# POLL 4 on a very simple argument especially designed for Sarkus

Discussion in 'General Philosophy' started by Speakpigeon, Jan 26, 2019.

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## Is the argument valid?

Poll closed Feb 25, 2019.

16.7%

50.0%

0 vote(s)
0.0%

33.3%
1. ### SpeakpigeonValued Senior Member

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I think I already replied to that.
"May be p" just means it is possible that p, or, we don't know that not p.
So, if we don't know that not p, it is possible that p, and so it is true that may be p, and therefore "may be p" is true.
Or else, if we do know that not p, it is not possible that p, and therefore it is false that may be p, and thus "may be p" is false.
Thus, "may be p" is either true or false.
As we should all know.
EB

3. ### arfa branecall me arfValued Senior Member

Messages:
7,832
So like I say, this represents a third logical value which is neither "true", nor "false", but "may be true or false"?

Shit, I'm not going to have to pull some more teeth am I?
Where are those electrician's pliers?

5. ### iceauraValued Senior Member

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30,994
They do not work for modal logic. The OP provider insists he is working in modal logic, and the attempts to axiomatize three valued modal logic are routinely described as "unsuccessful" and "failed" and the like.
https://en.wikipedia.org/wiki/Three-valued_logic
(Language that rings a bell here, no?).

If we reject the modal logic translation of this muddle, in order to go with three valued logic, we end up as before - with true premises
but a false conclusion.

7. ### DaveC426913Valued Senior Member

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18,897
Why don't you make your best attempt to show that truth table as it applies ere?

8. ### arfa branecall me arfValued Senior Member

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I'll just quote a reference instead:
"extended truth functions" here means what Speakpigeon has done with T/F.
Or, "it may be true" that Speakpigeon has extended 2-valued to 3-valued logic.
On the other hand . . .

And it means trawling through 19,683 truth functions to see if he has.

Last edited: Jan 30, 2019
9. ### SpeakpigeonValued Senior Member

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1,123
"May be p" is either true or false. Two values. Obviously.
And it is trivially true, for all p which are either true or false, that "p may be true or false".
Since "May be true or false" is true for all p, it is true and therefore not a third logical value as you claim.
You don't have teeth. Nothing you could use those pliers on.
EB

10. ### SpeakpigeonValued Senior Member

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1,123
Consider the possibility your assumption is false to begin with.
Just to save you time.
EB

11. ### iceauraValued Senior Member

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30,994
What assumption?
That your argument (as posted and as described by you, "modal") is invalid, because it presents us with true premises leading to a false conclusion, was a deduction.

12. ### arfa branecall me arfValued Senior Member

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7,832
That's some circular reasoning. If say, p is less than or equal to q, then p is either less than q, or p is equal to q. Not both.
Similarly, if p is true or false, then p is either true, or p is false. Not both.
However, it still is true that p is true, or p is false.

But the logic does not say p is true; the logic does not say p is false. Therefore the truth value of p is unknown, even though the domain of p is known. What you need is a function that decides which value p has.

But you do.

Last edited: Jan 30, 2019
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13. ### SpeakpigeonValued Senior Member

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Statements like "May be x is y" are standard form in modal logic and taken to be true or false.
The point of my thread is validity. That we disagree about that doesn't change the fact that modal logic takes the kind of may-statements I used in my arguments to have straightforwardly a truth value which is either "true" or "false".
EB

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1,123
EB

15. ### arfa branecall me arfValued Senior Member

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If that's the case, then prove it.
Stop asking everyone here to assume you've got it right. I simply can't do that, because I'm convinced you haven't described a modal logic, but a 3-valued logic.

Like in QM, you can say three things about the spin of a particle: "the spin is up", "the spin is down", "the spin is either up or down".
But the first two are true only when spin is measured.

16. ### arfa branecall me arfValued Senior Member

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7,832
But "may be true or false" does not map a variable to true or to false, it only says it's "maybe" possible to determine one or the other--in QM this mapping is measurement. In logic it's a function, and you're describing a "modal function" that doesn't do this.

FFS

17. ### arfa branecall me arfValued Senior Member

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7,832
In 2-valued T/F logic, every variable is "either true or false", but not "maybe true or false". The first option says a variable must be one of {T,F}.
Your "modal logic" has three states for a variable, {T, F, T/F} as you've defined it here.

You can't have it both ways, either every variable is one of {T,F} in binary logic (at some given time), or if you include "maybe", then every variable is one of {T,F,T/F}, and not binary.

Again, binary logic has two states for a variable (at any given time). Giving each variable the possibility to be either (at any given time) means each variable has three states, and you're skating right past the fact that a logic maps variables to states.
So if you're saying "maybe" T or "maybe" F maps each variable to one of {T,F}, then you're defining binary logic. You haven't defined how the third state is mapped, or maybe you just assume it happens somehow (it doesn't, you need to define it).

Last edited: Jan 30, 2019
18. ### iceauraValued Senior Member

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30,994
Nobody else knows what you were trying to talk about.
Your argument here yields false conclusions from true premises, whenever x and y are in mutually exclusive parts of B. That settles the "validity" question.
Next - - -

19. ### arfa branecall me arfValued Senior Member

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Speakpigeon's "argument" is exactly like this:

An electronics student is given an assignment, which involves the student checking that the binary logic gates in the circuit he has to build, are all working properly.
So the student (who appears somewhat bird-like), thinks, "Ok, say I think I have an OR gate, I've connected it to two inputs each of which may be T or F. So the output may be T or F!
And my work here is done."

Then the tutor walks in and asks the student if all the logic gates have been tested, and the student says "Well, they may have been" . . .

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20. ### SarkusHippomonstrosesquippedalo phobeValued Senior Member

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10,313
To me it seems more like a student defending their dissertation. He enters stage left, brimming with a cocksure attitude. Unfortunately there is little to no working in his thesis, no proofs, and he has no answer to the simplest of questioning. Student exits stage right.
"How did you do?" asks a friend.
"I may have aced it!"

21. ### SpeakpigeonValued Senior Member

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1,123
In the following syllogism, it is the statements P1, P2 and C which are either true or false:

P1 - x may be in B;
P2 - y is in B;
C - Therefore, x may be y.

P1 reads as: Possibly¨, it is true that x is in B.
This reading is equivalent to: It is true that x may be in B.
As such and by definition of "may", P1 will be true if we don't know that x is not in B. That's what "x may be in B" means.
P1 will be false if we know that x is not in B.
So, the statement "x may be in B" is either true or false since we either know or we don't that x is not in B.
EB

22. ### SpeakpigeonValued Senior Member

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1,123
Lots of people do, you don't.
Wrong assumption.
EB

23. ### arfa branecall me arfValued Senior Member

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This isn't an example of the "maybe" logic you've been selling us. This is formal , it says x can be a member of the set B, then if y is a member of B, then x can be y. "Can be" is a relation between x and y. Maybe it's transitive? Is it reflexive? is it symmetric?

If the answer to any of those three questions is not yes or no, it's not a relation. (defined on sets)
Note y is fixed by being in B, and this isn't a minor detail.