What's the converse of this statement?

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In summary: But if we do interpret it as an implication of a quantified statement, then the converse would be (by DeMorgan's law) if G is not A-magic, then either A is not a subgroup of B or G is not B-magicHowever, if G is a constant, then the converse would beif G is not A-magic, then A is not a subgroup of B and G is not B-magicIn summary, the statement "if A is an additive subgroup of the additive group B and G is B-magic then G is A-magic" does not make mathematical sense, so its converse cannot be properly defined. However, if we interpret the statement as an implication of a quant
  • #1
Adeimantus
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if A is an additive subgroup of the additive group B and G is B-magic then G is A-magic

In general, I know what a converse is. But I'm having trouble figuring out what the converse of this statement is. The literal converse, as far as I can tell, does not make any sense.

if G is A-magic, then A is an additive subgroup of the additive group B and G is B-magic

(i.e., Who said anything about B?). Is there an accepted convention about converses when you have a compound antecedent, like you do here?
 
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  • #2
Adeimantus said:
if A is an additive subgroup of the additive group B and G is B-magic then G is A-magic
The first statement is a description of the world in which the second statement takes place. It is a given: ##(A,+) < (B,+)##.

claim: ##G## is ##B-##magic ##\Longrightarrow \,G## is ##A-##magic

converse: ##G## is not ##A-##magic ##\Longrightarrow \,G## is not ##B-##magic

which means: ##(A,+) < (B,+)\,\,\wedge G \,\lnot \,A-##magic ## \Longrightarrow\,\,G \,\lnot \,B-##magic
 
  • #3
Okay, thank you. That is how I interpreted it, too--that the subgroup relation is "background knowledge" to be taken for granted. (Except that I think you have A and B switched in the converse). But a classmate who is smarter than me thought the converse would be something like "If G is both A- and B-magic, then one of the groups is a subgroup of the other." So I had doubts.
 
  • #4
Adeimantus said:
Okay, thank you. That is how I interpreted it, too--that the subgroup relation is "background knowledge" to be taken for granted. (Except that I think you have A and B switched in the converse).
Probably, as I wasn't sure what "converse" should mean. I negated the statement without changing its truth value. If you meant "opposite" then things are messier, since then all occasions in which ##A \nleq B ## are also in the opposite space. However, this doesn't make a lot of sense mathematically, as the set of pairs of groups with this condition is unbelievably big.

The statement is that a group being ##B-##magic implies magic for all subgroups, i.e. ##(A,+) \leq (B,+)\,\wedge \,G ## is ## B-##magic ##\Longrightarrow \,\, G## is ##A-##magic .

This means in sets: ##\{\,(A,+) \leq (B,+)\,\wedge\, G ## is ## B-##magic ##\,\} \subseteq \{\,(A,+) \leq (B,+)\,\wedge\,G ## is ## A-##magic##\,\}.##

Now the opposite is: ##\{\,(A,+) \leq (B,+)\,\wedge \,G ## is ##B-##magic ##\,\} \nsubseteq \{\,(A,+) \leq (B,+)\,\wedge\,G## is ##A-##magic##\,\}##, which means there is a group ##G## and subgroups ##A \leq B## such that ##G## is ##B-##magic but not ##A-##magic.
But a classmate who is smarter than me thought the converse would be something like "If G is both A- and B-magic, then one of the groups is a subgroup of the other." So I had doubts.
O.k. I have no idea what magic groups are, but this statement looks wrong in any case. It would require that there can't be groups with two or more types of magic, only subgroup-magic. A statement which hasn't much to do of what has been claimed, as the original statement only says something about subgroup-magic, not multi-magic.

So all in all, it depends on what "converse" is supposed to mean, and a little bit on what magic is.
 
  • #5
Converse is meant just in the sense of flipping an implication around: P->Q becomes Q->P, which is the same as not P -> not Q. So it's not really an opposite exactly, just "the other direction". And it's true I didn't define any of the terms here, or say what G is. G is a graph. It's edges are labeled with elements of B-{0}. For each vertex, compute the sum of the edges incident to it. If such a labeling of the edges can be found so that this sum, or weight, is the same for all vertices, then graph G is said to be B-magic.

Edit: what you did is the "contrapositive". It has the same truth value as the original.
 
  • #6
In this case, it depends on whether you make ##A \leq B## a part of ##P## or not. I interpret the statement as being part of both, ##P## and ##Q##. ## \lnot P## then means either not ##B-##magic or not a subgroup. ##\lnot Q## then is either not ##A-##magic or no subgroup. But how to deal with the implication? ##P \Longrightarrow Q## means ##P \subseteq Q##, means ##X-P \supseteq X-Q##, but ##\lnot (P \Longrightarrow Q)## means there is an element of ##P## which is not in ##Q##, or ##P \nsubseteq Q##.

But as said ##A \nleq B## is a vast set of groups. In your case with the graph, it doesn't make any sense to consider them, as there are just too many. E.g. if ##B=\mathbb{Z}_8## then ##A=\mathbb{Z}_3## is no subgroup and so are arbitrary many others, but how does it make sense to mention them? However, "no subgroup" might play a role for testing. Are there graphs possible, which are ##\mathbb{Z}_8## magic and ##\mathbb{Z}_3## magic at the same time? Probably not if ##G## is connected. In this case all ##\mathbb{Z}_3## labels automatically disqualify in a ##\mathbb{Z}_8## magic graph. In an algorithm, this could play a role.
 
  • #7
I'm not sure if there are graphs that are both [itex]\mathbb{Z}_8[/itex] and [itex]\mathbb{Z}_3[/itex] magic. My friend found a graph that was both [itex]\mathbb{Z}_5[/itex] and [itex]\mathbb{Z}_3[/itex] magic. And yes the graphs are connected and simple (no loops). But from what you are saying now, and what you said in your first post, it sounds like we are in agreement that it doesn't make much sense for the subgroup relation to be the consequent in the converse. It seems to be serving the role of background assumption.
 
  • #8
Adeimantus said:
if A is an additive subgroup of the additive group B and G is B-magic then G is A-magic

The statement (by itself) doesn't make mathematical sense, so we shouldn't expect its converse to make sense!

The statement doesn't make it clear whether "G" is a constant a constant or a quantified variable. It also does not make it clear whether "A-magic" is an arbitrary property or whether it has a definite relation to the group "A". The statement does not require that G be a group.

In general, I know what a converse is.
We'd have to know how the converses of quantified statements are defined.

To determine the converse of a statement requires:
1) Interpreting the logical quantifiers to be used in the statement unambiguously
2) Consulting authorities about how the converse of a quantified statement is defined.

The usual interpretation of a unquantified variable in mathematical writing is to assume it is universally quantified.
E.g. "If x is an even integer then x + 1 is an odd integer" is interpreted as "For each x, if x is an even integer then x+1 is an odd integer".

However, there can be contexts where a letter represents an unquantified constant or an existentially quantified variable.

You probably intend the property "x has y-magic" to relate two things. So we could try:

a(x): x is an additive group
s(x,y): x is a subgroup of y
M(x,y): x has magic of the type associated with y

##( \forall y ( \forall x \big{[}( a(y) \land a(x) \land s(x,y)) \implies (\forall g[ M(g,y) \implies M(g,x) ] )\big{]} ) )##

Then the authoritative definition of converse must be consulted.

I'd vote for reversing the second implication.

##( \forall y ( \forall x \big{[}( a(y) \land a(x) \land s(x,y)) \implies (\forall g[ M(g,x) \implies M(g,y) ] )\big{]} ) )##
 
  • #9
Stephen Tashi said:
The statement (by itself) doesn't make mathematical sense
Of course it does make sense. I have some doubts your response does, at least I don't understand a word.

Say B operates trivially on G and we call this B-magic, then the statement makes perfectly sense.
 
  • #10
fresh_42 said:
Say B operates trivially on G and we call this B-magic, then the statement makes perfectly sense.

I agree that you can add information to the statement and create a new statement that makes sense.

The statement, as it stands, does not even say that G is a set.
 
  • #11
Stephen,

It sounds like you also interpret the converse the same way I did, reversing the second implication. Where does one find the "authoritative" definition of converse that would settle something like this?

Also, I intentionally did not define any of the terms, or say what letter stands for what, because I was hoping the question could be answered based on form alone. I didn't want to get bogged down in the content of it, or whether it is true or not. My question is mostly about syntax and mathematical convention.

But if you're interested, G is a connected graph with no loops. I described what "B-magic" means in post #5.
 
  • #12
Adeimantus said:
Also, I intentionally did not define any of the terms, or say what letter stands for what, because I was hoping the question could be answered based on form alone.
It can.
I didn't want to get bogged down in the content of it, or whether it is true or not. My question is mostly about syntax and mathematical convention.
Convention is the description ##A \leq B## as definition of the space, where the statement takes places. To allow the entity of all possible statements as the defining space doesn't make sense. In order to set a property ##P(A)## which depends on ##A## and a property ##P(B)## which depends on ##B## into respect, you have to define ##A## and ##B## beforehand, since otherwise you cannot define ##P##.

You said there is an example, for which ##G## is ##\mathbb{Z}_3-##magic as well as ##\mathbb{Z}_5-##magic. This makes the statement of your classmate wrong, but this doesn't imply it is the opposite of the original statement. That opposite would be: There is a group ##G## and groups ##A \leq B## such that ##G## is ##B-##magic but not ##A-##magic. The statement describes the inclusion of two sets, so the opposite is "not included".

But here's a different statement, which covers the alternative given by your classmate.

Let ##\mathscr{G} := \{\,A\,|\,G \text{ is }A- \text{magic } \,\}##, then ##\mathscr{G}## is / is not totally ordered by inclusion.
It is not, as the example shows. But if ##B \in \mathscr{G}## then all subgroups ##B \geq A\in \mathscr{G}##, so it is partially ordered.
 
  • #13
fresh_42 said:
Convention is the description A≤BA≤BA \leq B as definition of the space, where the statement takes places. To allow the entity of all possible statements as the defining space doesn't make sense. In order to set a property P(A)P(A)P(A) which depends on AAA and a property P(B)P(B)P(B) which depends on BBB into respect, you have to define AAA and BBB beforehand, since otherwise you cannot define PPP.

Yes, what you say here makes the most sense to me, too.

Just to add a little context, it turns out that the original statement is true (don't ask me for the proof). But we were supposed to show that the converse is false by providing a counterexample. But not everybody had the same idea about what the converse was.
 
  • #14
Adeimantus said:
Yes, what you say here makes the most sense to me, too.

Just to add a little context, it turns out that the original statement is true (don't ask me for the proof). But we were supposed to show that the converse is false by providing a counterexample. But not everybody had the same idea about what the converse was.
But isn't every example for the statement automatically a counterexample for the converse, without even trying to determine what is meant by converse?
 
  • #15
No, because converse isn't the same as opposite, or even negation. For example, a true statement is: all prime numbers (greater than 2) are odd. Converse would be, all odd numbers are prime. Opposite would be (I suppose), all prime numbers are even. Negation would be, some prime numbers (greater than 2) are even.

Edit: Another way of saying it, it is the "if" part of an "if and only if" proof.
 
  • #16
Adeimantus said:
No, because converse isn't the same as opposite, or even negation. For example, a true statement is: all prime numbers (greater than 2) are odd. Converse would be, all odd numbers are prime. Opposite would be (I suppose), all prime numbers are even. Negation would be, some prime numbers (greater than 2) are even.
I don't think so. Now you offered two versions by the parentheses.

All primes greater than two are odd.

Negation: An even prime is less or equal two.
Opposite: There is a prime greater than two which is even.
Converse: (a) All primes greater than two are even. (+/-)
Converse: (b) All primes less or equal two are odd. (-/+)
Converse: (c) All primes less or equal two are even. (-/-)

All primes are odd.

Negation: An even number isn't prime.
Opposite: There is an even prime.
Converse: All primes are even.
 
  • #17
Adeimantus said:
Where does one find the "authoritative" definition of converse that would settle something like this?
I don't know. A textbook on symbolic logic might offer a rigoruous definition But it may be that "converse" is only defined unambiguosly for statements that have a simple form - only one use of an implication between two statements.

Various sources on the web say that the converse of ##\forall x (p(x) \implies q(x))## is ##\forall x ( q(x) \implies p(x))##. That's an example of "##\implies##"" occurring between two statement-functions rather than between two statements.

An interesting technical question is whether "converse" is defined on the basis of semantics or on the basis of syntax. For example, is the converse of ##P \lor Q## defined? ##P \lor Q## is logically equivalent to ##\lnot P \implies Q##, so one could say the converse of ##P \lor Q## is ##Q \implies \lnot P##. That would define "converse" in terms of semantics. However, one could restict the definition of "converse" so it is based on syntax and say that "converse" is undefined for statements expressed without using the symbol "##\implies##".
 
  • #18
fresh_42 said:
I don't think so. Now you offered two versions by the parentheses.

All primes greater than two are odd.

Negation: An even prime is less or equal two.
Opposite: There is a prime greater than two which is even.
Converse: (a) All primes greater than two are even. (+/-)
Converse: (b) All primes less or equal two are odd. (-/+)
Converse: (c) All primes less or equal two are even. (-/-)

All primes are odd.

Negation: An even number isn't prime.
Opposite: There is an even prime.
Converse: All primes are even.

I'm sorry, I don't follow this at all, except perhaps for the first two of the second group of statements. We seem to have different ideas about what the words mean. Converse and negation are pretty standard, and I thought I was using them properly. I'm less sure what the formal interpretation of "opposite" is, although I use the word all the time in everyday language.

@Stephen Tashi ,

That is a good question about whether it depends on the actual symbols used. I think I would argue that it makes sense to talk about the converse of (p or q), since it has the logically equivalent implication that you gave. Back to my original question--I was sort of assuming that since the literal converse of the statement did not make sense, "converse" must be interpreted in some informal sense. So I was wondering if people who are, unlike me, steeped in math literature know the correct informal interpretation in this type of situation.
 
  • #19
To me, the opposite of a statement A is non-A. So if an implication is viewed as inclusion, then the opposite is: not included, or does not follow. Opposite skips the truth value, negation does not.

And converse is undefined to me.
 
  • #20
fresh_42 said:
To me, the opposite of a statement A is non-A. So if an implication is viewed as inclusion, then the opposite is: not included, or does not follow. Opposite skips the truth value, negation does not.

And converse is undefined to me.

Okay, well it's just a terminology issue, then. I'm sure you have written dozens or hundreds of proofs that one thing holds if and only if some other thing holds. Call them P and Q. The typical format is to assume P and show that this implies Q. That's the "only if" part, or "necessity". Then, once you have done that, say something like "Conversely, assume Q..." and then show it implies P. That's the "if" part, or "sufficiency". So you have used the converse of a statement many, many times, even if you have a different word for it.

So, formally, the converse of P->Q is just Q->P. For logical formulas with that simple form, there is no controversy about what converse means. But we are speculating here about whether "converse" can be extended to more complicated formulas that involve more than one implication.
 
  • #21
Adeimantus said:
Okay, well it's just a terminology issue, then. I'm sure you have written dozens or hundreds of proofs that one thing holds if and only if some other thing holds. Call them P and Q. The typical format is to assume P and show that this implies Q. That's the "only if" part, or "necessity". Then, once you have done that, say something like "Conversely, assume Q..." and then show it implies P. That's the "if" part, or "sufficiency". So you have used the converse of a statement many, many times, even if you have a different word for it.
Maybe, and English isn't my native language although the words are the same, so I'd wonder if they had a different meaning in English.

In the above, I think you should swap "if" and "only if", since If P Then Q is the if-part (necessity of Q) and Only if Q Then P is the only if part (sufficiency of Q). It is seen from the standpoint of P, but the more often I read it the dizzier it makes me. At this point @Stephen Tashi's demand for quantifiers makes a lot of sense. Langauge isn't unambiguous and the quantifiers are there to overcome this.
 
  • #22
fresh_42 said:
Maybe, and English isn't my native language although the words are the same, so I'd wonder if they had a different meaning in English.

That may very well be the case. The "con" in "converse" makes it sound like it should have some sense of disputing the original claim, or contradicting it. But it doesn't. The "con" just refers to going in the "other direction".

And as for the "if" versus "only if", I agree that it sort of seems backwards, but I think I have it right in my post. Maybe someone else can confirm. "If P, then Q" means the same thing in English as "P only if Q". And "If Q, then P" means the same thing as "P if Q", although you wouldn't usually say it the second way. So, together those make up both parts of the "if and only if" statement.
 
  • #23
Adeimantus said:
That may very well be the case. The "con" in "converse" makes it sound like it should have some sense of disputing the original claim, or contradicting it. But it doesn't. The "con" just refers to going in the "other direction".
As said, the words are the same. The crux is another: "in the other direction" isn't well defined! See my response to your prime number example. The other direction are simply many directions, not one, as the opposite is: exactly the opposite direction, and thus well-defined. And negation is just a reformulation expressed by not's.
And as for the "if" versus "only if", I agree that it sort of seems backwards, but I think I have it right in my post. Maybe someone else can confirm. "If P, then Q" means the same thing in English as "P only if Q". And "If Q, then P" means the same thing as "P if Q", although you wouldn't usually say it the second way. So, together those make up both parts of the "if and only if" statement.
I never have considered it this way, nor have I seen it. However, I try to avoid this wording - in either language, as it depends on what is subject and what is object of the sentence, and as both are equivalent, this can only result in confusion.

Nevertheless, let me tell you where you're wrong. Imagine we only have P implies Q, no equivalence. Would you say you have to prove an IF-THEN statement, or an ONLY-IF-THEN statement? I bet you will have trouble to find the latter anywhere in a book.
 
  • #24
fresh_42 said:
As said, the words are the same. The crux is another: "in the other direction" isn't well defined! See my response to your prime number example. The other direction are simply many directions...

Fair enough. Sometimes it is hard to tell exactly what "the other direction" is because it is not clearly defined. That was the point of my original post.

fresh_42 said:
Nevertheless, let me tell you where you're wrong. Imagine we only have P implies Q, no equivalence. Would you say you have to prove an IF-THEN statement, or an ONLY-IF-THEN statement? I bet you will have trouble to find the latter anywhere in a book.

Right, because there's no such thing as ONLY-IF-THEN. There's no THEN on then end. In English, at least, IF-THEN and ONLY-IF mean the same thing. They are just two different ways of phrasing it. Which one gets used depends on the context. "If and only if" has a certain tradition to it, so we say that when things are logically equivalent. When the implication goes in one direction only, it is less common to hear "P only if Q". But you do occasionally, and it means the same as "If P, then Q".
 
  • #25
Adeimantus said:
IF-THEN and ONLY-IF mean the same thing
No it does not. If P then Q is not the same as only if P then Q, simply because if R then Q cannot be excluded. It is not the language barrier here. There are many ways to get Q and P is only one of them. If then is from left to right, only if is from right to left. I'm sure.
 
  • #26
fresh_42 said:
No it does not. If P then Q is not the same as only if P then Q, simply because if R then Q cannot be excluded.

I agree with you. And I didn't say that. I said "if P then Q" is the same as "P only if Q". The "only if" is applying to the Q, not the P.
 
  • #27
Adeimantus said:
I said "if P then Q" is the same as "P only if Q". The "only if" is applying to the Q, not the P.
Langauge is generally a difficulty here, no not English, language in general. If you like to use it this way, so be it. However, you will probably encounter many confusions as all others use it differently. Of course it's a convention. If it rains, the road will be wet. (left to right) Nobody says it rains only if the road is wet. But if we pretend for a moment, that rain is the only cause for a wet road, then we say: Only if the road is wet, it rains. (right to left) So the language is used as I said, although you can turn the sentences upside down, just that nobody else does it this way. Is it a proper use? Probably not. That's why I don't like it.
 
  • #28
I think we can agree that there are logically correct formulations that sound so awkward and pointless that nobody would bother to say them that way. The majority of logically correct statements are not very informative, and sound kinda ridiculous.
 
  • #29
Adeimantus said:
I think we can agree that there are logically correct formulations that sound so awkward and pointless that nobody would bother to say them that way. The majority of logically correct statements are not very informative, and sound kinda ridiculous.
Meanwhile I have a question for you:
Given
Adeimantus said:
IF-THEN and ONLY-IF mean the same thing
how can it ever be used to distinguish two different directions?

I have to think about how mathematicians, resp. computer scientists define language ...
 
  • #30
If you use the IF-THEN construction, you change the order of P and Q to distinguish directions. If you use the IF and ONLY-IF constructions, you leave P and Q in the same order. In case this is a point of confusion, "P if Q" is not the same as "if P, then Q". Instead, it's actually the same as "if Q, then P". To recapitulate,

"if P, then Q" is the same as "P only if Q"
"if Q, then P" is the same as "P if Q"

Putting them together,

"(if Q, then P) AND (if P, then Q)" is the same as "P if and only if Q"

I hope this is clear. If not, you might google it. If you find that I have it backwards, show me, and I will gladly take back everything I said.
 
  • #31
Adeimantus said:
If you use the IF-THEN construction, you change the order of P and Q to distinguish directions. If you use the IF and ONLY-IF constructions, you leave P and Q in the same order. In case this is a point of confusion, "P if Q" is not the same as "if P, then Q". Instead, it's actually the same as "if Q, then P". To recapitulate,

"if P, then Q" is the same as "P only if Q"
"if Q, then P" is the same as "P if Q"

Putting them together,

"(if Q, then P) AND (if P, then Q)" is the same as "P if and only if Q"

I hope this is clear. If not, you might google it. If you find that I have it backwards, show me, and I will gladly take back everything I said.
Yes, but allow me not to follow such a - in my view artificial - construction. To attach meaning to the order of where P and where Q has to stand would imply to make the entire wording even more context sensitive as it already is given by the language itself. I don't think it is actually used this way. I think it's the other way around.

I would search for an example, but before I put effort in it, since I still think your position has to be proven not mine, I like to ask you whom would you accept as an authority to decide this question?

We say "P is valid then and only then if Q is valid" so in this case there is actually a difference in language.

Edit: You may be right. I haven't found a direct example, as the authors of my books were smart enough to avoid a reference to "the if-part" or "to the only-if-part". They either use arrows, or simply recall the condition they start with. The closest I came on a quick search was the following proposition stated by Peter Hilton:
Statement: ## \mu## is monomorphic if and only of it is injective.
Proof: If ##\mu ## is injective then ... Conversely, suppose ##\mu## monomorphic

It doesn't finally answer the question, as I said, they avoid ambiguities, but it tends a little bit towards your interpretation.
 
Last edited:
  • #32
Here's the first sentence of the Wikipedia article on "if and only if". https://en.wikipedia.org/wiki/If_and_only_if
Wikipedia said:
In logic and related fields such as mathematics and philosophy, if and only if (shortened iff) is a biconditional logical connective between statements.

In that it is biconditional (a statement of material equivalence),[1] the connective can be likened to the standard material conditional ("only if", equal to "if ... then") combined with its reverse ("if"); hence the name.
 
  • #33
Adeimantus said:
Here's the first sentence of the Wikipedia article on "if and only if". https://en.wikipedia.org/wiki/If_and_only_if
Yes, I do not doubt the linguistic, but what is which direction?

Let me summarize our question: ##P## if and only if ##Q##.

Me: if case: ##P \Rightarrow Q## ; only-if case: ##Q \Rightarrow P##
You: only-if case: ##P \Rightarrow Q## ; if case: ##Q \Rightarrow P##

Is that correct? Not that we mean the same after all. Have you read my edit in the previous post? I guess those authors (I checked two native speakers, a British and an American) did well to avoid the entire situation by clearly repeating what they assume, resp. use ##\Rightarrow## or ##\Leftarrow## which seems more and more to be a good advice.
 
  • #35
Adeimantus said:
The "con" in "converse" makes it sound like it should have some sense of disputing the original claim, or contradicting it.
The prefix would be "contra" to negate what it applies to, not "con".

Adeimantus said:
"if P, then Q" is the same as "P only if Q"
This is what the wikipedia page says, and I agree somewhat, but their explanation isn't as clear as it should be, because it neglects to mention that P and Q have to be equivalent statements. Consider the example they give in the section on Euler diagrams. In the first diagram of this section, where A is a proper subset of B, we have A = {1, 9, 11} and B = A ∪ {4, 8}.

Let P be the statement ##x \in A##, and let Q be the statement ##x \in B##.

"If P then Q" is the implication ##x \in A \Rightarrow x \in B##, which is clearly true. Every element of A is also an element of B.
The statement "P only if Q" is not true for all elements of B. As a counter example, let x = 4, an element of set B that is not also an element of set A. Here P and Q are not equivalent statements, being related to two sets, one of which is a proper subset of the other.

To my way of thinking "if and only if" implications apply only to equivalent sets. "P if and only if Q" can be resolved into "P if Q", which is the same as "If Q then P", and "P only if Q", which is the same as "If P then Q", provided that we're dealing with equivalent statements or sets that are identical.

BTW, their example of Madison's taste in fruits does more to confuse than to clarify.
 

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