just to clarify, does A and B --> C mean the same as (A and B) --> C? Like the statement is NOT A, also B --> C? The parenthesis existing sometimes but not always is a bit hard to wrap one's head around, especially with any background in math. thank you!
Hmm, i understand and have no questions but feel that: if one tries this on the exam, then one will use lots time. if one uses a lot of time, then one will fail the test.
Commenting to come back to this if I need to. To make sure I do, here's a question people can respond to: if you could eliminate one type of question from the logical reasoning section, what would it be? I would chuck all the assumption questions out the window because I'm terrible at making assumptions in real life, so on tests I really struggle with this!
I think I've figured out where I'm struggling with this: how do we know we're dealing with an embedded conditional? Is it by identifying multiple conditional relationships indicated by structural indicators in the same sentence or even multiple conditional relationships without conditionals?
When it comes to embedded conditionals is there a table explaining variations of
A → (B → C) becomes A and B → C?
How is "B or C" translating to "/B → C?" (and vice versa).
I am trying to understand how to "pull the inside sufficient condition out and make it a sufficient conjunct in the outside conditional" .
Is there a section / lesson online explaining how this works? and all the various permutations and combinations of "if then" conditionals and how they translate to AND (&) OR (or) relationships?
For example I know that if I negate (A&B) I get :
(A&B) = /A or /B
So my question is what are the rules for opening up the brackets of a conditional as in A → (B → C) ?
Let me try it here:
Given
(A-> B)= (/A or B) (1)
(A-> B)= /(A & /B) (2)
How would you bring B in A → (B → C) out of the brackets?
A-> (/B or C) (from 1)
converting the remaining conditional
/A or (/B or C) (from 1)
opening up the brackets
/A or /B or C
bracketing /A or /B:
(/A or /B) or C
converting the right most or into a conditional
/(/A or /B) -> C
distributing the outermost not
(A & B ) -> C
Ok so I was able to derive that after all that effort but its not obvious, is it?
I don't think this is a necessary method to use, personally I have an interesting method that groups every sentence into either a hypothetical sentence, or an actual sentence.
If one lives in NYC then one must either file a city tax or pay a penalty.
NYC -> FT or PP
This version is a hypothetical situation, we the important part is that we DO NOT commit to either possibility of FT or PP.
Whereas
NYC + /FT -> PP,
NYC + /PP -> FT
these are actual statements, where you commit to a possibility. Hope this helps with conceptualising this transformation.
don't like this strategy. it just added confusion after I was already starting to grasp it all. Like others have said, using the same examples for this new technique confused me.
it is a little frustrating to use the same example over and over again for these techniques instead of new ones. i think already having an understanding of the rule of this example obscures the usefulness of the technique
Some conditional statements can be very confusing - especially if there's an "or" or "either/or" in the necessary condition!
Either/Or in the necessary can be super confusing when there's also a negation in the statement (ex: either this, or not that)
But we can simplify using Lawgic to understand what's going on!
Identify the Lawgic first. If there is an "or," write that in the necessary condition. Don't worry if it doesn't make sense yet.
Ex: A --> B or /C
Zero-in on the necessary condition of the Lawgic statement. (pretend the stuff to the left of the arrow doesn't exist right now).
Ex: B or /C
Apply the Group 3 translation rule to the necessary condition. Take either element, negate that element, and then make that element the sufficient condition. But wait! We're still working on the right side of the arrow. So, bracket your new Lawgic condition in parentheses.
(C --> B)
Here, I negated /C, which makes it a yes-C lol.
Now, we have 3 arrows. Bring on back that stuff from the left side of the arrow.
Ex: A --> (C--> B)
Scoot the parenthetical sufficient element out to the proper sufficient section. Have the element jump across the main arrow. Then, marry the sufficient elements together with "and"
Ex: A and C --> B
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138 comments
can someone please explain how it went from B10+ -> (R or /OpNo) to B10+ (OpNo -> R)? Did I forget a rule? That part is tripping me up.
If Jack walks he will go to the store or the movies.
Jack walks → store or movie
It becomes
Jack walks → /store → movies
Jack walks → store →/movies
just to clarify, does A and B --> C mean the same as (A and B) --> C? Like the statement is NOT A, also B --> C? The parenthesis existing sometimes but not always is a bit hard to wrap one's head around, especially with any background in math. thank you!
if you are reading this, you got this!! I believe in you!
A team who won the NBA finals must have scored more during regular time or scored more in overtime.
Won the NBA Finals--> (Scored more during Regular time OR scored more in Overtime)
Won the NBA Finals--> (/Score more during Regular time--> Scored more in Overtime)
Won the NBA finals-->(/Score more in Overtime--> Scored more in Overtime)
Pulling the embedded sufficient condition would be
Won the NBA Finals AND /Score more during Regular time--> Scored more in Overtime
Won the NBA Finals AND /Score more in Overtime--> Scored more during Regular time
This method makes it easier to identify the options that lead to a conclusion.
Hmm, i understand and have no questions but feel that: if one tries this on the exam, then one will use lots time. if one uses a lot of time, then one will fail the test.
TT > UT
UT > FT
TT > UT > FT
i haven't been confused until parenthesis started being involved. like what the heck even is this
Would you also be able to say:
NYC and PP -> /FT
NYC and FT -> /PP
Are you guys memorizing all of this? Who has made a cheat sheet? Haha and also who made a cheat sheet and found it useful for the actual exam?
this is literally algebra atp 😀
when would I have to use this?
It just keeps getting more complicated
Commenting to come back to this if I need to. To make sure I do, here's a question people can respond to: if you could eliminate one type of question from the logical reasoning section, what would it be? I would chuck all the assumption questions out the window because I'm terrible at making assumptions in real life, so on tests I really struggle with this!
does this still hold that B or C in A->B or C are jointly necessary?
I think I've figured out where I'm struggling with this: how do we know we're dealing with an embedded conditional? Is it by identifying multiple conditional relationships indicated by structural indicators in the same sentence or even multiple conditional relationships without conditionals?
he said peepee lol
I don’t know how I missed it: can someone point me to the lesson that talks specifically about what goes inside parenthesis? I feel dumb
When it comes to embedded conditionals is there a table explaining variations of
A → (B → C) becomes A and B → C?
How is "B or C" translating to "/B → C?" (and vice versa).
I am trying to understand how to "pull the inside sufficient condition out and make it a sufficient conjunct in the outside conditional" .
Is there a section / lesson online explaining how this works? and all the various permutations and combinations of "if then" conditionals and how they translate to AND (&) OR (or) relationships?
For example I know that if I negate (A&B) I get :
(A&B) = /A or /B
So my question is what are the rules for opening up the brackets of a conditional as in A → (B → C) ?
Let me try it here:
Given
(A-> B)= (/A or B) (1)
(A-> B)= /(A & /B) (2)
How would you bring B in A → (B → C) out of the brackets?
A-> (/B or C) (from 1)
converting the remaining conditional
/A or (/B or C) (from 1)
opening up the brackets
/A or /B or C
bracketing /A or /B:
(/A or /B) or C
converting the right most or into a conditional
/(/A or /B) -> C
distributing the outermost not
(A & B ) -> C
Ok so I was able to derive that after all that effort but its not obvious, is it?
I don't think this is a necessary method to use, personally I have an interesting method that groups every sentence into either a hypothetical sentence, or an actual sentence.
If one lives in NYC then one must either file a city tax or pay a penalty.
NYC -> FT or PP
This version is a hypothetical situation, we the important part is that we DO NOT commit to either possibility of FT or PP.
Whereas
NYC + /FT -> PP,
NYC + /PP -> FT
these are actual statements, where you commit to a possibility. Hope this helps with conceptualising this transformation.
don't like this strategy. it just added confusion after I was already starting to grasp it all. Like others have said, using the same examples for this new technique confused me.
it is a little frustrating to use the same example over and over again for these techniques instead of new ones. i think already having an understanding of the rule of this example obscures the usefulness of the technique
I don't think I like this strategy. May not use it
If one lives in NYC, then one must either file a city tax return or pay a penalty.
Isnt this a disjunction?
Once we write the conjunction, can we then negate the conjunction?
Here's my spin on this lesson.
Some conditional statements can be very confusing - especially if there's an "or" or "either/or" in the necessary condition!
Either/Or in the necessary can be super confusing when there's also a negation in the statement (ex: either this, or not that)
But we can simplify using Lawgic to understand what's going on!
Identify the Lawgic first. If there is an "or," write that in the necessary condition. Don't worry if it doesn't make sense yet.
Ex: A --> B or /C
Zero-in on the necessary condition of the Lawgic statement. (pretend the stuff to the left of the arrow doesn't exist right now).
Ex: B or /C
Apply the Group 3 translation rule to the necessary condition. Take either element, negate that element, and then make that element the sufficient condition. But wait! We're still working on the right side of the arrow. So, bracket your new Lawgic condition in parentheses.
(C --> B)
Here, I negated /C, which makes it a yes-C lol.
Now, we have 3 arrows. Bring on back that stuff from the left side of the arrow.
Ex: A --> (C--> B)
Scoot the parenthetical sufficient element out to the proper sufficient section. Have the element jump across the main arrow. Then, marry the sufficient elements together with "and"
Ex: A and C --> B