I just want hints to make sure I'm going in the right direction.

  1. All S are ~ P. I got true
  2. Some S are not ~ P. I got false
  3. No P are S. I got false
  4. Some P are ~ S. I got true
  5. No S are ~ P. I got false
  • You might want to add what you have tried so this question does not get closed. I offered some hints in an answer. – Frank Hubeny Sep 17 '19 at 0:05
  • Have you tried to find the truth values of these statements? If you made the attempt, was there a problem? – Mark Andrews Sep 17 '19 at 0:27
  • I am confused on trying to find the truth values so I was looking for hints on how to answer these questions. – user40964 Sep 17 '19 at 0:28
  • The path to resolve this type of question requires knowledge of some inference rules. Propositions are related by the square of Opposition & the inference rules. They are conversion, obversion & contraposition in classical logic not the Mathematical version. Many people confuse the two distinct inferences by the same name. The Mathematical version is actually called transposition as contraposition does not always work. – Logikal Sep 17 '19 at 3:13
  • You should actually state explicitly if this is Mathematical logic or not. Some of the alleged propositions are not formulated correctly to be standard form categorical propositions. You cant just put any mobo jumbo & call it logic nor call any kind of statement a proposition. This is frequently done in math. There is no such thing as ALL s are not p. There is no such thing as NO s are not p. You cant have a negative quantifier with a not at the same time. You cant have any NOT appear in any universal proposition. They are ambiguous when used that way. You can have the answer undetermined. – Logikal Sep 17 '19 at 3:20

Assume that Some S are P is true. Here are some thoughts about the five statements to help you decide for yourself whether they are true, false or not able to be determined.

  1. All S are ~P. By assumption there is at least one S that is P.

  2. Some S are not ~P. Assuming double negation elimination, "not ~P" would be the same as P.

  3. No P are S. By assumption there is at least one S that actually is P.

  4. Some P are ~S. With the assumption that some S are P, it is possible that every P is also an S.

  5. No S are ~P. All we know is that some S are P not that all S are P.

Here is the square of opposition which may be helpful in thinking about these or at least being aware of when considering such categorical propositions.


I suppose S and P are meant to stand for 1-place predicates. Let [S] denote the set of objects that S is true of. Then we can simply take 'Some S are P' to be true if and only if there is some object o such that o is both in [S] and in [P]. I take it that [~P] is the complement of [P], i.e. the set of objects P is not true of and that [P] and [~P] are disjoint and exhaustive (i.e. every object either is in [P] or [~P]).

  1. All S are ~P iff no S is not ~P iff there is no object o such that o is in [S] and o is not in [~P] iff there is no o with o in [S] and o in [P]. But by assumption, some S are P, so there is some object o' that is both in [S] and [P]. So, 1. is false.

  2. No S are ~P iff there is no o with o in [S] and o in [~P] iff there is no o with o in [S] and o not in [P] iff all S are P. Since we only know that some S are P, we lack information to evaluate 'no S are ~P' for truth or falsity.

The other cases can be treated analogously.

  • This is correct, but I suspect it presumes too much comfort with the language style of logic on the part of the questioner. If you're confused about whether "Some S's are P" entails "all S are ¬P", then "the set of objects P is not true of and that [P] and [~P] are disjoint and exhaustive" is likely an intimidating sentence. I'm assuming this since the exercise being asked about is one that's often given right at the start of a logic course, as a bridge towards being able to reason in such an abstract style. I'd just like to signpost to the OP that he doesn't need to find this intuitive yet. – Daniel Prendergast Sep 17 '19 at 14:18
  • The term 'disjoint and exhaustive' is explained in the text. But of course I see the basic point. On the other hand, what my little proofs amount to is no more than playing around with baby set theory and that should actually be the beginning of a decent logic course. But especially philosophy departments' logic courses seem to get worse, while at the same time philosophical work is getting more and more technical. – sequitur Sep 17 '19 at 17:33
  • I do hope it wasn't taken as a criticism. It's just that the questioner is clearly a stage below that, and i didn't want him to think he had to be reasoning in that style in his head with any fluidity. Yet. Ime, you just kind of get an ear for it over time. As a point of interest, what makes you say that philosophy is getting more technical? It seems, and to my disappointment, that several less than technical fields are now taking over from the metaphysics/logic stuff. Also, I wonder if this strikes you as bad: my 1st yr logic: prop/pred, 2nd: set theory and completeness proofs – Daniel Prendergast Sep 17 '19 at 18:08
  • No problem, as I said your point about my answer is quite accurate. To my mind the increasing technicality is clearly evident in the recent discussions about substructural proof theories in connection with semantic paradoxes, or recent heavily technical discussions about hyperintensionality. But simultaneously there's the tendency to focus on issues that are not only lacking any rigorous formulation (which in itself is not bad since logic is not relevant everywhere), but which are uninteresting expressions of zeitgeist, e.g. phil of love or misunderstood gender issues. – sequitur Sep 17 '19 at 23:00
  • "are uninteresting expressions of zeitgeist" such a good away of putting it – Daniel Prendergast Sep 18 '19 at 14:33
  1. All S are ~ P. I got true

  2. Some S are not ~ P. I got false

  3. No P are S. I got false

  4. Some P are ~ S. I got true

  5. No S are ~ P. I got false

1) Must be false. We have the statement "there's some x, such that Sx and Px". This disproves the claim that all the x's that are S, are also ¬P: we've just asserted that at least one of them is P. I think you've just made one of those silly errors here. We all make them. More formally, ∃x(Sx & Px), by assumption. Now suppose there's no S's that are Ps: ¬∃x(Sx & Px). That's a straight contradiction.

Note: we use x as a generic object, and say that "this object is S". I won't go into the details of why, you don't need to know yet. This is just to clarify how an x got involved in a conversation about Ss and Ps; what's it got to do with x! (you might think). Well, an x is an object, capitalised letters are properties of the object.

2) True: ¬¬∃x(Sx & Px) is equivalent to ∃x(Sx & Px) by double negation elimination.

I've given you the first two answers because i think it's a lack of confidence with formalism that's tripping you, so using the symbols for the easy questions might help (not that these are good enough for proofs, but i don't think you're at the proof stage yet in your studies).

Following the other commenter, I'll give you hints for the rest:

3) It's certainly not demonstrably true given the assumption "Some S are P". But does that mean its false? can we show that "no Ps are S" from "some S are Ps"? Try it to find out.

4) As above.

5) We have an existential quantification (backwards E), and it says there's at least one object in our domain of discourse (fancy term for "all the objects we're talking about") that's an S and a P. To show that all the objects that are S are also P, we'd need a universal quantifier "∀" over every object in the domain of discourse, saying that they're all both S and P. The symbolism for second claim, "∀x(Sx --> Px)" proves "∃x(Sx & Px)", does it the other way around? I suspect you know the answer to this.

Hope this helps. Sorry if it's early days, and you're not symbolising things yet. If that's the case, i hope what I've done by symbolising some statements of the reasoning will help aid the intuitive connection between the symbols and English statements

  • Your answer to question 2 is false and your inference is invalid. Are you telling me that if we are given some tenured professors are PhD holders is TRUE that the proposition some tenured professors are not PhD holders is also true? This is blatantly false. What do you think? All I did was substitute some s is p and the other proposition some s is not p. Told you concepts help. Understanding helps. – Logikal Sep 17 '19 at 20:48
  • @Logikal You've missed the double negation – Daniel Prendergast Sep 18 '19 at 10:48
  • Double negation or not there is no way to guarantee the truth of a particular from another true particular in deductive reasoning. If anything both could be true or only one of the two can be true. Consider this: some of marbles in the bag are blue. This you are given. How can you guarantee some of the marbles are not blue is also true? You would have to pull more marbles from the bag correct. That would be experimental and not deductive. You can't make this inference & be correct 100 percent of the time. Such an inference is invalid even if you end up being correct that both are true. – Logikal Sep 18 '19 at 11:20
  • @Logikal The assumption the OP is given: Some S are P. Question 2: Some S are ¬¬P. So it's true; it's literally the same statement. ¬¬P and P are the same, and every other word in the question is the same. – Daniel Prendergast Sep 18 '19 at 11:27
  • After reviewing the statements again, you are correct I did seem to miss a double negation. But notice in your last comment you were way more detailed and gave justification to show my error. This is the type of reasoning worthwhile. You have to detail and show things instead many people assume the other person knows certain details that are not really clear. In my answer I analyzed only some s is not p which is not what is given as a question. – Logikal Sep 18 '19 at 14:53

If the questions are specific to categorical syllogisms and the original square of Opposition then you can use the inference rules of conversion, obversion, and contraposition (philosophy way not the math definition).

LET'S recap the original proposition is SOME S ARE P which we are given a True. Here are the answers that I came up with without showing the work:

1 this one is tricky because it is not standard categorical form. This form of "All . . . Are NOT. . . " can have two translations. One is an E proposition Such as no s are p. The other is an O type proposition such as some s are not p.

1 a. So in the case you use the translation No s is p, this is blatantly false to the original proposition above. Check the square of Opposition literally. What relationship do you see? There are at least five specific proposition relationships that have specific names to them. Know that and you have the reason why the answer is false.

1 b. Using the translation some s is not p for the proposition all s are not p there is no inference rule that allows you to go from the truth of a particular proposition to another particular with no further details. You can't use the square of Opposition to derive a conclusion from two particular propositions that are true alone. Even if we know one is true we can't 100 percent derive the other particular. In order to 100 percent derive particulars we must be given what is FALSE. A special relationship governs this on the square of Opposition. If you know one particular is false we can figure out the entire square. This relationship concept defines why there cannot be two false particular propositions. Both can be true but both cannot be false simultaneously. But here we are given truth. Find the name of that relationship and that inference will justify why there is no way to determine the answer here. The answer is UNDETERMINED as written. We need more details to go further.

  1. some s is not p is what I just covered in 1 b above. The answer is identical: undetermined. UPDATED EDIT: The given question has a dual negation of the predicate that I missed. So the answer is NOT undetermined as I stated above. However in standard categorical syllogism form you would not write the word NOT twice in a proposition like that. It would read "Some s are not non-p." In this case the inference rule named obversion would be applied to get the original proposition given. The point here I am making is that there are inference rules that are taught to resolve this type of question. It is not about me memorizing anything. I don't memorize the rules. I understand them and as a result I can recall or walk through them to answer the question. I am not concerned with just the result but quality & the result later. Substances wiith good quality usually comes with good results.

  2. No p is s. In comparison to the original proposition this is on the square of Opposition. The answer is false. You must state what rule of inference proves this or state the relationship needed to know this proposition is false.

  3. Some p is not s. This is again a particular. We cannot 100 percent guarantee that if some s is p is true that always some s is not p is true 100 percent of the time. For instance, some male voters are democratic voters happens to also be accidentally true with some male voters are not democratic voters. Both claims are true. Here is a counter example, some bachelors are single males. This happens to be true as there is at least one man who is single and a bachelor. But some bachelors are not single males is false. One is true while the other is false. You can't determine particular propositions that are given as true. If I know one is false I can figure out the other using the rules of inference and relationships of the square of Opposition.

  4. No s are not p. This one is not in standard categorical form either. This requires translation. Literally speaking double negatives in this sentence grammatically should stand out. You would not write this is an English paper would you? But none the less we see double negation occurs here. The problem here is the relationships of the square of Opposition do not work well when we have a true particular. As I stated above knowing a particular is true does not guarantee the universals hold true or other particulars hold true. You would need to be told the particular is false to go somewhere. Truth flows down the square of Opposition-- not upwards. The answer here is undetermined. Perhaps this is accidentally true or accidentally false. We know we cant rely on whatever answer you find out 100 percent no matter the example. There will be a counter example so this can't be a valid inference even if you found out later the proposition was true or even false.

The understanding is important here not the results. Understand the concepts, terminology and relationships at hand and that will guide your thinking method. Many people are scared to make mistakes because you pay for each mistake the hard way by getting a lower grade. Get too many wrong and you get lower grades all at the same time you are allegedly learning (being taught). Seems like being between a rock and a hard place to some people. I have been there.

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