for "Some Before All" and "Most Before All" does it have to be before? can't it be an all statement, and then a some/most statement and it still works??
If all implies most and most implies some, cant we just convert all these most arrows to some arrows making it easier for us to conclude that some A are E or whatever else we want to conclude?: A --> B -m-> C -m-> D -m-> E
While I have come to appreciate "Lawgic", it was a bit tricky to follow in the beginning. I wish the curriculum was structured so that we got this entire list (in this format) prior to learning about each type of valid argument so that as we were learning we could see how each component contributes to the overall argument, rather than the structure of the argument being revealed in Lawgic after the fact.
Not that it would make a huge difference to the overall learning, but I think it would have helped me follow along a bit easier.
Visualizing these arguments is helping me learn them:
The Conditional Argument:
Imagine a conveyor belt (→) where if A items come in, B items come out. If you put something on the conveyor belt, you will get the corresponding outcome.
The Contrapositive Argument:
Visualize a seesaw with A on one end and B on the other. If B goes down, then A must go up to balance it out.
Conditional Chaining:
Picture a train moving from A to B to C. If the train goes from A to B and then from B to C, logically, it's like it directly went from A to C.
Some Before All:
Imagine a group of people entering a room. If some people (A) are in a room with all of B and some of B are in a room with all of C, then logically, some of A must also be in the room with all of C.
Most Before All:
Think of a river flowing from A to B to C. If most of the water flows from A to B and then from B to C, it means that there's a significant portion of the water from A that ends up in C.
Two Mosts:
Visualize two buckets, one labeled B and the other labeled C. If most of the items in bucket A can fill both buckets B and C, then logically, there must be some overlap between the contents of buckets B and C.
One of the things that helped me with visualizing conditionals with quantifiers will be below. IDK if I am right so I welcome feedback. This is based on my understanding from the lessons on here and two videos I saw on YouTube.
With chaining the quantifiers is that when it is a quantifier before all, we are trying to see if there is a relationship between the two supersets. These must be connected by a common subset. And in all cases but one, one of the conditional relationships must be all:
X's → Y's
(all/most/some) X's → Z's
the inference will always be X's ←s→ Z's because we do not how big Y or Z are.
The one exception is most
Most X's → Y's
Most X's → Z's
Inference: Some Y's
The other one is a straight chain like A→B→C. Like we learned about the subset → superset lessons, I will call B the sub-superset because it acts as both and IDK a proper word for it. The reason why the quantifier needs to be before the all in the chain ( how its actually written does not matter) is because if we say all of A is B and most of B is C, we do not know how big A or B are.
However, since all of B is in C, even if A is just 0.1% in B, A is the same amount in C. Therefore, the relationship between the subset and the superset will be the same as the subset sub=superset. So, the sub-superset needs to be all in in the superset. If this relation unfolds, any relationship between A and B will be the same between A and C so long as all of B is in C.
what does the “(x)” mean for the conditional & contrapositive ? I feel pretty solid on those but I don’t remember seeing that in the original version of this syllabus (just switched to this version so maybe I missed that).
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40 comments
Definitely book marking!
Simple examples for anyone struggling:
1. Conditional:
Basketball players are athletes
Tom plays basketball
---
Tom is an athlete
2. Contrapositive:
Basketball players are athletes
Tom is not an athlete
---
Tom does not play basketball
3. Conditional Chaining:
Kittens are cats. Cats are cute
---
Kittens are cute
4. Some before all:
Some cats are pets. (All) Pets are kind
---
Some cats are kind
5. Most before all:
Most cats are pets. (All) Pets are kind
---
Most cats are kind
6. Two Mosts:
Most cats are pets. Most cats are kind.
---
Some pets are kind
Love this breakdown summary probably most important page in this section
I'm still a bit confused about when we use all these formal arguments
#help for formal argument #6 Two Mosts:
Is it correct to say that if most of A are B, and most of A are C, that (instead of saying most of B are C) some of A are B and C?
I don't know how to comprehend this argument without thinking of it like that. Here's an example I've been using:
Most dogs have leashes.
Most dogs have collars.
Therefore, some dogs have leashes and collars.
But (correct me if I'm wrong) this^ argument in lawgic form would look more like:
Premise: A -m-> B
Premise: A -m-> C
Conclusion: A B and C
Am I just overcomplicating things? Or is my version incorrect?? Idk I'm getting confused if anyone can help me on this.
Whoever is studying and reading this keep crushing it.
Stick with it you owe it to yourself.
over and outttieeee.
for "Some Before All" and "Most Before All" does it have to be before? can't it be an all statement, and then a some/most statement and it still works??
Some before all example
Some dogs love kibble. All dogs that love kibble are energetic.
dogs ←s→ love kibble → energetic
therefore we can conclude......
some dogs are energetic
Most before all example
Most people who drink water are healthy. All people who are healthy are athletic.
drink water ‑m→ healthy → athletic
therefore we can conclude...
Most people who drink water are athletic.
Two mosts example
Most students are smart. Most students have planners.
therefore we can conclude....
Some smart students have planners.
Correct me if any of these are incorrect.
If all implies most and most implies some, cant we just convert all these most arrows to some arrows making it easier for us to conclude that some A are E or whatever else we want to conclude?: A --> B -m-> C -m-> D -m-> E
#feedback in the skill builders, this form appears quite a lot:
All As are Bs
All As are Cs
Therefore, some Bs are Cs.
Shouldn't this also be part of the list then?
#help Is this valid? A→B, A→C, therefore B←s→C?
Is there a form that follows:
A‑m→B
A←s→C
--------
A←s→C?
While I have come to appreciate "Lawgic", it was a bit tricky to follow in the beginning. I wish the curriculum was structured so that we got this entire list (in this format) prior to learning about each type of valid argument so that as we were learning we could see how each component contributes to the overall argument, rather than the structure of the argument being revealed in Lawgic after the fact.
Not that it would make a huge difference to the overall learning, but I think it would have helped me follow along a bit easier.
#feedback
Can we get this in a diagram?
what does the (x)/B mean?
Visualizing these arguments is helping me learn them:
The Conditional Argument:
Imagine a conveyor belt (→) where if A items come in, B items come out. If you put something on the conveyor belt, you will get the corresponding outcome.
The Contrapositive Argument:
Visualize a seesaw with A on one end and B on the other. If B goes down, then A must go up to balance it out.
Conditional Chaining:
Picture a train moving from A to B to C. If the train goes from A to B and then from B to C, logically, it's like it directly went from A to C.
Some Before All:
Imagine a group of people entering a room. If some people (A) are in a room with all of B and some of B are in a room with all of C, then logically, some of A must also be in the room with all of C.
Most Before All:
Think of a river flowing from A to B to C. If most of the water flows from A to B and then from B to C, it means that there's a significant portion of the water from A that ends up in C.
Two Mosts:
Visualize two buckets, one labeled B and the other labeled C. If most of the items in bucket A can fill both buckets B and C, then logically, there must be some overlap between the contents of buckets B and C.
Should I memorize this or work on being familiar with the language surrounding these arguments?
How should I be learning this concept?
One of the things that helped me with visualizing conditionals with quantifiers will be below. IDK if I am right so I welcome feedback. This is based on my understanding from the lessons on here and two videos I saw on YouTube.
With chaining the quantifiers is that when it is a quantifier before all, we are trying to see if there is a relationship between the two supersets. These must be connected by a common subset. And in all cases but one, one of the conditional relationships must be all:
X's → Y's
(all/most/some) X's → Z's
the inference will always be X's ←s→ Z's because we do not how big Y or Z are.
The one exception is most
Most X's → Y's
Most X's → Z's
Inference: Some Y's
The other one is a straight chain like A→B→C. Like we learned about the subset → superset lessons, I will call B the sub-superset because it acts as both and IDK a proper word for it. The reason why the quantifier needs to be before the all in the chain ( how its actually written does not matter) is because if we say all of A is B and most of B is C, we do not know how big A or B are.
However, since all of B is in C, even if A is just 0.1% in B, A is the same amount in C. Therefore, the relationship between the subset and the superset will be the same as the subset sub=superset. So, the sub-superset needs to be all in in the superset. If this relation unfolds, any relationship between A and B will be the same between A and C so long as all of B is in C.
Just a general comment... I come from a mathematics background and all of this reminds me of calculus but with the English language.
Very complex structuring. Making sure we understand English very well, I can see, will help immensely on the exam!
Best of luck everyone!
Also (previous lesson):
A → B
A —m→ C
B ←s→ C
#help#feedback should the very last argument here have a conclusion of "B ←s→ C"?
Isn't the last one (Two Mosts) wrong? it should be B←s→C
#help (Added by Admin)
#help
what does the “(x)” mean for the conditional & contrapositive ? I feel pretty solid on those but I don’t remember seeing that in the original version of this syllabus (just switched to this version so maybe I missed that).