Review

Question 1

Qualifiers will change the logic of any chained condition:

Most Before All vs All Before Most:

if the all arrow shows up first in the chain and then you see the most arrow, there are no valid conclusions to be drawn via the chain.

A —m→ B → C yields valid conclusions via the chain. (Conclusion: A —m—> C)
A → B —m→ C yields no valid conclusions via the chain. (Conclusion: none)

Answer 1

A —m→ B → C:
We know that most of the A set falls into the B set; we also know that ANY member of the B set will also be in the C set.
This means that any of the A set (those covered by the “most”) that fall into the B set will simultaneously be in the C set by virtue of membership in the B set.

A → B —m→ C:
Even if all of A set falls into B set, we cannot guarantee that it will overlap with ‘most of the B set’ since A set could be tiny relative to the B set.
For example, even if ALL of A set’s members are in B set but that only accounts for 1% of the B set, MOST of the B set falling into the C set could easily not include anyone in the A set.

Question 2

Some before All vs. All before Some:

A ←s→ B → C yields valid conclusions via the chain. (Conclusion: A <— s —> C)

A → B ←s→ C yields no valid conclusions via the chain. (Conclusion: none)

Answer 2

From the lesson on the relationships between the quantifiers, you know that the “most” arrow implies the “some” arrow which means that “A —m→ B” implies “A ←s→ B.” 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.

Another way you can see this is to think about how strong your premises need to be to support a conclusion. B —m→ C is stronger than B ←s→ C. If the stronger premise can’t even support the conclusion, then of course the weaker premise cannot either.

This argument also looks similar to a valid formal argument where a some arrow precedes the all arrow. In that argument, we can draw a valid conclusion via the chain.

Question 3

A —m→ B —m→ C yields no valid conclusions via the chain.
We cannot make any inferences about the relationship between A and C via this chain.

We CANNOT infer that A ←s→ C

Notice that this NOT the same as:
A —m→ B
A —m→ C
———
B ←s→ C

Answer 3

We know from the logic of ‘All Before Most (A → B —m→ C)’ that even when ALL of A set falls into B set, the relationship between B set and C set (most of B set falls into C set) cannot be cross-applied between the A set and C set. Because:

1. We cannot guarantee the A set is the same size as B set and C set; the A set may be tiny relative to the latter two.
2. Even if we drop everyone from the A set into the B set, the B set is still overwhelmingly made up of members not from the A set.
3. When the “most” scoop comes to pick members from B to C, it may or may not capture any As.

Therefore, it is even less likely to be valid that ‘most of A set falls into B set’ would yield any valid conclusions between A set and C set.

https://7sage.com/lesson/most-before-most/

Question 4

A ←s→ B ←s→ C yields no valid conclusions via this chain.
We CANNOT infer A ←s→ C.

Answer 4

From the logic of ‘All Before Most’ and ‘Some Before Most’:

1. We know no valid conclusions can be drawn when we cannot guarantee ALL members of the two sets are 1:1 in the first relationship (i.e., EVERY member of A is simultaneously in the B set)
2. Both ‘All Before Most’ and ‘Most Before Most’ do not yield any valid conclusions.
3. As ‘All’ already covers ‘Most’ and ‘Some’, we can already know that ‘Some Before Most’ cannot yield any valid conclusions.

ALL > MOST > SOME. If ALL of A is in B, MOST and SOME are covered too.

https://7sage.com/lesson/some-before-some/

Question 5

Causal logic is not formal. It’s informal. That means arguments that use causal logic will never be valid. Their premises, even when true, will never guarantee the truth of their conclusions.

But, that doesn’t mean causal arguments are necessarily weak. In fact, causal arguments can be very strong. The fact that they can never be valid merely sets a ceiling on how strong the support relationship can be.

Question 6

Psuedo-sufficient Assumption question.

The camp has a non-negotiable three-strike policy for late attendance; however, if a reason is given on the third strike whereby any reasonable person who is unaware they might be late would be unable to be on time given those circumstances, the third strike can be forgiven. Jay’s late attendance was forgiven.

What best supports this conclusion?

(A) Jay was unaware he might be late and any reasonable person could see he could not have made it on time. One of the counselors mentioned Jay could be on his third strike.
(B) Jay’s reason for being late left no doubt that no reasonable person could have made it on time.
(C) Jay is on his third strike; given the circumstances, Jay could not have foreseen he would be late and anyone can reasonably see he could not have made it on time.
(D) Jay did not know he was going to be late and no one could reasonably expect him to be on time given his circumstances.
(E) Jay is on his third strike. Though he reasonably expected he might be late, he could not be responsible for it since any reasonable person could see he could not have made it on time.

Answer 6

Answer = (C)

Remember there is a hierarchy in any condition/assumption/logical relationship.

Here, the top priority is whether this is Jay’s third strike; if it’s not, none of the other conditions/assumptions would follow OR even matter if they are true.
‘The third strike’ dictates the domain in which the rule begins to be applicable.

The prompt says there is forgiveness for the third strike but that doesn’t mean it applies to all late attendance.

(C) is stronger than (A) in that the whether Jay is on his third strike in (A) is actually not fully supported. We know a counselor is OF THE OPINION that Jay is on his third strike.
In (C), it is stated as fact that Jay is on his third strike.

Question 7

Remember there is a big difference:
A —m→ B —m→ C
VS
A—m→ C
A—m→ B

The first one yields NO valid conclusions.
The latter yields the conclusion:
B ←s→ C

Answer 7

We know most of A set falls into B set; we know most of A set simultaneously falls into C set.

As ‘Most’ means any range more than ‘Some’ and less than ‘All’ (can assume it covers more than 50% of A Set), this means there will be a non-zero overlap (at least one) between B set and C set by virture of MOST of A set falling into B set and C set at the same time.

For example: 51% of A set falls into B set and 51% of A set falls into C set. Even if we assume the lowest possibility, there will at least be 1% of A set that has membership in B set and C set simultaneously.

Question 8

What are the formal negations for the below?

1. ALL A are B
2. SOME A are B
3. MOST A are B

Answer 8

1. ALL A are B, negated: SOME A are NOT B.
   Original: All A are B
   Negated: Some A are not B
   Original: A → B
   Negated: A ←s→ /B
2. SOME A are B, negated: NO A are B.
   Original: Some A are B
   Negated: No A are B
   Original: A ←s→ B
   Negated: A → /B
3. MOST A are B, negated: It’s NOT the case that most A are B.
   Original: Most A are B.
   Negated: It’s not the case that most A are B.
   Negated: Anywhere from none to exactly half of A are B.