Linda says that, as a scientist, she knows that no scientist appreciates poetry. And, since most scientists are logic...

Skylar May 24, 2020

@Anna2020, happy to help!

Our diagram from the first sentence of the passage is:
S -> NOT P
P -> NOT S

Our diagram from the first part of the second sentence of the passage is:
S - most - L
L - some - S

We know that we can combine an S->N statement with a quantifier when they share the Sufficient condition so that the arrow is pointing away from the quantifier. This is the case with "L - some -S" and "S -> NOT P." The statements overlap with "S."

This gives us: L - some - S -> NOT P
We can simplify this to: L - some - NOT P

You ask about concluding the statement "S - most - L -> NOT P." This would be incorrect. Although we have the statement "S - most - L," we do not have the statement "L -> NOT P." Therefore, there is no S->N statement that can be combined with the quantifier that shares L as the Sufficient condition.

Does that make sense? Please let us know if you have any other questions!

Anna20 May 25, 2020

Hi @Skylar - thanks so much for this, really appreciate it. Would you also be able to help on the Argument Completion Drills?

I am struggling with how to come to the conclusion for the following:
P: A --> B
QS: Missing
C: B - some - C
How do you combine these using the transitive property for the quantifier statement?

Please could you also run through the steps for the following:
P: C - most - A
P: Missing
C: C - most not - B

P: C --> not D
P: Missing
C: not D - some - X

P: Z --> not C
P: Missing
C: not Z - some - A

I think I am not really understanding the strategy for completing these - and for setting these out in the transitive to find what the missing link is. Would be grateful for support on this! Many thanks.

Skylar May 28, 2020

@Anna2020, glad that makes sense! Of course, I'm happy to help with any other questions you may have.

Let's walk through the first Argument Completion Drill you listed. We start with:
P: A -> B
P: ?
C: B-some-C

Our first step should be to make any possible deductions that we can. For S->N statements, this means reversing and negating to find the contrapositive. For most or some statements, this means finding the reverse some statement. Doing so gives us:
P: A -> B
not B -> not A
P: ?
C: B-some-C
C-some-B

Now, our next step is to look for a way to connect our given premise to the conclusion. We should notice that we have a new variable in the conclusion- C- that is not mentioned in our given premise. Similarly, our given premise is an S->N statement but our conclusion is a some statement. This indicates that our missing premise will be a some statement with the variable C. Remember, we can connect S->N statements with quanitifer statements when they share the Sufficient condition so that the arrow points away from the quantifier. What is the Sufficient condition in our given premise? A.
So, let's make our missing premise: C-some-A.
We can now connect our missing premise (C-some-A) to the given premise (A->B) to get: C-some-A->B
We can simplify this further to: C-some-B, which is reversible to B-some-C and matches the conclusion.

So, our final answer is as follows (please note that C-some-A is equivalent in meaning to A-some-C, so either is correct as the missing premise):
P: A -> B
not B -> not A
P: C-some-A
A-some-C
C: B-some-C
C-some-B

Does that make sense? We can take a similar approach to answer the other drills you asked about.

The second one you listed was:
P: C-most-A
P: ?
C: C-most-not B

Again, let's start by making deductions from our given statements. Remember, most statements are reversible into some statements:
P: C-most-A
A-some-C
P: ?
C: C-most-not B
not B-some-C

Now, we need to figure out how to connect the premise to the conclusion. We should note that a new variable is mentioned in the conclusion- not B. This means that we should likely introduce not B in our missing premise. How can our given premise and our conclusion be the same most statement with only the right side different? We must use the transitive property to connect the quantifier with an S->N statement. We know we want the variable "A" to change to the variable "not B" from the given premise to the conclusion, so our missing premise should be: A -> not B. The contrapositive of this is: B -> not A.
We can combine the most statement in the given premise (C-most-A) with this missing premise (A -> not B) to get: C-most-A->not B.
We can simplify this further to get: C-most-not B, which is the conclusion.

So, our final answer is as follows (again, please note that A -> not B is equivalent in meaning to B -> not A, so either is correct as the missing premise):
P: C-most-A
A-some-C
P: A -> not B
B -> not A
C: C-most-not B
not B-some-C

At this point I recommend that you try to solve the remaining two drills on your own following the steps in this post. Practicing is the best way to master this skill. If you run into any questions while working through them, please let me know and I would be happy to help!