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Conditionality

jessicamorehead
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Hi there, I just wanted to confirm my understanding on some conditionality rules. I am reviewing the "Mapping vs Grouping" video for lesson 6 and am working through the June 1995 game.

I totally get that we can only diagram absolute rules, rather than possible ones.
On the main diagram, we have the following:

R :dblline: S :dblline: U :arrow: T (w/ another arrow from U :arrow: R - just not sure how to show that on here).

Since we know that U :arrow: R and R :dblline: S, we can combine those rules to produce U :dblline: S.

Where I get confused is understanding if that logic works in the other direction. For example, we know S :dblline: U and U :arrow: T. Can we combine that into the following S :dblline: U :arrow: T and simplify it to S :dblline: T?? Why or why not? I figured we can NOT simplify it into S :dblline: T because of the two reasons:

1) When S is in, U is out. And when U is out, we fail the sufficient condition for U :arrow: T, so that rule falls away. So T could be in or out.

2) When U is in, S is out. And U :arrow: T still holds true.

But I am still a little confused on when you can condense rules vs when you cannot. Can someone please clarify?
Jamena Pirone
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Hi Jessica,

Great question.

Since we know that U :arrow: R and R :dblline: S, we can combine those rules to produce U :dblline: S.


That is absolutely correct. We can combine those rules because the Rs overlap. They are both non-negated, regular Rs.

we know S :dblline: U and U :arrow: T. Can we combine that into the following S :dblline: U :arrow: T and simplify it to S :dblline: T?


These two rules CANNOT be combined, as you correctly stated, because there are no overlapping variables. Be very careful when working with the bi-conditional arrow. In S :dblline: U it may look like that U is a non-negated U, but in reality it's ̶U̶. The negation is embedded within the arrow itself!

When working with biconditional arrows and condensing rules, it is very important to be aware of whether there is a negation embedded in the arrow, which could make your intended overlap points effectively opposites. In general, you can attach conditional statements leading into a biconditional statement without worry, but you must be vigilant when attempting to connect statements to the end of a biconditional arrow.

Hope that helps!
jessicamorehead
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Jamena,

I think I need to stay away from the biconditional arrow, as it generally confuses me when I am trying to think quickly while solving a problem. In terms of singular arrows:

U :arrow: R
R :arrow: ~S
combines into U :arrow: ~S since the first rule has R in the necessary and the second has R in the sufficient.

However,
U :arrow: T
S :arrow: ~U
CANNOT combine since they have no variables in common.
The contrapositive of the second rule would be U :arrow: ~S. However, since U :arrow: T and U :arrow: ~S both have U in the sufficient side, we cannot create a chain.

Thank you for your clarification, yet again!
Francis O'Rourke
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since U :arrow: T and U :arrow: ~S both have U in the sufficient side, we cannot create a chain.


That is perfect reasoning! We cannot infer a chain from two conditional rules that only share a sufficient condition. All that we know is that when a plane is flying in Zone U, it must also be in Zone T, and it cannot be in Zone S. There are two things that we know about any plane flying in Zone U, but this is not a chain.

It sounds like you are working through this concept very well. Keep at it and let us know if you have any additional questions! :-D