one way - all down to most down to some. Wouldn't we transform the previous argument into this one (A —m→ B implies A ←s→ B) and not the other way around? #feedback
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54 comments
A--->B
B<s>C
A<s>c
this is invalid. but if there is at least 1 B in the set of C
then shouldnt there be at least 1 A in the set of C. Since all A's are B's?
Here's my attempt at explaining:
All of A are B: all humans are mammals (A --> B)
Some B are C: some mammals have tails (B <s> C)
Therefore, some humans have tails (A <s> C)
We cannot conclude that some humans have tails just because all humans are mammals. I tried to use an example that was obvious, I hope this helps!
who else is confused :p
Here's my attempt at an invalid and valid argument. Someone correct me if I'm going about it wrong!
Invalid: All peaches have pits. Some pits are rotten. Therefore, some peaches are rotten.
The reason why this is invalid is because we don't know the peach is rotten. It could be because the air had weird chemicals in it, or maybe a bug was hiding in the peach. Simply knowing that some peach pits are rotten isn't enough to know that the peach itself is rotten.
Valid: Some peaches have pits. All pits are rotten. Therefore, some peaches are rotten.
This is valid because we know ALL of B is true. If all pits are rotten and we know some peaches have pits, we at least know some of the peaches are rotten.
What I have found that trips me up is thinking 'all A are B'. To me, it makes it sound like they are interchangeable ideas, when really it just means all of A is in the B group. It does not guarantee that if some (or most) of B is part of the C group, that that part of B has any A in it at all.
My understanding of this is that all A have B, but just because they have B doesn't mean that A will take on C. Because it can't be confirmed that just because you have B you also have C. The key indicator is "some". A is sufficient for B but does not guarantee that B brings about C.
why is there no video for the important subjects :'(
Ok here is my attempt at an invalid argument.
All Options at Marys dessert shop is frozen yogurt, some frozen yogurt is mint flavored; therefore some of the Options at Marys shop is mint.
this in incorrect because the statement about mint is about all frozen yogurt, it doesn't tell us anything about if frozen yogurt at Marys dessert shop. It is just as likely she only serves vanilla and chocolate. we cannot disprove or prove she has mint, just that it is possible.
Here's an example I made for myself if it helps:
Trap 5: Swapping "some" and "all" arrows:
Remember: a valid conclusion about intersecting sets can only be made if the some arrow appears before the all arrow.
Ex.
Artists in the master class are painters. Some people who call themselves painters suck at art. Therefore, some artists in the master class suck at art.
Art's -> Pnt's <-s-> Suck@it
____
Art's <-s-> Suck@it
Thus is NOT valid
Again, just because the first set shares a characteristic with the second (painters), does not mean you can equate the two.
Ex.
Some artists in the master class are painters. All people who paint suck at art. Therefore, some artists in the master class suck at art.
Art's <-s-> Pnt's -> Suck@it
____
Art's <-s-> Suck@it
Here, we know 100% of the subset (painters) ALL share the quality we are discussing (sucking at it). So, we can easily conclude that the portion of artists in the class that are painters definitely suck at it.
im confused on the quantifiers. If we say all A's are B's, and we equate themselves logically, can they be interchangeable? A B and All B's are C's (or A = B), we cans say A B/C (because theyre interchangeable .
Why cant it work with some? All A's are B's (or A = B) and B C, therefore A/B C.
Is it all just contextual basing it on what sets youre talking about?
So what about the following argument? Is this valid?
A→B
A ←s→ C
B←s→ C
I think it helps to think about it like order of operations.
If you dump some of you A's into your B's
then dump all of your B bucket into your C's, then you will definitely have some A's in your C's.
But if you dump all of your A's into your B's
and then dump some of your B's into your C's, then you won't necessarily have any A's in your C's, you could, but its not guaranteed. If it is not guaranteed then it is not valid.
I think this reasoning also work for the most rule as well.
I found that drawing the circles for the subsets and supersets make it much easier to understand.
im going to need a video explanation on this one chief im starting to dive myself insane
So basically, the smaller amount needs to come before the larger amount in order for an argument to be true
Why B —m→ C is stronger than B ←s→ C? And what does it mean it is stronger?
Okay so pretty much in order for this to be valid is by this:
A→B←s→C
----------------
B←s→C
And this would be valid because A is not affecting the conclusion??
Invalid:
A→B←s→C
-------------------
A←s→C
X-
This could also be valid if it said Some before All obviously.....
Can someone please explain this in English and not Lawgic?
So in this and the last one, to put it very simply, if there is an arrow before the ←s→ or ‑m→ it is invalid? The ←s→ and ‑m→ must be first.
I have the same issue with this lesson as the previous. In V1 we were taught that EVERY which is the same as ALL before SOME is a valid argument structure as if ALL A are B and SOME A are C then Some B are C is a VALID ARGUMENT with the use of ALL before SOME. Can someone please clarify why the following is incorrect?
Okay so if this is invalid:
A → B ←s→ C
_
A ←s→ C
can this be valid??
A → B ←s→ C
_
B ←s→ C
So to make it work it would be B that goes to A that goes to C? #feedback
#feedback
I think it would really be helpful to use the scoops example from previous lessons across these as well. It started to click for me when I read someones comment about scoops!
.
Am I getting tripped up on this sentence?
"Given that the previous argument is invalid and we can transform this argument into the previous one, then this one must be invalid as well."
I thought inferring only went one way - all down to most down to some. Wouldn't we transform the previous argument into this one (A —m→ B implies A ←s→ B) and not the other way around? #feedback