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example 1

Suppose we define $A_1$ as $A_1=B$ and $A_2$ as $A_2=-B$.

Next suppose $B|A_1=-5$ and $B|A_2 =5$ (where $y|x=c$ denotes $y$ conditional on $x$ is equal to $c$).

Is it valid to claim that given the system above $A_1 \Leftrightarrow A_2$ and that they are equivalent definitions, or would they be considered two distinct definitions?

I intuitively think that they should be equivalent definitions since given the definition of $B$ both $A_1$ and $A_2$ are equal to $-5$ ($A_1=A_2=-5$). Hence, the claim $A_1\Leftrightarrow A_2$ should be true claim given the system and they can be considered same definitions.

However, an argument could be made that since $A_1=B$ and $A_2=-B$ they are not equivalent definitions because regardless of what $B$ is the definitions are not exactly the same.

example 2

Also the first example above might be convoluted so let me provide another probably clearer example:

suppose there are two scientists work in the same field. They both independently realize that $F(x,y)=c$ can be totally differentiated as: $F_x'dx+F_y′dy=0$.

Now each of the scientists manipulates the expression differently (I assume $F(x,y)$ is such function that $dx$, $dy$ can be manipulated algebraically - I know its not always appropriate but let us assume the functions used in that field are always such that $dx$ and $dy$ can be algebraically manipulated).

Now scientist 1 manipulates the expression as $\frac{F_x'}{F_y'}=-\frac{dy}{dx}$ and since it is important discovery in a field decides that from now on the $\frac{F_x'}{F_y'}$ will be definition of concept $A_1$ (so it can be always referenced in future proofs), then scientist 2 manipulates the expression as $-\frac{F_x'}{F_y'}=\frac{dy}{dx}$ and also recognizes the relationship important, but since the scientist manipulated the expression differently second scientist defines this important concept using definition $A_2$ as $\frac{F_x'}{F_y'}$.

Could we logically claim that the definitions are equivalent $A_1 \Leftrightarrow A_2$? Clearly both scientists are discovering the same scientific principle, but at the same time the definitions use different 'wording'. This being said in logic two propositions can be logically equivalent even if they are worded differently.

What would be the correct answer here?

PS:

Are examples 1 and 2 even illustrating the same concept? I am not fully certain they do.

1muflon1
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    What are $B, A_i$? Are they integers? Something else? If they are just some sort of formal objects, then what does $-B$ mean? What does $B(A_i)$ mean? – lulu Jun 29 '22 at 00:23
  • @lulu I am not really sure how to say it rigorously, but here by definition I mean definition as used in science, for example in physics work is W=Fs where F is force and s displacement – 1muflon1 Jun 29 '22 at 00:33
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    I didn't ask for the definition of "definition". I asked what sort of things $B,A_1,A_2$ were. And I asked what $B(A_i)$ meant. Is $B$ a function? – lulu Jun 29 '22 at 00:34
  • @lulu but that's what I mean would it be valid to say they are functions, I am not sure if there is any subtle difference between function y(x)=x+1 and definition of y=x+1 – 1muflon1 Jun 29 '22 at 00:34
  • but they are functions in a sense that work is also function of Fs so we can say $W(F,s)=Fs$ – 1muflon1 Jun 29 '22 at 00:35
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    I don't understand. If $B$ is a function, what is its domain? And what sort of things are $A_1, A_2$? Are they functions too? I expect you have something in mind here, but it is very hard to guess what it is. – lulu Jun 29 '22 at 00:36
  • @lulu I am not 100% sure, suppose that I would want to express the idea that distance is equal to rate times time, am I defining a function or simply an equation? I am bit unsure about this to be honest – 1muflon1 Jun 29 '22 at 00:41
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    I think this is too vague to answer. Or, rather, I just don't understand what you are asking. Good luck. – lulu Jun 29 '22 at 00:42
  • @lulu I edited the question, does for the answer matter whether A and B are both equations or functions? – 1muflon1 Jun 29 '22 at 00:47
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    As I say, I have no idea what you are asking. I still don't understand your notation. $B,|,(A_1)$ doesn't have any universal meaning I am aware of. As it stands, I think this is all too vague to answer. In any case, I don't think I can help here. – lulu Jun 29 '22 at 00:52
  • @lulu B|A_1=-5 is B conditional on A_1 is -5 – 1muflon1 Jun 29 '22 at 00:53
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    But, then what does "B conditional on $A_1$" mean? Can you be more literal in your question, rather than getting embroiled in figurative or vague language? – paul garrett Jun 29 '22 at 01:03
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    From the literal terminology of your question, you are asking whether the truth of an assertion $A_1$ is equivalent to the truth of an assertion $A_2$. But neither of your A's is an assertion. Please be more literal and precise. Also, again, $A|B$ does not have a clear sense without an adequate context... – paul garrett Jun 29 '22 at 01:05
  • @paulgarrett could you please help me be more precise? This is the idea I am trying to express: suppose we define some concept in science A_1=-B (equivalent of physics defining work as force times displacement), then B is further defined to be some constant c. Next suppose some other scientists defines second concept A_2=B and the scientist defines B as -c. Are the two definitions equivalent? How can I express the idea more rigorously? – 1muflon1 Jun 29 '22 at 01:20
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    I'll comment tomorrow... :) if no one else does. – paul garrett Jun 29 '22 at 01:24
  • @paulgarrett also I am bit confused by you saying A|B does not have a clear sense without an adequate context. In statistics, A|B is used often to denote outcome or probability of one variable based on another, in general math this does not hold? – 1muflon1 Jun 29 '22 at 01:24
  • @paulgarrett thanks I appreciate your help – 1muflon1 Jun 29 '22 at 01:24

1 Answers1

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I think there are genuine, worthwhile questions here, but entangled with notational troubles and terminological troubles.

First, as others have commented, reading $B|A$ as "$B$, given $A$", does not occur in the fashion here, in standard use. Rather, as in probability, it would be more like $P(B|A)$, which would be the probability of $B$ given $A$.

This is entangled here with the (unsymmetrical) use of equality as assignment, rather than as an assertion of equality of two pre-existing objects/quantities/functions/whatever. And entangled with the question of assignment by value, or by reference!?!

Mathematicians don't overtly talk about those latter issues, but basic computer science does. That is, if $B$ is a thing, and we assign/define $A=B$, and we change $B$, does $A$ keep the original value, or does it (magically?) take on the new value?

And, there's the question of what one might mean by "equivalent definitions". In practice, mathematicians and physicists seem to have an informal notion of this, insofar as normalization details (!!!) are not considered serious differences.

In the example at hand, $A_1$ is defined to be $B$, and $A_2$ is defined to be $-B$, but/and (this is why I wonder about passing values versus passing references) $B$ seems to be $5$ when it is used to define $A_1$, and is $-5$ when it is used to define $A_2$. So $A_1$ turns out to be the same number as $A_2$, but by somewhat different routes.

I would say that these are equal values, but not (superficially) equivalent definitions of the $A_1,A_2$, because for a given single $B$, they give different values.

Nevertheless, perhaps $B$ is just an auxiliary intermediate gadget, itself "defined" somewhat artificially, and in different ways depending on conventions (not on fundamental facts). Then it might indeed be that $A_1$ and $A_2$ refer to the outcomes of two different-but-equivalent stories, so their numerical equality reflects that.

One example of what I mean is Fourier transform on the real line. There are several different normalizations/formulas, but, nevertheless, in the end, we still do have Fourier Inversion and Plancherel's theorem... and people do not view these different normalizations/conventions/formulas as fundamentally different. "Equivalent", perhaps.

paul garrett
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  • thank you for your answer I find it insightful and it helped me to clarify my own thinking. But now I think my use of constants was maybe misleading (sorry for that). What I meant was definition in a way how its used more in basic science. For example, suppose there are two scientists work in the same field. They both independently realize that F(x,y)=c can be totally differentiated as: $F'_xdx+F'ydy=0$. Now each of the scientists manipulates the expression differently (I assume F(x,y) is such function that dx, dy can be manipulated algebraically - I know its not always appropriate but let us) – 1muflon1 Jun 29 '22 at 20:20
  • assume that it can be done in this case). Now scientist 1 manipulates the expression as $\frac{F'_x}{F'_y}= -\frac{dy}{dx}$ and since it is important discovery in a field decides that the $\frac{F'_x}{F'_y}$ will be definition of concept $A_1$, then scientist 2 manipulates the expression as $-\frac{F'_x}{F'_y}= \frac{dy}{dx}$ and also recognizes the relationship important, but since the scientist manipulated the expression differently second scientist defines this important concept using definition $A_2$ as $-\frac{F'_x}{F'_y}$ – 1muflon1 Jun 29 '22 at 20:26
  • Clearly, both scientists work on the same idea, it is valid to say their definitions are equivalent? Is this different from question in my original post? (if so I might open it as a new question) I don't want to rewrite my current question now you posted very insightful answer, since you clearly put a lot of thought and work into it, or does your answer apply to this clarification as well? – 1muflon1 Jun 29 '22 at 20:27
  • I'd say that their two conceptions are equivalent, and that my discussion covers that case. It's not literally what you originally wrote, but much clearer as an example. You might edit your original question to include that example... – paul garrett Jun 29 '22 at 20:39
  • Thanks I will edit it to include it. Also, by the part that is covered you mean the part about Fourier transformation? – 1muflon1 Jun 29 '22 at 20:40
  • Yes, for example, by the classic example of Fourier transform. – paul garrett Jun 29 '22 at 20:41
  • thanks for all the help, I will wait a bit to see if there are any other answers if not I will eventually accept yours, I really appreciate you took the time despite my original question being vague. Thank you for your patience and sorry for the poor state of question. – 1muflon1 Jun 29 '22 at 20:54