Leontiskos wrote: ↑August 10th, 2021, 12:26 pm
-0+ wrote: ↑August 10th, 2021, 8:37 amLeontiskos wrote: ↑August 6th, 2021, 11:33 am
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)?)
Leontiskos wrote: ↑August 10th, 2021, 12:26 pm
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?