Is Absence of Evidence ever Evidence of Absence?

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2 separate functions can be defined to better distinguish each interpretation of Evidence:
(1) R-Evidence(proposition) is "Evidence" relevant to the proposition
(2) P-Evidence(perceptible,domain) is "perceptual experience" of perceptible in domain
[...]
However, "X" can be expanded into "P-Evidence(Y,D)", where Y is perceptible thing (eg, "milk") and D is domain (eg, fridge), and the central "is" can be turned into function "Is-Equivalent(A,B)", resulting in:

Is-Equivalent(
Absent(R-Evidence(Present(P-Evidence(Y,D)))),
Present(R-Evidence(Absent(P-Evidence(Y,D))))
)

This probably needs to be revised ...

P-Evidence could potentially fit in there as part of X to help tie in D but this may be adding something to the semantics that wasn't there ...

The truth of proposition "Y is Present in D" doesn't ever need to be known. P-Evidence can be excluded from the semantic expression.

Parameter D may belong more appropriately in Present() and Absent():
- Present(perceptible,domain) means perceptible is present in domain
- Absent(perceptible,domain) means perceptible is absent in domain

These complementary functions are applied to 2 different things: Y in D (eg "milk in fridge"); and Evidence in ???

What is the domain of Evidence to tell if it is present there or not? The Subject's mind? This may not be very important so long as the domain is same for Present(Evidence,D) and Absent(Evidence,D). Subject S can be used for now.

Revised semantic expression:

Is-Equivalent(
Absent(R-Evidence(Present(Y,D)),S),
Present(R-Evidence(Absent(Y,D)),S)
)

[Leontiskos]

Case 3:

1c. P <-> O
2c. ~O
3c. ∴ ~P

This is a sound syllogism, and it is the only case where absence of evidence is evidence of absence. In this case (2c) does count as absence of evidence, for we know from (1c) that (O -> P). Therefore O counts as evidence and ~O counts as absence of evidence. Further, evidence of absence (~P) really does follow from absence of evidence (~O). This case succeeds. An example of (1c) would be, "There is milk in the refrigerator if and only if I can see it." Note that (1c) could also be written (O <-> P).

??
There can be milk in the fridge without having the ability to see it. How does this qualify as an example of "if and only if" (especially "only if")?

Sure, if the investigator is blind or if by "milk" we mean a microscopic molecule of milk. See my post to Pattern-chaser here (link).

My point is that Sherlock Holmes and the mediocre detective could examine the same physical/perceptual evidence and yet come to different conclusions or at least different levels of certitude because Sherlock's deductive capabilities far exceed the mediocre detective's. The point is that 1) The deductive capabilities are not perceptual evidence,

If they have the same relevant perceptual data, they could weigh this up differently. It is not uncommon for judges and jurors who are exposed to the same evidence in court to reach different conclusions. All sorts of things may come into play that lead to different conclusions. These things could ultimately be based on perceptual evidence or on other things like wishful thinking.

Then you've agreed with me (ignoring the fact that your first sentence contradicts your last--either they have the same perceptual evidence or they don't).

The more inclusive "evidence" becomes, the easier it may be for something to qualify as "evidence of absence", but the harder it becomes for something to qualify as "absence of evidence" (and vice versa the more exclusive "evidence" becomes).

I don't see why that would follow.

How can there be evidence of absence of milk without scoping this to a finite domain (eg, a fridge) that milk is absent from?

I think you're making this more difficult than it is. Is there such a thing as an infinite physical domain? Only in a logical sense, for our empirical evidence suggests that no such thing actually exists. Again, see my post to Pattern-chaser here (link).

A proposition and a perceptible thing are semantically quite distinct. A proposition can be True or False. A perceptible thing is an object that can be included in a proposition.

It may not seem that there is much difference between perceptible "milk" in (2) and proposition "milk is present" in (1). But their complements are very different.

The complement of proposition "milk is present" is proposition "milk is absent". If perceptible "milk" is the set of of all perceptible things that qualify as milk, then perceptible "~milk" is the set of all perceptible things that don't qualify as milk (eg: beer implies ~milk; ~milk doesn't imply beer; ~beer doesn't imply milk; ~milk doesn't imply "milk is absent".)

No, propositions contain terms, not perceptual objects. There is no such thing as a perception apart from its conceptual representation when we are communicating. You can conceptually define "milk" however you like. If the concept is defined the same way, the complement will be the same.

Attempting to use different datatypes like these interchangeably is likely to cause problems.

Did you see my question about your approach in this post (link)?

To go your route, we could clarify each term in the sentence, "Absence of evidence of P is evidence of absence of P."

Expanding this to make "presence" explicit where this is implicit:
"Absence of evidence of presence of P is presence of evidence of absence of P"...

I think my explanation was much simpler:

My contention is that this is how we must interpret the sentence if it is to be true:

• [Absence] of [evidence of P] is [evidence of absence of P].
• [Privation] of [the sort of thing that would have counted as evidence for P's presence] is [a justified reason to believe that P is absent].

Basically this boils down to:
A: Absent(Evidence(P))
B: Present(Evidence(Q))
C: Is-Equivalent(A,B)

A and B can both be True, but A doesn't imply B, so C is not True.

Alternative semantic interpretations are welcome.

I already answered this objection:

The point isn't that A and B are merely related, but rather that they are identical. This is even implied if you do the substitution you suggested. According to your definitions these two statements are equivalent:

• Absence of evidence for P is presence of evidence for Q.
• A = Absence of evidence for P
• B = Presence of evidence for Q
• A is B (A = B)

Here is an example:

P: "There is milk in the fridge"
Q: "There is no milk in the fridge"
Z: "I see milk in the fridge"
A: "I do not see any milk in the fridge"
B: "I do not see any milk in the fridge"

Note that Z represents presence of evidence for P, which is precisely what is absent in A (and B).

This is an example of Case 3 from this post.

If A didn't imply B we would have Case 2. I gave Case 3, not Case 2. See this post (link).

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The absence of a deduced conclusion is evidence for the falsity of that conclusion.

I think the absence of a deduced conclusion is evidence for the uncertainty of that conclusion, as deduction is the only bulletproof form of logical reasoning. To declare an uncertain conclusion "false" may be going a bit too far, don't you think?

In the contextual case of Sherlock Holmes which I gave, the fact that Sherlock did not deduce, "A crime has occurred," counts as evidence for the claim that no crime occurred. That said, it is true that we are talking about evidence rather than proof & certainty. I'm not saying we have certainty.

----------

Again, I am NOT indulging the logorrhea that people have on boards like this. People write horrible, rambling, unclear posts that poorly broach tons of issues. If they'd learn to write better, I'd read and respond to longer posts. But I'm not about to indulge horrible writing.

And then if you write something longer in reply, questions/points etc. are ignored, regardless of how long the response is.

I've no interest in that. People need to learn how to communicate better.

That's all true, but some of us aren't able to engage in sentence-by-sentence discussions, nor do I take that to be the purpose of such forums. At least a paragraph of some elucidation would be expected.

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What you are failing to see is that absence of evidence for one thing can be presence of evidence for a different thing.

How can absence of evidence for P (one thing) be presence of evidence for Q (a different thing)?
Yes, A (absence of evidence for P) and B (presence of evidence for Q) can both be True at the same time, but how can A ever imply B?
If P and Q are independent variables then A doesn't reveal anything about B (B may be True or False).
If Q is dependent on P (eg, Q is ~P, or some other function of P) then A reveals there is also absence of evidence for Q (B is False).

The point isn't that A and B are merely related, but rather that they are identical. This is even implied if you do the substitution you suggested. According to your definitions these two statements are equivalent:

• Absence of evidence for P is presence of evidence for Q.
• A = Absence of evidence for P
• B = Presence of evidence for Q
• A is B (A = B)

A and B are clearly not identical (textually, syntactically, semantically).

A and B could be equivalent if A always implies B and B always implies A (for any values of P, Q, and any evidence). Only one example is needed where A doesn't equal B to show that A and B aren't equivalent.

P: "There is milk in the fridge"
Q: "There is a pie in the oven"

Absence of evidence for P does not imply presence of evidence for Q. There may or may not be evidence for Q. A and B are not equivalent.

The next question is: Can A ever be/imply B? (Are there any values of P and Q where A implies B (regardless of evidence)?)

Here is an example:

P: "There is milk in the fridge"
Q: "There is no milk in the fridge"
Z: "I see milk in the fridge"
A: "I do not see any milk in the fridge"
B: "I do not see any milk in the fridge"

Note that Z represents presence of evidence for P, which is precisely what is absent in A (and B).

?? How does A and B in above quote along with P, Q, and Z in example lead to A and B in this example?

In order to answer Yes to "Can A ever be B?" the following is needed [and square brackets say what has already been provided]:
(1) Expressions for A and B [provided in top quote above]
(2) Example values for any variables in A and B [example provides proposition values for P and Q: these are the main values of P and Q discussed in this topic; Q is ~P; evidence For P is evidence Against Q (and vice versa)]
(3) A variety of cases of hypothetical evidential data to test this [one case provided in example (Z: "I see milk in the fridge"); more cases are needed]
(4) Methods for consistent evaluation of A and B that anyone can apply and get the same results [not yet provided]
(5) Evaluations of A and B that show A = B for all test cases of Z [not yet provided]

A distinction can be made between individual high-level perceptual evidence events ("PE-Event") like "I see milk in fridge", and collective evidence as a dataset of zero or more PE-Events like, Z: {"I see milk in fridge", "I see pie in oven"}.

Before continuing, it must be noted that "Evidence for P" is not identical to "Evidence of P" (although some people may use these phrases interchangeably at times).

"Evidence for P" suggests evidential support for P. Each PE-Event can be evaluated individually, and collective evidence can be evaluated collectively, with respect to P. Evaluation of evidential support for P can be Positive, Negative, or Zero, and can include strength of support (eg, a number between -100 and 100).

"Evidence for P" is ambiguous: "evidence" can mean "collective evidence" (a dataset of PE-Events) or "evaluation of evidential support" (a numeric value); 'for' can mean "with regard to" (relevant to; including negative and positive) or For as opposed to Against (only including positive). In the absence of a specific interpretation, multiple interpretations can be analysed in turn, including these three ...

EF1: (dataset) accumulate all PE-Events that individually evaluate to Positive or Negative support for P
EF2: (number) simple evaluation of EF1: sum of individual evaluations of all relevant PE-Events with respect to P
EF3: (dataset) accumulate all PE-Events that individually evaluate to Positive support for P

For EF1 and EF3:
Absent(dataset): True if count of PE-Events in dataset is Zero
Present(dataset): True if count of PE-Events in dataset is NonZero

For EF2:
Absent(number): True if number is Zero
Present(number): True if number is NonZero (greater than Zero or less than Zero)
(This assumes acceptance that Absent(number) translates okay to Zero(number) and Present(number) translates okay to NonZero(number). If don't accept this then EF2 can be eliminated from study.)

EF1 and EF2 include all PE-Events that are relevant to P. EF3 excludes relevant PE-Events that negatively support P. This is highly questionable, like only admitting evidence that positively supports a theory and ignoring any evidence that negatively supports it. However, EF3 is not an unreasonable interpretation of "Evidence for/of P", and this may be the interpretation that comes closest to supporting "A can be B" so this can be included in study.

There are at least 3 EF-Cases: {EF1,EF2,EF3}

A way is needed to individually evaluate how much a PE-Event positively supports propositions P and Q. This way can be the same for each EF-Case, and each EF-Case may process individual evaluations differently. Here are some simplistic evaluation guides for 3 different categories of PE-Events (and 1 invalid type) with respect to P and Q (precise evaluations are not necessary here) ...
z1: "I see milk in fridge" (strong positive support for P; strong negative support for Q )
z2: "I see [non-milk] in fridge" (weak negative support for P; weak positive support for Q)
z3: "I see [something] outside fridge" (zero support for P; zero support for Q)
z4: "I don't see [X]" (**NOT A VALID PE-Event**)

Here are 5 example cases of Z:
Z1: {z1} (see milk in fridge")
Z2: {z2} (see beer in fridge)
Z3: {z3} (see pie in oven)
Z4: {z2,z1} (see beer in fridge; see milk in fridge)
Z5: {} (empty set; zero PE-Events)
Z-Cases: {Z1,Z2,Z3,Z4,Z5}

Rephrasing A and B:
A: Absent(Evidence-For(Z,P))
B: Present(Evidence-For(Z,Q))

This procedure can be followed ...
For each Evidence-For in EF-Cases:
- For each Z in Z-Cases:
-- Evaluate and compare A and B.

Here are the results ...
_____________________________________________

Case EF1: (collection of PE-Events that positively or negatively support proposition)
A: Absent(EF1(Z,P))
B: Present(EF1(Z,Q))

Case Z1: {z1} (see milk in fridge)
A = False; B = True; ~(A = B)

Case Z2: {z2} (see beer in fridge)
A = False; B = True; ~(A = B)

Case Z3: {z3} (see pie in oven)
A = True; B = False; ~(A = B)

Case Z4: {z2,z1} (see beer in fridge, see milk in fridge)
A = False; B = True; ~(A = B)

Case Z5: {} (empty set; zero PE-Events)
A = True; B = False; ~(A = B)

Summary of case EF1:
A doesn't equal B for any Z cases, therefore A is not equivalent to B.
A may be equivalent to ~B (unless a Z case is found which results in A = B).

_____________________________________________

Case EF2: (sum of individual evaluations of relevant PE-Events with respect to proposition)
A: Absent(EF2(Z,P))
B: Present(EF2(Z,Q))

Same results as EF1 for all Z-cases (no need to repeat this).

Case Z4 (see beer in fridge, see milk in fridge) is interesting. Individually there is both positive and negative support for P. Positive support is stronger than negative support so collective support is positive if sum up individual support evaluations of EV-Events. But the negative support of seeing beer may not reduce the strength of positive support if milk is seen, so a realistic collective evaluation may be more complex than EF2. Also, case Z4 of EF2 could evaluate A as True and B as False if Positive and Negative Support for P have the same absolute value. This is one way that EF2 results can differ from EF1. But ~(A = B) remains unchanged.
_____________________________________________

Case EF3: (collection of PE-Events that positively support proposition)
A: Absent(EF3(Z,P))
B: Present(EF3(Z,Q))

Case Z1: {z1} (see milk in fridge)
A = False; B = False; A = B

Case Z2: {z2} (see beer in fridge)
A = True; B = True; A = B

Cases Z3, Z4, and Z5 evaluate the same as EF1 and EF2:
~(A = B)

Summary of Case EF3:
A = B in cases Z1 and Z2, but A doesn't equal B in other cases.
Therefore A is not equivalent to B.

Here are some responses to questions that haven't been asked yet ...

> How to explain the two cases where A = B? What is EF3 doing differently from EF1 and EF2?
EF3 excludes relevant evidence (if this negatively supports P or Q). EF1 and EF2 exercise proper accounting of debits and credits. EF3 doesn't.

> Do results of cases Z1 and Z2 mean that A reveals something about B in case EF3?
No. There are 3 cases where A is True, but B is True in only one of these cases (Z2). Likewise there are 2 cases where A is False, but B is False in only one of these cases (Z1). A doesn't imply B. B doesn't imply A. Any equality of A and B is coincidental. (Actually, it is not a coincidence that A = B only results when presence of relevant (negatively supporting) evidence is overlooked, not accepted as relevant, or otherwise excluded.)

> How do these cases fit with claim that: "If Q is dependent on P (eg, Q is ~P, or some other function of P) then A reveals there is also absence of evidence for Q (B is False)"?
The claim doesn't totally fit and needs to be revised. It appears that it is the broader functions of Evidence-For(P) and Evidence-For(Q) which need to be dependent for this claim to fit. EF1(P) and EF1(Q) are mutually dependent. However EF3(P) and EF3(Q) are mutually independent. This independence allows Absent(EF3(P)) to equal Present(EF3(Q)) in some cases but this also prevents one of them from implying the other.
_____________________________________________

Can anyone provide methods for evaluating
A: "Absence of Evidence for P" and
B: "Presence of Evidence for Q"
which are reasonable interpretations of these expressions, and result in consistent evaluations of A and B that equal each other for all 5 Z cases provided here, using same values of P and Q?

[Leontiskos]

A and B are clearly not identical (textually, syntactically, semantically).

A and B could be equivalent if A always implies B and B always implies A (for any values of P, Q, and any evidence). Only one example is needed where A doesn't equal B to show that A and B aren't equivalent.

P: "There is milk in the fridge"
Q: "There is a pie in the oven"

Absence of evidence for P does not imply presence of evidence for Q. There may or may not be evidence for Q. A and B are not equivalent.

The next question is: Can A ever be/imply B? (Are there any values of P and Q where A implies B (regardless of evidence)?)

This thread is about whether absence of evidence is ever evidence of absence, not whether absence of evidence is always evidence of absence. My point was that yes, it sometimes is evidence of absence, and this occurs precisely when a number of different criteria are met, including, vacuously, the identity of A and B.

How can absence of evidence for P (one thing) be presence of evidence for Q (a different thing)?

Yes, A (absence of evidence for P) and B (presence of evidence for Q) can both be True at the same time, but how can A ever imply B?

If P and Q are independent variables then A doesn't reveal anything about B (B may be True or False).

If Q is dependent on P (eg, Q is ~P, or some other function of P) then A reveals there is also absence of evidence for Q (B is False).

Here is an example:

P: "There is milk in the fridge"
Q: "There is no milk in the fridge"
Z: "I see milk in the fridge"
A: "I do not see any milk in the fridge"
B: "I do not see any milk in the fridge"

Note that Z represents presence of evidence for P, which is precisely what is absent in A (and B).

?? How does A and B in above quote along with P, Q, and Z in example lead to A and B in this example?

In order to answer Yes to "Can A ever be B?" the following is needed [and square brackets say what has already been provided]:
(1) Expressions for A and B [provided in top quote above]
(2) Example values for any variables in A and B [example provides proposition values for P and Q: these are the main values of P and Q discussed in this topic; Q is ~P; evidence For P is evidence Against Q (and vice versa)]
(3) A variety of cases of hypothetical evidential data to test this [one case provided in example (Z: "I see milk in the fridge"); more cases are needed]
(4) Methods for consistent evaluation of A and B that anyone can apply and get the same results [not yet provided]
(5) Evaluations of A and B that show A = B for all test cases of Z [not yet provided]

I assigned the variables in relation to the inferential context you gave in this post (link). We can lay out the entire example:

You asked, "How can absence of evidence for P (one thing) be presence of evidence for Q (a different thing)?"

This is how:

P: "There is milk in the fridge"
Q: "There is no milk in the fridge"
Z: "I see milk in the fridge"
A: "I do not see milk in the fridge"
B: "I do not see milk in the fridge"

Z is presence of evidence for P (Z -> P)
A is absence of evidence for P (A = ~Z)
B is presence of evidence for Q (B -> Q)
A and B are equal (A = B) (absence of evidence for P is presence of evidence for Q)


This occurs because (P <-> Z) and (P = ~Q) - See Case 3 in this post (link).

(P <-> Z) holds because, if there is milk in the fridge then I will see it; and if I see milk in the fridge then there is milk in the fridge.

Q.E.D.

"Evidence for P" suggests evidential support for P.

That's right, and Z is evidence for P because (Z -> P) ("If I see milk in the fridge then there is milk in the fridge.")

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A and B are clearly not identical (textually, syntactically, semantically).
The next question is: Can A ever be/imply B? (Are there any values of P and Q where A implies B (regardless of evidence)?)

This thread is about whether absence of evidence is ever evidence of absence, not whether absence of evidence is always evidence of absence.

Yes, this is the topic question ...
A = A1: Absence of Evidence
B = B1: Evidence of Absence
Is A ever B? (Can A ever be B? Are there any contexts where A is B?)

My point was that yes, it sometimes is evidence of absence, and this occurs precisely when a number of different criteria are met, including, vacuously, the identity of A and B.

If A and B have not been specified (instantiated), then A can be identical to B. However, A and B (relabelling now as A2 and B2) were already specified as ...

A = A2: absence of evidence for P
B = B2: presence of evidence for Q

A2 and B2 are not identical. (A1 and B1 are not identical either. A2 and B2 are possible interpretations of (but not identical to) A1 and B1.) Some other criteria is needed to show that A2 can be B2.

If A or B is ambiguous, it can be argued that: A can be B if there is an interpretation of A that is identical to an interpretation of B. With freedom of interpretation, anyone can argue that any A can be any B. No semantic analysis or logic is required. "A can be B because sometimes I use these expressions interchangeably." But this lacks substance without (unambiguous) semantic expressions for A and B to base this on.

Even if unambiguous semantic expressions for A and B are provided, the answer may be far from obvious, especially if A and B include variables like P and Q. This may be like trying to solve simultaneous equations. What values of P and Q (if any) allow A to be B (especially where Q is ~P)?

You asked, "How can absence of evidence for P (one thing) be presence of evidence for Q (a different thing)?"

This is how:

P: "There is milk in the fridge"
Q: "There is no milk in the fridge"
Z: "I see milk in the fridge"
A: "I do not see milk in the fridge"
B: "I do not see milk in the fridge"

?? This seems no less perplexing than it was before ... Is this meant to be an example of how "A can be B" if specify that A is identical to B, and this A and B is not intended to represent specified A and B (A2 and B2) above? If yes, then P, Q, and Z are irrelevant. If this A and B are intended to represent specified A and B above, then it appears that A and (especially) B have mutated significantly.

[Sticking with specified A and B (A2 and B2) above and this P, Q, and Z while continuing to read ...]

Z is presence of evidence for P (Z -> P)
A is absence of evidence for P (A = ~Z)
B is presence of evidence for Q (B -> Q)
A and B are equal (A = B) (absence of evidence for P is presence of evidence for Q)

This occurs because (P <-> Z) and (P = ~Q) - See Case 3 in this post (link).

(P <-> Z) holds because, if there is milk in the fridge then I will see it; and if I see milk in the fridge then there is milk in the fridge.

Ah, it appears this argument is based on P <-> Z, which doesn't really apply to this milk in fridge example (P -> Z clearly isn't true and Z -> P is questionable). However, there may be other examples where P <-> Z really does apply (or nearly applies), and it may be interesting to explore this in more detail from at least 2 angles:
(1) What is required for an example to qualify as P <-> Z?
(2) If, hypothetically, an example does qualify as P <-> Z, how sound is this argument?

Responses to (1):

What if P is identical to Z, eg ...
P: "I see milk in the fridge" ?

P is meant to represent an assertion of the presence of something (in a domain). In this example, "something" can be visual experience (of milk in the fridge), and the domain can be the mind/consciousness of "I". Z can be an instance of this visual experience. Spelling this out more ...

P: "Visual experience of 'milk in the fridge' is present in my mind"
Q: "Visual experience of 'milk in the fridge' is absent from my mind"
Z: "I presently have visual experience of 'milk in the fridge'"

P may need to refer to the presence of something that is self-evident. Can this ever be something physical, or can this only be some form of subjective experience? Does subjective experience qualify as perceptible?

Responses to (2):

If Z <-> P and Q is ~P then:
~Z <-> Q (no need for any evidence to conclude Q directly from ~Z)
~Z <-> A, but ...
~Z doesn't imply B (Q can be true without having any evidence for this)
~(A <-> B)

B -> Q is questionable, partly because: how can there be evidence that positively supports Q?
The only evidence that is relevant to P and Q is Z.
Z is conclusive positive evidence for P (can conclude P is True).
Z is conclusive negative evidence for Q (can conclude Q is False).
Any other evidence is irrelevant to P and Q.
~Z is absence of relevant evidence (~Z may or may not include irrelevant evidence).

[Leontiskos]

P: "There is milk in the fridge"
Z: "I see milk in the fridge"

...This occurs because (P <-> Z) and (P = ~Q) - See Case 3 in this post (link).

(P <-> Z) holds because, if there is milk in the fridge then I will see it; and if I see milk in the fridge then there is milk in the fridge.

Ah, it appears this argument is based on P <-> Z, which doesn't really apply to this milk in fridge example (P -> Z clearly isn't true and Z -> P is questionable).

You aren't providing an argument for your claim. You are merely asserting it. Here are the two relevant conditionals:

• If there is milk in the fridge then I will see it.
• If I see milk in the fridge then there is milk in the fridge.

These statements are true. This isn't rocket science. Our eyes perceive visible objects, and milk is a visible object.

[-0+]

P: "There is milk in the fridge"
Z: "I see milk in the fridge"

...This occurs because (P <-> Z) and (P = ~Q) - See Case 3 in this post (link).

(P <-> Z) holds because, if there is milk in the fridge then I will see it; and if I see milk in the fridge then there is milk in the fridge.

Ah, it appears this argument is based on P <-> Z, which doesn't really apply to this milk in fridge example (P -> Z clearly isn't true and Z -> P is questionable).

You aren't providing an argument for your claim. You are merely asserting it. Here are the two relevant conditionals:

• If there is milk in the fridge then I will see it.
• If I see milk in the fridge then there is milk in the fridge.

These statements are true.

Are these "true" statements meant to be hypothetical with a higher-level conditional: if the two relevant conditionals are valid (if presence of "milk in the fridge" really implies "I will see it" and vice versa) then P <-> Z?

If P -> Z and Z -> P then P <-> Z but the validity of the 2 conditionals in this example is questionable.

If there is milk in the fridge then John may or may not see it. If he is blind or he doesn't open the fridge door then he is unlikely to see it. Many other examples can be provided where milk is in the fridge but John doesn't see it. Only one example is needed to invalidate P -> Z.

If John sees milk in the fridge then this is positive evidential support for him to believe there is milk in the fridge. John's model of reality may include milk in the fridge, but there may not really be milk in the fridge if, for example, what he perceived as milk is actually non-milk. Additional evidence may lead him to realise this, and that Z doesn't imply P.

This isn't rocket science. Our eyes perceive visible objects, and milk is a visible object.

Very young children may believe that if they don't see an object behind an obstacle then the object isn't there, but they soon learn that ~Z doesn't necessarily mean ~P.

[Leontiskos]

P: "There is milk in the fridge"
Z: "I see milk in the fridge"

...This occurs because (P <-> Z) and (P = ~Q) - See Case 3 in this post (link).

(P <-> Z) holds because, if there is milk in the fridge then I will see it; and if I see milk in the fridge then there is milk in the fridge.

Ah, it appears this argument is based on P <-> Z, which doesn't really apply to this milk in fridge example (P -> Z clearly isn't true and Z -> P is questionable).

You aren't providing an argument for your claim. You are merely asserting it. Here are the two relevant conditionals:

• If there is milk in the fridge then I will see it.
• If I see milk in the fridge then there is milk in the fridge.

These statements are true.

Are these "true" statements meant to be hypothetical with a higher-level conditional: if the two relevant conditionals are valid (if presence of "milk in the fridge" really implies "I will see it" and vice versa) then P <-> Z?

Yes, that's what (P <-> Z) means.

If there is milk in the fridge then John may or may not see it. If he is blind or he doesn't open the fridge door then he is unlikely to see it. Many other examples can be provided where milk is in the fridge but John doesn't see it. Only one example is needed to invalidate P -> Z.

If John sees milk in the fridge then this is positive evidential support for him to believe there is milk in the fridge. John's model of reality may include milk in the fridge, but there may not really be milk in the fridge if, for example, what he perceived as milk is actually non-milk. Additional evidence may lead him to realise this, and that Z doesn't imply P.

But as I guessed in this post, these are strawmen par excellence. Why would you assume that I am talking about a blind man or a man who has not opened the refrigerator? How are these intentional misrepresentations adding to the conversation whatsoever?

Very young children may believe that if they don't see an object behind an obstacle then the object isn't there, but they soon learn that ~Z doesn't necessarily mean ~P.

Why would you assume I am talking about children who haven't grasped object permanence? If a child hasn't grasped object permanence then they aren't capable of abstract reasoning at all.

What you are doing here is not philosophy or critical reasoning. You've essentially argued that if four month-old infants can't grasp my argument then it isn't valid. If that were the case then no philosophical argument in the history of mankind would be valid. These sorts of "counterarguments" are sophistry in the extreme.

[RJG]

Leon and -0+, you guys are overcomplicating the matter. If you would back up and use Simple Logic, you would more easily get your answer. According to Simple Logic, the "absence-of-evidence" could NEVER be the "evidence-of-absence".

The axioms of Simple Logic:

• X=X
  X=~X is logically impossible
  X<X is logically impossible

Evidence = X
Absence-of-evidence = ~X
Note: it does not matter what you have evidence "-of-". Evidence is still evidence [X=X] regardless.

"Absence-of-evidence" [~X] = "evidence-of-absence" [X] is logically impossible.
~X=X is logically impossible.

Therefore, the "absence-of-evidence" could NEVER be the "evidence-of-absence".

****
To put it simply --- Either you have evidence or you don't. And if you don't, then you don't!

****
Also, the absence-of-evidence A, can NEVER logically be the evidence-of-the-absence of B. The absence of any evidence, is still no-evidence! [~X=~X].

[AgentSmith]

E = Absence of evidence is not evidence of absence.

If E were true, there must exist something, say D, that exists but there's no evidence for D: D exists & There's no evidence for D.

We'll focus on the 2^nd conjunct (There's no evidence for D). Possibilities:

1. There's evidence for D, it's just that we haven't found it (E is true).

2. There's no evidence for D at all. It's not that there's evidence which we haven't been able to trace; au contraire, no such evidence exists. Even after a thorough search of all possible evidence, none support the existence of D (E is false).

[RJG]

E = Absence of evidence is not evidence of absence.

If E were true, there must exist something, say D, that exists but there's no evidence for D: D exists & There's no evidence for D.

Why must D exist? - if you don't have evidence, then you don't have evidence. Period. This does not logically imply that (some other) evidence must exist.

If you don't have evidence that I robbed the bank, then this does not logically mean that there is "some other" evidence that links me to robbing the bank that I didn't rob.

*******
The simple solution is:

P1. "Absence of evidence is evidence of absence" translates into "~X=X".

P2. And if we know the axioms of Simple Logic, then we know this [~X=X] is logically impossible.

C1. Therefore, the "absence-of-evidence" could NEVER be the "evidence-of-absence" [~X could NEVER be X].

[AgentSmith]

E = Absence of evidence is not evidence of absence.

If E were true, there must exist something, say D, that exists but there's no evidence for D: D exists & There's no evidence for D.

Why must D exist? - if you don't have evidence, then you don't have evidence. Period. This does not logically imply that (some other) evidence must exist.

If you don't have evidence that I robbed the bank, then this does not logically mean that there is "some other" evidence that links me to robbing the bank that I didn't rob.

*******
The simple solution is:

P1. "Absence of evidence is evidence of absence" translates into "~X=X".

P2. And if we know the axioms of Simple Logic, then we know this [~X=X] is logically impossible.

C1. Therefore, the "absence-of-evidence" could NEVER be the "evidence-of-absence" [~X could NEVER be X].

If you don't have evidence, then

1. There's evidence, it's just that you haven't found it (the absence of evidence is not evidcence of absence is true).

OR

2. There's no evidence at all (the absence of evidence is not evidence of absence is false).

[RJG]

If you don't have evidence, then

1. There's evidence, it's just that you haven't found it (the absence of evidence is not evidcence of absence is true).

OR

2. There's no evidence at all (the absence of evidence is not evidence of absence is false).

Agreed. If 'we' don't have evidence, it does not necessarily mean that there is no evidence elsewhere.

BUT, the meaning of "absence-of-evidence" does not logically imply that there might be evidence elsewhere. The "absence-of-evidence" means "no evidence", translating to ~X. And the phrase "Absence of evidence is evidence of absence" translates into "~X=X", which is a clear and obvious logical impossibility.

Therefore, the "absence-of-evidence" could NEVER be the "evidence-of-absence" [~X could NEVER be X].

[AgentSmith]

If you don't have evidence, then

1. There's evidence, it's just that you haven't found it (the absence of evidence is not evidcence of absence is true).

OR

2. There's no evidence at all (the absence of evidence is not evidence of absence is false).

Agreed. If 'we' don't have evidence, it does not necessarily mean that there is no evidence elsewhere.

BUT, the meaning of "absence-of-evidence" does not logically imply that there might be evidence elsewhere. The "absence-of-evidence" means "no evidence", translating to ~X. And the phrase "Absence of evidence is evidence of absence" translates into "~X=X", which is a clear and obvious logical impossibility.

Therefore, the "absence-of-evidence" could NEVER be the "evidence-of-absence" [~X could NEVER be X].

I've been mulling over this off and on for quite some time now.

If there's no evidence at all[/i] for a claim P, does that mean ~P e.g. suppose you've searched high and low (the entire universe) for God and you come up empty-handed, does it mean that God doesn't exist (in the universe)? Absence of evidence, in this case, is evidence of absence. God is nowhere to be found; how then can God exist?

[RJG]

If there's no evidence at all for a claim P, does that mean ~P…

No, it doesn't mean "~P". It means "~E"; i.e. "No Evidence" (of claim P).

...e.g. suppose you've searched high and low (the entire universe) for God and you come up empty-handed, does it mean that God doesn't exist (in the universe)?

No, all this means is that we have no evidence of God's existence. This does not mean that "God does not exist".

Absence of evidence, in this case, is evidence of absence. God is nowhere to be found; how then can God exist?

The "Absence-of-Evidence" means Evidence does not exist". It does not mean that God does not exist.

For example, if a pea is hiding under a shell and we have no evidence of it (i.e. we can't see, hear, smell, or detect it any way), this does not mean that the "pea does not exist". It only means that we have no evidence (one way or the other). The pea might exist under the shell, or it might not.