Pangs, Pings and Pongs

From the first statement we can draw the following venn diagram:

From second statement two cases arises which are shown as:

Hence we are not sure that which is true as nothing is said .Hence the correct option is " False, there might not be a pang that is a pong"

"All pangs are pings." Alright, straightforward, but this does not imply the converse: all pings are not necessarily pangs.

"Some pings are pongs." Alright, but which pings exactly?

"Some pangs are pongs." Wait a minute, that doesn't have to be true! All the pongs could be pings that are not pangs!

This was basically what ran through my head when I was solving this problem. I realize it's not formal at all, but more of intuitive reasoning, which seems like what your friend wants to investigate.

I'm curious to hear his hypothesis and whether or not this experiment supported it!

All cows eat grass. Some that eat grass are sheep. Therefore some cows are sheep?

$\boxed{\text{False, there might not be a pang that is a pong}}$

I just substituted pangs for humans, pings for apes, and pongs for gorillas.

All humans are apes. Some apes are gorillas. But no humans are gorillas, therefore the statement is false.

I used the following example:

All humans are animals. Some animals are zebras. Therefore some humans are zebras.

After this example, I realized that the answer HAD to be False. The second part was a bit more complex. We know that humans are definitely NOT zebras, but the example limits the set. Given another set, there is a possibility that a "human" could be a "zebra", maybe in a parallel universe or something!

In first order logic, an argument is valid if the premise implying the conclusion is a tautology. However, this is not a tautology, making the argument False.

To show that the argument is not a tautology, it suffices to construct one model where the conclusion is false, the premises being true.

We also need to show that the argument's validity contingent. To show this, we need to construct a model where both the premises and the conclusion are true.

I will complete these proofs more formally when I come back from school.

Using Aristotelian Logic,

$\text{All }\underline{\text{PANGS}} \text{ are PINGS} \rightarrow \text{SaP Statement},~+\\ \text{Some PINGS are PONGS} \rightarrow \text{SiP Statement},~+$

Asserted conclusion: $\text{Some PANGS are PONGS}\rightarrow\text{SiP Statement},~+$

Required Conditions

• (1) If all the statements are $+$ ve, then the conclusion must be positive $~\checkmark$
• (2) Middle term must be underlined $\times$
• (3) Any term distributed in the conclusion must be distributed in any of the statements $~\checkmark$

So the conclusion given in the question is wrong. Since Condition (2) is not satisfied in any case, we cannot say to certain what is true. Hence the answer.

According to me, the syllogism is as below..

The case of any pang being pong should not arise as it has not been specified by the question.

(I'm new here, and since I do not know Latex, I shall type what I thought) Let P be the set of all Pangs, P1 be the set of all Pings and P2 be the set of all Pongs. Clearly, P is strictly a subset of P1. Some elements of P1 can be expressed as an intersection of P1 and P2. Now, these elements may or may not belong to P. Therefore, there might not be a Pang that is a Pong.

"All pangs are pings"

Assume pings is a set of 4 numbers, 1 of which is a pang.

"Some pings are pings"

Assume half of the pings are pongs. There are ways to chose 2 elements of the set Ping so that none or 1 is a pang.

Thus only some (if any) are pangs.

Btw, I'm very interested as to your friends hypothesis on how different people solve this problem. Sounds like a very cool thing to study.

Ping large O
Pang Hatched O
Pong Cross Hatched
Hence The statement is true only for Case B

This is a pretty simple logic problem, So we have three groups, a smaller group that is pangs, all of them are a sub-sect of pings, so then we have the third group pongs, some of the ping group overlap the pong group, but we don't have enough information to determine if any of the pang grouping is actually part of the pong grouping, while there MAY be some that are we can't say for sure, where the answer "true, some pangs are pongs" may be true we don't know that for certain, HOWEVER, the "False, there might not be a pang that is a pong." we don't know some might be but there MIGHT NOT be, so this statement is the best solution it does not say that none of them are, it just says there might not be any pangs that are pongs, which is true, without further data we can not determine if any pangs are pongs.

The shortest version to try to explain this is, a hypothetical we have 15 people as represented in this line

| --------------- |

Now we know that the smaller group, Pangs are all inside this group so let's make a division, say 5 of them

|----- // ---------- |

Now let's make cases for the third group, some of group 2 are part of group three. The members that are part of both groups signified by x.

|----- // --- // xxx // ---- |

Case two:

| ---xx // --- // xxx // --- |

either case could be accurate so in this vein we don't know for certain so we should take the answer that indicates that while it may not be true, the key here is in that less than solid phrasing, Might not, does mean can not, it does not mean that any are, it simply states that they may not be, which also means some may possibly be. Might is non-definitive, and without more information we can not give a definitive answer.

Here are three distinct solutions that come to mind.

Solution 1

These are two of the options: "True, some pangs are pongs", and "False, there might not be a pang that is a pong". The latter can be translated to "it is false that some pangs are pongs", so these two are complementary. Thus one of them must be true, reducing our options to two. We can then use Venn diagram or anything of the sort to show that the answer is not true, so the answer has to be False, there might not be a pang that is a pong .

Solution 2

We solve this with two constructions.

• Suppose there is no pang at all. This causes the first premise to be true vacuously, and it's easy to make the second premise true (there is exactly one ping, which is itself a pong). However, the conclusion is clearly false; there is no pang at all, how can there be a pang that is also a pong? This makes the True answer incorrect.
• Suppose there are two pangs, both pings. One pang is also a pong; the other isn't. This makes two of the False answers incorrect: "no pang is pong" is made incorrect, and "all pang is pong" is also made incorrect.

This leaves the sole option that has the uncertainty: False, there might not be a pang that is a pong .

Solution 3

Three of the four options actually assert syllogisms . The true option asserts the syllogism IAI-1; the "no pang is pong" option asserts the syllogism IAE-1; the "all pang is pong" option asserts the syllogism IAA-1. The last option asserts the negation of IAI-1. We can see that none of the three syllogisms are valid (there are only 24 valid syllogisms; just find some list of them), so the correct option is the one asserting the negation of IAI-1: False, there might not be a pang that is a pong .