What is the symbol for "not necessarily imply"

[swampwiz]

(I should say that I have never done proper coursework in introductory proofs like a mathematics major would.)

I figure there must be a way to denote a situation in which, from sloppy intuition, etc., a certain proposition might (erroneously) imply a result, but that that result could happen, just that it is not guaranteed to happen.

A simple example would be the case of the product of a pair of matrices as LHS & RHS being equal to the sides being swapped.

[ A ] [ B ] = [ C ] does not necessarily imply [ B ] [ A ] = [ C ], but nor does it imply [ B ] [ A ] != [ C ], as is could very well luck out that [ A ] [ B ] = [ B ] [ A ]

 

[Orodruin]

$A \nRightarrow B$ is "A does not imply B". This is probably what you are looking for since "A does not imply B" does not imply that B is necessarily false if A is true - just that B can be false if A is true.

 

[fresh_42]

I would simply write $\nRightarrow$ or $\stackrel{\nRightarrow}{i.g.}\,$

 

[member 587159]

While I have seen the notations mentioned in the other posts, I prefer the $\neg(p \implies q)$.

 

[SSequence]

I think there is (perhaps?) an issue here, in a manner of speaking. Below I describe informally what it is.

It seems to me that when we normally use propositional logic like statements (in maths), a lot of the times we have essentially a sort of "possibility space" in mind. I am thinking of a statement like (over the integers for example):
(x=2) → (x²=4)

Below I will assume that the possibility space is non-empty.

So, for example:
p -- q -- p∨q -- occurence-map(of p∨q)
(i) T -- T -- T -- O/NO
(ii) T -- F -- T -- O/NO
(iii) F -- T -- T -- O/NO
(iv) F -- F -- F -- NO

Here "O" means that a possibility definitely occurs (over the entire "possibility space") ... given that p∨q is true. "NO" means that a possibility never occurs ... given that p∨q is true. "O/NO" means that anything is possible w.r.t. this possibility (it may or may not have occurred over the entire "possibility space") ... given that p∨q is true.

Basically we can place "O/NO" in place of "T" for any truthtable and similarly replace "F" with "NO". The last column in the above truth table of p∨q, for example, shows what it takes for "p∨q" to be true in terms of occurrence or non-occurrence of the individual (simultaneous) truth values of p and q.

We can "always" think of this as dividing into "concrete" separate possibilities. For case above, for example, the "concrete" possibilities are:
(i) O (ii) NO (iii) NO (iv) NO
(i) NO (ii) O (iii) NO (iv) NO
(i) NO (ii) NO (iii) O (iv) NO
(i) O (ii) O (iii) NO (iv) NO
(i) O (ii) NO (iii) O (iv) NO
(i) NO (ii) O (iii) O (iv) NO
(i) O (ii) O (iii) O (iv) NO

Similarly:
p -- q -- p∧q -- occurence-map(of p∨q)
T -- T -- T -- O/NO
T -- F -- F -- NO
F -- T -- F -- NO
F -- F -- F -- NO

Here, interestingly, we can replace the "O/NO" in the first row with "O" (since one possibility must have at least occurred). Hence we "really" have just one concrete possibility.

=============================

Now let's look "p implies q" and its negation:

p -- q -- p→q -- occurence-map(of p→q)
T -- T -- T -- O/NO
T -- F -- F -- NO
F -- T -- T -- O/NO
F -- F -- T -- O/NO

p -- q -- ~(p→q) -- occurence-map(of ~(p→q))
T -- T -- F -- NO
T -- F -- T -- O/NO
F -- T -- F -- NO
F -- F -- F -- NO

But as we observed (in the first half of the post) in the case of p∧q that when proposition is true "only" for one row alone we can replace O/NO with O for just one concrete possibility. Hence we can write:

p -- q -- ~(p→q) -- occurence-map(of ~(p→q))
T -- T -- F -- NO
T -- F -- T -- O
F -- T -- F -- NO
F -- F -- F -- NO

Now coming back to the OP, "p doesn't necessarily imply q" (writing it as "r" as shorthand) might "mean" the following three occurrence possibities:
(a)
p -- q -- r
T -- T -- O
T -- F -- O
F -- T -- F -- O/NO
F -- F -- F -- O/NO

(b)
p -- q -- r
T -- T -- O/NO
T -- F -- O
F -- T -- F -- O/NO
F -- F -- F -- O/NO

(c)
p -- q -- r
T -- T -- NO
T -- F -- O
F -- T -- F -- O/NO
F -- F -- F -- O/NOIt seems perhaps possibility (b) can be thought of as allowing both the concrete scenarios in possibility (a) and possibility (c) (similar to division into concrete possibilities for which I gave an example in the first half of the post).

Possibility (c) doesn't seem the correct interpretation "p doesn't necessarily imply q" to me (given OP's example I suppose). The better interpretations seems to me (a) or (b). Both (a) and (b) are incompatible with ~(p→q). Furthermore since both these meanings carry an "O" in front, I am not clear whether this wording can be expressed just with propositional logic like statements (over possibility space)?
In case of example given in OP, possibility (a) seems to be a good fit.

Perhaps I have missed something trivial entirely?

 

[Mark44]

I think there is (perhaps?) an issue here, in a manner of speaking. Below I describe informally what it is.

It seems to me that when we normally use propositional logic like statements (in maths), a lot of the times we have essentially a sort of "possibility space" in mind. I am thinking of a statement like (over the integers for example):
(x=2) → (x²=4)

There is no need to add anything about limiting the equations above to the integers. If x is an integer, x² necessarily will also be an integer.

Below I will assume that the possibility space is non-empty.

So, for example:
p -- q -- p∨q -- occurence-map(of p∨q)
(i) T -- T -- T -- O/NO
(ii) T -- F -- T -- O/NO
(iii) F -- T -- T -- O/NO
(iv) F -- F -- F -- NO

I didn't read the rest of your very long post. For an implication such as the one discussed here, the truth table is very simple.

p   q   ## p \Rightarrow q##
T   T     T
T   F     F
F   T     T
F   F     T

In other words, the only combination of truth values for p and q for which the implication is false is when p is true and q is false. One can also easily verify that $p \Rightarrow q$ is eqivalent to $\neg(p \wedge \neg q)$

 

[SSequence]

There is no need to add anything about limiting the equations above to the integers. If x is an integer, x² necessarily will also be an integer.

Yes, that part wasn't really meant to be important. I was just trying to say that we usually have some domain in mind when writing these kind of equations (naturals, integers, rationals, reals, complex nos. etc. when consider numbers for example).

You will have to read the previous post in detail to try to understand what I am trying to say. Admittedly, there might be some minor confusions because I was trying to keep the post to manageable lengths. In short, here is what I am trying to say:
Quite often when we say, for example, something like "p implies q" or "p or q" (in maths), we are informally referring actually to the following statements:
$\forall x [p(x) \rightarrow q(x)]$
$\forall x [p(x) \vee q(x)]$
and perhaps even, depending on context when we might have more variables, something like:
$\forall x \forall y [p(x,y) \rightarrow q(x,y)]$
$\forall x \forall y [p(x,y) \vee q(x,y)]$

Informally the domain of quantification can be thought of depending on the context I suppose. Now consider the example in OP:

I figure there must be a way to denote a situation in which, from sloppy intuition, etc., a certain proposition might (erroneously) imply a result, but that that result could happen, just that it is not guaranteed to happen.

A simple example would be the case of the product of a pair of matrices as LHS & RHS being equal to the sides being swapped.

[ A ] [ B ] = [ C ] does not necessarily imply [ B ] [ A ] = [ C ], but nor does it imply [ B ] [ A ] != [ C ], as is could very well luck out that [ A ] [ B ] = [ B ] [ A ]

One way I can think of formalising the statements "AB=C does not necessarily imply BA=C" is something like:
$\forall A,B,C [AB=C$ doesn't necessarily imply $BA=C]$
Here the quantification is assumed over all matrices (for simplicity, with further implicit qualification that the sizes of A and B are assumed to be suitable for both AB and BA to be defined).

Now if you take $p(X,Y,Z)\equiv [XY=Z]$ and $q(X,Y,Z)\equiv [YX=Z]$. Now if you replace the statement "AB=C doesn't necessarily imply BA=C" with:
$\sim (p(A,B,C) \rightarrow q(A,B,C))$
$\sim (AB=C \rightarrow BA=C)$
Quite clearly the following statement is not correct:
$\forall A,B,C [\sim (AB=C \rightarrow BA=C)]$
because we can easily find instances where the first and second equation are both correct ... that is A and B do commute for these instances (which will lead the implication to be true and whole expression to be false).

Towards the end of last post I described what seemed to me was the intended meaning of "AB=C does not necessarily imply BA=C" (ofc it can be put explicitly by using quantifiers I think ... but the expression will be longer).

For this kind of wording where we are using wording of propositional logic but in essence thinking about an actual domain ... in the previous post ... I was just describing a simple way to think terms of symbols O, NO, O/NO and furthermore in terms of idea of "concrete possibilities". Furthermore, I described how compound expressions (that are worded like propositional logic but in essence quantified statements) ... can be thought of in very simple terms using the same occurrence symbols.
Of course one can retort back to quantified logic if it is really needed, but still I find it easier to think in terms of truth tables of occurrences and concrete possibilities (in simple cases where it is possible).

 

[Mark44]

Towards the end of last post I described what seemed to me was the intended meaning of "AB=C does not necessarily imply BA=C" (ofc it can be put explicitly by using quantifiers I think ... but the expression will be longer).

I have never seen a single linear algebra textbook that uses quantifiers like this when talking about matrix multiplication. Instead, what they usually say is that matrix multiplication is not generally commutative, for either of the following reasons.

1. BA might be defined, while the product AB is not defined (due to the matrices not being conformable for multiplication in this order).
2. BA is defined, and AB is defined, but BA ≠ AB.

What more needs to be said?

For this kind of wording where we are using wording of propositional logic but in essence thinking about an actual domain ...

Sometimes the domain is implied, but other times it is explicitly given, for example as "Let A be an m x n matrix, and B be an n x p matrix. Then AB is an m x p matrix whose entries are ..."

in the previous post ... I was just describing a simple way to think terms of symbols O, NO, O/NO

Simple? I don't think so. Mapping two symbols to three isn't simplification. How is it advantageous to replace T and F by a much more cumbersome notation of O/NO (for T) and NO (for F)? If there was a rationale for doing this in your "wall o' text" it escaped me.

 

[SSequence]

I have never seen a single linear algebra textbook that uses quantifiers like this when talking about matrix multiplication. Instead, what they usually say is that matrix multiplication is not generally commutative, for either of the following reasons.

1. BA might be defined, while the product AB is not defined (due to the matrices not being conformable for multiplication in this order).
2. BA is defined, and AB is defined, but BA ≠ AB.

What more needs to be said?

Yes, we can say same things formally (possibly at different levels of distinction) and informally. For example, we often write something like:
$2x+1=3$
$2x=2$
$x=1$
The way I think of it (and possibly see it little more formally) is as:
$\forall x([2x+1=3] \Leftrightarrow [2x=2] \Leftrightarrow [x=1])$
where the domain of quantification can be thought of as ℝ. Personally, I have never seen this kind of notation either.

In particular let's take the last two lines (it will also illustrate what I am trying to say better I suppose):
$\forall x([2x=2] \Leftrightarrow [x=1])$

p -- q -- p iff q -- occurrence-map
(i) T -- T - T - O/NO
(ii) T -- F - F - NO
(iii) F -- T - F - NO
(iv) F -- F - T - O/NO

Now if we list "all" the allowed concrete possibilities we have:
(1) (i) O (ii) NO (iii) NO (iv) NO
(2) (i) NO (ii) NO (iii) NO (iv) O
(3) (i) O (ii) NO (iii) NO (iv) O

Now it is the possibility(3) that actually happens to be the case. The quantified statement above would be true for both possibility(1) and possibility(2) too.

And that's a precursor to the more general point. I don't know know of anyway to place only logical connectives (and, or, not etc.) inside statement of the form (taking p(x) to be [2x=2] and q(x) to be [x=1]):
$\forall x ($ combination of connectives and p(x) and q(x) $)$
so that the given statement is true ONLY for (concrete) possibility(3) and no other possibility.

Simple? I don't think so. Mapping two symbols to three isn't simplification. How is it advantageous to replace T and F by a much more cumbersome notation of O/NO (for T) and NO (for F)? If there was a rationale for doing this in your "wall o' text" it escaped me.

The symbol "O/NO" is really a bit of a short-cut. The real point is this:
Suppose we have two propositional functions p(x) and q(x) over some domain and a quantified statement of the form:
$\forall x ($ combination of connectives and p(x) and q(x) $)$

This is strictly weaker than allowing an arbitrary subset of concrete possibilities. It seems to me that certain subsets of concrete possibilities simply can't be expressed (in equivalent form) by a statement of the form above. As for what I mean by concrete possibility:
p q
(i) T T
(ii) T F
(iii) F T
(iv) F F
All strings of length-4 (with the alphabet set as {O,NO}) count as concrete possibilites here, except the string (NO,NO,NO,NO). Now as I explained in the second half of post#7 that the example given in the OP is better handled by thinking in terms of a subset of concrete possibilities.

That's all there is to it really. It wasn't meant to be a deep (or complex for that matter) point, but just as a simple observation (that I personally found useful way to think in simple scenarios). And one can always retort back to full-fledged quantified descriptions if needed.

 

[Mark44]

Yes, we can say same things formally (possibly at different levels of distinction) and informally. For example, we often write something like:
$2x+1=3$
$2x=2$
$x=1$
The way I think of it (and possibly see it little more formally) is as:
$\forall x([2x+1=3] \Leftrightarrow [2x=2] \Leftrightarrow [x=1])$
where the domain of quantification can be thought of as ℝ. Personally, I have never seen this kind of notation either.

I sometimes write things like this:
$2x+1=3$
$\Leftrightarrow 2x=2$
$\Leftrightarrow x=1$
-- with the understanding, either implicit or explicit that we're talking about real numbers.

p -- q -- p iff q -- occurrence-map
(i) T -- T - T - O/NO
(ii) T -- F - F - NO
(iii) F -- T - F - NO
(iv) F -- F - T - O/NO

Why?
All you need for equivalence is the usual truth table

p   q   p <=> q
T   T      T
T   F      F
F   T      F
F   F      T

In short, p and q are equivalent if both are true or both are false. Why complicate something that's straightforward with excess baggage of O/NO, NO, etc.?

Now if we list "all" the allowed concrete possibilities we have:
(1) (i) O (ii) NO (iii) NO (iv) NO
(2) (i) NO (ii) NO (iii) NO (iv) O
(3) (i) O (ii) NO (iii) NO (iv) O

I don't see any value in this listing.

 

[SSequence]

Why?
All you need for equivalence is the usual truth table

p   q   p <=> q
T   T      T
T   F      F
F   T      F
F   F      T

In short, p and q are equivalent if both are true or both are false. Why complicate something that's straightforward with excess baggage of O/NO, NO, etc.?

Yeah sorry the notation in my post was somewhat careless. Instead of writing p or q, it would be better to write p(x) or q(x), where x denotes an arbitrary instance from the assumed domain.

I don't see any value in this listing.

Now if we list "all" the allowed concrete possibilities for p(x) iff q(x) we have:
(1) (i) O (ii) NO (iii) NO (iv) NO
(2) (i) NO (ii) NO (iii) NO (iv) O
(3) (i) O (ii) NO (iii) NO (iv) O

The point was that if you take an arbitrary instance "a" from the assumed domain then:
(i) p(a) true , q(a) true
(ii) p(a) true , q(a) false
(iii) p(a) false , q(a) true
(iv) p(a) false , q(a) false

Now with concrete possibilities I just meant that when you search over "ALL" of the domain and for each instance evaluate the values of p(--) and q(--), which "combination" of possibilities you will find for at least one instance (these are ones marked with "O") and which "combinations" you will never find through the whole domain (these are ones marked with "NO").
The second half of my previous post should be reasonably clear after this. Whether it can be thought of as useful or not is a subjective matter.