Problem with categorical syllogisms

  • #1
honestrosewater
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Eh, if this question seems odd, it's tied in with another thread...
D1. Some thoughts and memories are observables.
D2. If thoughts and memories are observables, then the evidence for them is observable.
D3. Phenomenal contents (i.e. qualia) provide evidence for observable kinds of thoughts and memories.
D4. Therefore, qualia are observable.
I want to know if it follows from the above that "All observables are qualia" is false. If not, I want to know what additional info I need.

Certainly some thoughts and memories are not qualia, otherwise his argument is circular (assume he knows better), yes? IOW, if his argument is not circular, then "All thoughts and memories are qualia" is false. If "All thoughts and memories are qualia" is false, then "Some thoughts and memories are not qualia" is true, yes (A and O statements are contradictory)? If he uses qualia only as observed (I'm told he does use qualia this way), I can interpret "Qualia are observables" as "All qualia are observables", yes? This may not actually follow from the above- so I'm adding it on. :approve:
So the true premises I have are
1. Some T(houghts and memories) are not Q(ualia). (an O statement)
2. Some T are O(bservables). (I)
3. All Q are O. (A)
"Some T are O" and "Some O are T" are equivalent (they have equivalent Venn diagrams), yes? So I also have
4. Some O are T. (I)
From 1, 2, 3, or 4, I want to conclude
5. Some O are not Q. (O)
Running through the valid syllogism forms, I don't see how I can conclude 5 from 1, 2, 3, or 4 without
6a. No Q are T. (from EIO-2, EIO-4)
6b. No T are Q. (EIO-1, EIO-3)
6c. All Q are T. (AAO-2)
6d. All T are O. (OAO-3)
6e. Some O are not T. (AAO-2)
And I don't see any way to get any 6 from 1, 2, 3, or 4. All valid forms with E conclusions (6a and 6b) have an E premise. All valid forms with A conclusions (6c and 6d) have two A premises. I only have an O, an A, and two Is. 6e requires additional premises
6f. Some Q are not T. (OAO-3)
6g. Some O are not Q. (AOO-2)
6f requires additonal premises
6h. Some Q are not O. (AOO-2)
6i. All T are Q. (AOO-2)
6j. Some O are not T. (OAO-3)
6h is false (by 3). 6i is false (by 1). 6j is 6e, which I cannot conclude from 1, 2, 3, or 4. 6g is 5, which I cannot conclude from 1, 2, 3, or 4.
Did I miss something? (I hope I don't need to take a longer route.) Could I get 5 from a "more powerful" system (than categorical syllogisms)? It seems categorical syllogisms should be able to handle it (but what do I know). I will look again but would appreciate some help; I'd especially like to know if what I've done so far is right- I've never really worked with syllogisms before. :blushing:
 
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Answers and Replies

  • #2
honestrosewater
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Pretty please. I know some of you know syllogisms. [cough]Mattson[/cough]
Do I have to try to prove my method works and ask someone to check my proof?
 
  • #3
hypnagogue
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I don't know about your syllogisms, but how about this.

1. Some T are O and not Q.
2. Therefore, not all O are Q.
 
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  • #4
AKG
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D1. Some thoughts and memories are observables.

There exists some thoughts (and memories) such that those thoughts are observable.

[tex]\exists t\,O(t)[/tex]

D2. If thoughts and memories are observables, then the evidence for them is observable.

If a thought is observable, then the evidence for that thought is observable.

[tex]O(t) \Rightarrow O(e_t)[/tex]

D3. Phenomenal contents (i.e. qualia) provide evidence for observable kinds of thoughts and memories.

If a thought is observable, the evidence for that thought is a quale.

[tex]O(t) \Rightarrow Q(e_t)[/tex]

D4. Therefore, qualia are observable.

All qualia are observable.

[tex]\forall x\,Q(x) \Rightarrow O(x)[/tex]

D4. does not follow, assuming I've interpreted the argument correctly. All we can conclude is that there are some observable thoughts, and the evidence for these observable thoughts is itself observable, and are called qualia, in other words, D4 would have to be replaced with the much weaker assertion [itex]\exists t\, (\exists x\, O(x) \wedge Q(x))[/itex]. His argument (if the premises are true) does show that there are qualia that are observable, but doesn't necessarily show that all qualia are observable.

"All observables are qualia" is most likely false given the argument above. The argument above asserts that some thoughts are observables. Unless thoughts are qualia, some observables are not qualia, since some of them are thoughts.

Note that "all qualia are observables" ([itex]\forall x\, Q(x) \Rightarrow O(x)[/itex]) is quite different from "all observables are qualia" ([itex]\forall x\, O(x) \Rightarrow Q(x)[/itex]).
 
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  • #5
honestrosewater
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Eh, first off, I thought of conversion, but not of the two other immediate inferences, obversion and contraposition. So things are more complicated than I had thought (I have more statements to begin with). I'm working on it- there's a lot to keep track of.
I'm actually now more interested in the general method of (or an algorithm for) working backwards from a conclusion to find all "argument paths" or whatever you'd call them. Syllogisms having been around for so long, I'd be surprised if no one has tried this before. I can work on it on my own though.
hypnagogue said:
I don't know about your syllogisms, but how about this.

1. Some T are O and not Q.
2. Therefore, not all O are Q.
Okay, you mean the same T are both O and not Q? I have to use a new term, otherwise 1 is just (Some T are O) and (Some T are not Q) which I already have. So call that subset of T, U.
All U are T.
All U are O.
No U are Q.
Clever, great. I'll try it and get back to you.
AKG said:
D3. Phenomenal contents (i.e. qualia) provide evidence for observable kinds of thoughts and memories.

If a thought is observable, the evidence for that thought is a quale.

[tex]O(t) \Rightarrow Q(e_t)[/tex]
That can't be true. Zombies have thoughts but not qualia (by definition). He must mean that qualia provide evidence for some but not all thoughts. Edit: He does say elsewhere that
[tex]O(x) \Rightarrow O(e_x)[/tex]
is true:
E1. If x has the status of being an observable, then the evidence for x must also have the status of being observable.
I don't know how to qualify this. Presumably, all observables must have evidence so he probably means
[tex]\forall x \exists e_x (O(x) \Rightarrow O(e_x))[/tex]
But it could be that some observables provide evidence for themselves, they are their own and only evidence, [itex]x \equiv e_x[/itex], and this would seem to be true of qualia. I asked this elsewhere but didn't get a reply.
Anyway, that qualia provide evidence for some thoughts (D3) is another proposition. I guess it is only true a posteriori.
D4. Therefore, qualia are observable.

All qualia are observable.

[tex]\forall x\,Q(x) \Rightarrow O(x)[/tex]

D4. does not follow, assuming I've interpreted the argument correctly. All we can conclude is that there are some observable thoughts, and the evidence for these observable thoughts is itself observable, and are called qualia, in other words, D4 would have to be replaced with the much weaker assertion [itex]\exists t\, (\exists x\, O(x) \wedge Q(x))[/itex]. His argument (if the premises are true) does show that there are qualia that are observable, but doesn't necessarily show that all qualia are observable.
But by the definition I think he is using, all qualia are observable. That's why I added it in.
"All observables are qualia" is most likely false given the argument above. The argument above asserts that some thoughts are observables. Unless thoughts are qualia, some observables are not qualia, since some of them are thoughts.
Do we know (are we given) that no thoughts are qualia? Not that I can see. Can we derive it from the other statements given? That's what I'm trying to figure out.
Note that "all qualia are observables" ([itex]\forall x\, Q(x) \Rightarrow O(x)[/itex]) is quite different from "all observables are qualia" ([itex]\forall x\, O(x) \Rightarrow Q(x)[/itex]).
Yes, I know.

I'll make some coffee and get to work. Hopefully I can straighten things out. Thanks for the help.
 
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  • #6
honestrosewater
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Ah, IE is the devil!!
hypnagogue,
Given:
1. All U are O.
2. No U are Q.
We can immediately infer:
3. Some U are O. (from 1, by subalternation, assuming U has at least one member)
So EIO-3 will do:
No U are Q. (2)
Some U are O. (3)
Therefore, Some O are not Q. :smile:

(Edit: I would ask what is so special about U, but :rolleyes: )
 
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  • #7
AKG
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That can't be true. Zombies have thoughts but not qualia (by definition). He must mean that qualia provide evidence for some but not all thoughts.
Yes, as I interpreted it, qualia provide evidence for observable thoughts, not necessarily all thoughts. Hence [itex]O(t) \Rightarrow Q(e_t)[/tex], and not [itex]\forall t\, Q(e_t)[/itex]*. Note that given D1 and D3, we get [itex]\exists t\, Q(e_t)[/itex], compare that to *.
But by the definition I think he is using, all qualia are observable. That's why I added it in.
If D4. is just another premise (if so, it shouldn't have started with "therefore"), then what is the argument actually trying to prove?
Do we know (are we given) that no thoughts are qualia? Not that I can see. Can we derive it from the other statements given? That's what I'm trying to figure out.
No, it doesn't give that information.
 
  • #8
honestrosewater
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AKG said:
Yes, as I interpreted it, qualia provide evidence for observable thoughts, not necessarily all thoughts. Hence [itex]O(t) \Rightarrow Q(e_t)[/tex], and not [itex]\forall t\, Q(e_t)[/itex]*. Note that given D1 and D3, we get [itex]\exists t\, Q(e_t)[/itex], compare that to *.
Zombies' thoughts are also observable. Also, other non-thought, non-quale observables exist in the same world as qualia (so even if zombies cannot exist in the same world as humans, observables that are qualia and observables that aren't qualia do exist in the same world). There is some difference between observables that are qualia and observables that are not qualia, but I beat this to death and didn't get the kind of answer or clarification I wanted. I was trying to see what premises I would need to put the two kinds of observables into the same world.
If D4. is just another premise (if so, it shouldn't have started with "therefore"), then what is the argument actually trying to prove?
D4 is the conclusion, but it doesn't say that all qualia are observables. He adds the "all" elsewhere, though he hardly needs to- I can't imagine how qualia could exist independently of our subjective experience (observation) of them. Anyway, the argument is meant to prove that humans have epistemic access to qualia. Apparently, some people try to deny that qualia are observables (that we have epistemic access to qualia). He argues that if you deny that qualia are observables, you must also deny that thoughts and memories are observables.
To avoid circularity, some thoughts and memories must not be qualia.
...
That is how I arrived at my three statements:
1. Some T are not Q.
2. Some T are O.
3. All Q are O.
 
  • #9
honestrosewater:

So the true premises I have are
1. Some T(houghts and memories) are not Q(ualia). (an O statement)
2. Some T are O(bservables). (I)
3. All Q are O. (A)
"Some T are O" and "Some O are T" are equivalent (they have equivalent Venn diagrams), yes? So I also have
4. Some O are T. (I)
From 1, 2, 3, or 4, I want to conclude
5. Some O are not Q. (O)


This argument fails in FOPL.

1. (Some T are not Q) = Ex(Tx & ~Qx)
2. (Some T are O) = Ex(Tx & Ox)
3. (All Q are O) = Ax(Qx -> Ox)
4. (Some Q are not O) = Ex(Qx and ~Ox).

Ex(Tx & ~Qx) & Ex(Tx & Ox) & Ax(Qx -> Ox) -> Ex(Qx and ~Ox), is not valid.

Do Venn diagrams show that it is true?
 
  • #10
honestrosewater
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Owen Holden said:
honestrosewater:

So the true premises I have are
1. Some T(houghts and memories) are not Q(ualia). (an O statement)
2. Some T are O(bservables). (I)
3. All Q are O. (A)
"Some T are O" and "Some O are T" are equivalent (they have equivalent Venn diagrams), yes? So I also have
4. Some O are T. (I)
From 1, 2, 3, or 4, I want to conclude
5. Some O are not Q. (O)


This argument fails in FOPL.

1. (Some T are not Q) = Ex(Tx & ~Qx)
2. (Some T are O) = Ex(Tx & Ox)
3. (All Q are O) = Ax(Qx -> Ox)
4. (Some Q are not O) = Ex(Qx and ~Ox).

Ex(Tx & ~Qx) & Ex(Tx & Ox) & Ax(Qx -> Ox) -> Ex(Qx and ~Ox), is not valid.
Yes, 3 and 4 are contradictory. But the desired conclusion is Ex(Ox & ~Qx). I'm iffy on the restrictions on instantiation- my book barely covered it at all, and I'm still waiting for another book to arrive. If I can eliminate the quantifiers properly, I should have no problem with the rest.
Note: "a" and "b" denote individual constants; "y" denotes an arbitrary constant.

1. Ex(Tx & ~Qx)
2. Ex(Tx & Ox)
3. Ax(Qx -> Ox) /.: Ex(Ox & ~Qx)
4. Ta & ~Qa [1, EI]
5. Tb & Ob [2, EI]
Correct?
I know I can infer
6. Qy -> Oy [3, UI]
but I don't know if or how that will help. But I can infer
6. Qa -> Oa [3, UI]
7. Qb -> Ob [3, UI]
Correct?
Okay, skipping ahead a bit, I checked
1. Ta & ~Qa
2. Tb & Ob
3. Qa -> Oa
4. Qb -> Ob /.: (Oa & ~Qa) v (Ob & ~Qb)
for invalidity. I went out on a limb (I've never actually been told how to do this) and chose [(Oa & ~Qa) v (Ob & ~Qb)] as my conclusion since either of those will give me Ex(Ox & ~Qx) by UG. This argument is invalid by truth-values:
Ta - T
Tb - T
Ob - T
Oa - F
Qa - F
~Qa - T
Qb - T
~Qb - F
I guess it's easier to see this way:
1. Ta(T) & ~Qa(T)
2. Tb(T) & Ob(T)
3. Qa(F) -> Oa(F)
4. Qb(T) -> Ob(T) /.: (Oa(F) & ~Qa(T)) v (Ob(T) & ~Qb(F))

Edit: Okay, I found something that tells me how to check arguments with quantified propositions for invalidity. It'll have to wait- I'm way too tired now.
 
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  • #11
honestrosewater:
So the true premises I have are
1. Some T(houghts and memories) are not Q(ualia). (an O statement)
2. Some T are O(bservables). (I)
3. All Q are O. (A)
"Some T are O" and "Some O are T" are equivalent (they have equivalent Venn diagrams), yes? So I also have
4. Some O are T. (I)
From 1, 2, 3, or 4, I want to conclude
5. Some O are not Q. (O)


Owen:
This argument fails in FOPL.
1. (Some T are not Q) = Ex(Tx & ~Qx)
2. (Some T are O) = Ex(Tx & Ox)
3. (All Q are O) = Ax(Qx -> Ox)
4. (Some Q are not O) = Ex(Qx and ~Ox).
Ex(Tx & ~Qx) & Ex(Tx & Ox) & Ax(Qx -> Ox) -> Ex(Qx and ~Ox), is not valid.

honestrosewater:
Yes, 3 and 4 are contradictory. But the desired conclusion is Ex(Ox & ~Qx).

My error. I copied your (5. Some O are not Q) incorrectly.

But, Ex(Tx & ~Qx) & Ex(Tx & Ox) & Ax(Qx -> Ox) -> Ex(Ox and ~Qx), is not valid either.

honestrosewater:
If I can eliminate the quantifiers properly, I should have no problem with the rest.

I don't know what you mean by eliminating the quantifiers here.

We can eliminate the quantifiers by expanding each proposition,
with a maximum of 2^(number of terms) individuals, i.e. 8 individuals and 24 distinct 2-valued propositions.
Truth tables can then be applied to decide your problem.

This tedious but effective procedure resolves all monadic formulas of FOPL.
 
  • #12
honestrosewater
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Owen said:
honestrosewater said:
If I can eliminate the quantifiers properly, I should have no problem with the rest.
I don't know what you mean by eliminating the quantifiers here.
Using Universal Instantiation and Existential Instantiation (aka Elimintation) to change the quantified propositions into compound propositions or -sorry, tired, whatever propositions like Pa, where "a" is a constant, are called. For instance, EI allows us to infer [Pa -> Qa] from [Ex(Px -> Qx)].

We can eliminate the quantifiers by expanding each proposition,
with a maximum of 2^(number of terms) individuals, i.e. 8 individuals and 24 distinct 2-valued propositions.
Yeah, I just found this or something like it. I'll have to read it later. I'm in no shape to think right now.
Truth tables can then be applied to decide your problem.

This tedious but effective procedure resolves all monadic formulas of FOPL.
I think truth trees can be used on arguments with quantified propositions. I know truth trees are effective for arguments with propositional statements. I haven't gotten to do much with quantifiers yet- having book problems. Sorry, I hope that made sense.

Oh, just out of curiosity, what proof method(s) do you use? Truth tables? I'm learning all the other ones I know of: natural deduction (not effective for constructing proofs but more fun), truth trees (effective- kind of like a selective truth table- the only challenge is making them as short or simple as possible), tableaux (don't know yet). Do you use any of these? Everyone seems to have a favorite.
 
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  • #13
honestrosewater said:
Oh, just out of curiosity, what proof method(s) do you use? Truth tables? I'm learning all the other ones I know of: natural deduction (not effective for constructing proofs but more fun), truth trees (effective- kind of like a selective truth table- the only challenge is making them as short or simple as possible), tableaux (don't know yet). Do you use any of these? Everyone seems to have a favorite.
I have devised a truth table method of dealing with monadic predicate logic and propositional logic within the same method...


From: Owen Holden (oorionus@yahoo.com)
Subject: Truth tables for monadic predicate logic
View: Complete Thread (3 articles)
Original Format
Newsgroups: sci.logic, sci.math
Date: 2002-08-10 02:21:50 PST


Given that propositions, p, q, r..etc., have two truth
values (T,F), logical truth (tautology) is confirmed if
the propositional expression has the value T for all
possibilities. It is contradictory if it has the
value F for all possibilities and it's contingent
if it has at least one T and at least one F for all
possibilities. My intension here is to extend the
concepts of truth by calculation to monadic predicate
logic. Truth by calculation for predicate logic in
determined domains reduces to propositional logic via
expansion of the quantifiers 'All' and 'Some'.
In undetermined or infinite domains I suggest the
following method of decision. A propositional form
with one free individual variable x, e.g. 'Fx' , is
monadic. ExFx means 'for some x Fx' . AxFx means
'for all x Fx'. The usual propositional logic operators
(not '~', or 'v', and '&', implies '->', equivalence '<->')
are maintained.

Fx ~Fx ExFx Ex~Fx AxFx Ax~Fx
__ ___ ____ _____ ____ _____
1 0 T 1 F 0 T 1 F 0
2 3 T 2 T 3 F 2 F 3
3 2 T 3 T 2 F 3 F 2
0 1 F 0 T 1 F 0 T 1

The following theorems are valid (true by deduction)
and tautologous (true by calculation).

(1) AxFx->ExFx (2) ~Ex~Fx<->AxFx (3) ~Ax~Fx<->ExFx
__________ _____________ _____________
T 1 T T 1 TF 0 T T 1 TF 0 T T 1
F 2 T T 2 FT 3 T F 2 TF 3 T T 2
F 3 T T 3 FT 2 T F 3 TF 2 T T 3
F 0 T F 0 FT 1 T F 0 FT 1 T F 0

(4)(AxFx & ExFx)<->AxFx (5) AxFx<->((ExFx->AxFx)<->ExFx)

(6)(AxFx v ExFx)<->ExFx (7) ExFx<->((ExFx->AxFx)<->AxFx)

(8)(Ex~Fx->Ax~Fx)<->(ExFx->AxFx)
etc.

Table 1. (these tables include propositional logic)

| ~ | E | A (not, some, all)
______________
T | F | T | T
F | T | F | F
1 | 0 | T | T
2 | 3 | T | F
3 | 2 | T | F
0 | 1 | F | F

Table 2.

v | T F 1 2 3 0 (or)
________________
T | T T 1 1 1 1
F | T F 1 2 3 0
1 | 1 1 1 1 1 1
2 | 1 2 1 2 1 2
3 | 1 3 1 1 3 3
0 | 1 0 1 2 3 0

The other tables result from the definitions.

p->q defined ~p v q (implies)

Table 3.

-> | T F 1 2 3 0
________________
T | T F 1 2 3 0
F | T T 1 1 1 1
1 | 1 1 1 1 1 1
2 | 1 3 1 1 3 3
3 | 1 2 1 2 1 2
0 | 1 1 1 1 1 1

p & q defined ~(~p v ~q) (and)

Table 4.

& | T F 1 2 3 0
_______________
T | T F 1 2 3 0
F | F F 0 0 0 0
1 | 1 0 1 2 3 0
2 | 2 0 2 2 0 0
3 | 3 0 3 0 3 0
0 | 0 0 0 0 0 0

p<->q defined (p->q)&(q->p) (equivalence)

Table 5.

<-> | T F 1 2 3 0
_________________
T | T F 1 2 3 0
F | F T 0 3 2 1
1 | 1 0 1 2 3 0
2 | 2 3 2 1 0 3
3 | 3 2 3 0 1 2
0 | 0 1 0 3 2 1

(8) Ax:Fx->ExFx (9) Ax:AxFx->Fx (10) Ax:(Fx->AxFx) v ExFx
___________ ___________ ____________________
T 1 1 T 1 T T 1 1 1 T 1 1 T 1 1 T 1
T 2 1 T 2 T F 2 1 2 T 2 3 F 2 1 T 2
T 3 1 T 3 T F 3 1 3 T 3 2 F 3 1 T 3
T 0 1 T 0 T F 0 1 0 T 0 1 F 0 1 F 0

(11) Ax:(ExFx->AxFx)->(ExFx->Fx)
(12) Ex:(Fx->AxFx) (13) Ex:(Fx->AxFx)
(14) Ex:((ExFx->Fx)->AxFx)<->ExFx)
(15) Ax:((ExFx->Fx)->AxFx)<->AxFx)
(16) Ax(Fx<->p)->(AxFx<->p)
(17) (AxFx<->p)->Ex(Fx<->p)
(19) Ex(p)<->p (20) Ax(p)<->p

Predicate logic tautologies with pure and prenex forms.

(20) Ex~Fx<->~AxFx (21) Ax~Fx<->~ExFx

(21) Ex(Fx v p)<->(ExFx v p)
_______________________
T 1 1 T T T 1 T T
T 2 1 T T T 2 T T
T 3 1 T T T 3 T T
T 0 1 T T F 0 T T
T 1 1 F T T 1 T F
T 2 2 F T T 2 T F
T 3 3 F T T 3 T F
F 0 0 F T F 0 F F

(22) Ax(Fx v p)<->(AxFx v p)
(23) Ex(Fx & p)<->(ExFx & p)
(24) Ax(Fx & p)<->(AxFx & p)
(25) Ex(p->Fx)<->(p->ExFx)
(26) Ax(p->Fx)<->(p->AxFx)
(27) Ex(Fx->p)<->(AxFx->p)
(28) Ax(Fx->p)<->(ExFx->p)

(29) Ex(Fx<->p)<->((AxFx->p)&(p->ExFx))
__________________________________
T 1 1 T T T 1 T T T TT T 1
T 2 2 T T F 2 T T T TT T 2
T 3 3 T T F 3 T T T TT T 3
F 0 0 T T F 0 T T F TF F 0
F 1 0 F T T 1 F F F FT T 1
T 2 3 F T F 2 T F T FT T 2
T 3 2 F T F 3 T F T FT T 3
T 0 1 F T F 0 T F T FT F 0

(30) Ax(Fx<->p)<->((ExFx->p)&(p->AxFx))
__________________________________
T 1 1 T T T 1 T T T TT T 1
F 2 2 T T T 2 T T F TF F 2
F 3 3 T T T 3 T T F TF F 3
F 0 0 T T F 0 T T F TF F 0
F 1 0 F T T 1 F F F FT T 1
F 2 3 F T T 2 F F F FT F 2
F 3 2 F T T 3 F F F FT F 3
T 0 1 F T F 0 T F T FT F 0

The last two theorems,(29) and (30), do not appear in
most logic texts. I believe they should appear to
complete the pure and prenex forms.

To deal with truth functions of two or more predicates,
which includes Boolean logic and Syllogistic logic,
we need to extend the truth tables. It will then become
evident that all of the axioms of propositional logic
and all of the axioms of general predicate logic are
tautologies.

Truth tables for two monadic predicates. These tables include
the above tables.

| ~ | E | A (not) (Some) (all)
_________________
T | F | T | T
F | T | F | F
15 | 14 | T | F
13 | 12 | T | F
11 | 10 | T | F
9 | 8 | T | F
7 | 6 | T | F
5 | 4 | T | F
3 | 2 | T | F
1 | 0 | T | T
0 | 1 | F | F
2 | 3 | T | F
4 | 5 | T | F
6 | 7 | T | F
8 | 9 | T | F
10 | 11 | T | F
12 | 13 | T | F
14 | 15 | T | F

v | T F 15 13 11 9 7 5 3 1 0 2 4 6 8 10 12 14 (or)
_________________________________________________________
T | T T 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
F | T F 15 13 11 9 7 5 3 1 0 2 4 6 8 10 12 14
15 | 1 15 15 13 11 9 7 5 3 1 15 13 11 9 7 5 3 1
13 | 1 13 13 13 9 9 5 5 1 1 13 13 9 9 5 5 1 1
11 | 1 11 11 9 11 9 3 1 3 1 11 9 11 9 3 1 3 1
9 | 1 9 9 9 9 9 1 1 1 1 9 9 9 9 1 1 1 1
7 | 1 7 7 5 3 1 7 5 3 1 7 5 3 1 7 5 3 1
5 | 1 5 5 5 1 1 5 5 1 1 5 5 1 1 5 5 1 1
3 | 1 3 3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1
1 | 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0 | 1 0 15 13 11 9 7 5 3 1 0 2 4 6 8 10 12 14
2 | 1 2 13 13 9 9 5 5 1 1 2 2 6 6 10 10 14 14
4 | 1 4 11 9 11 9 3 1 3 1 4 6 4 6 12 14 12 14
6 | 1 6 9 9 9 9 1 1 1 1 6 6 6 6 14 14 14 14
8 | 1 8 7 5 3 1 7 5 3 1 8 10 12 14 8 10 12 14
10 | 1 10 5 5 1 1 5 5 1 1 10 10 14 14 10 10 14 14
12 | 1 12 3 1 3 1 3 1 3 1 12 14 12 14 12 14 12 14
14 | 1 14 1 1 1 1 1 1 1 1 14 14 14 14 14 14 14 14

(32) Ax(Fx->Gx)->(AxFx->AxGx)
(33) Ax(Fx->Gx)->(ExFx->ExGx)
(34) Ax(Fx<->Gx)->(AxFx<->AxGx)
(35) Ax(Fx<->Gx)->(ExFx<->ExGx)
(36) Ax(Fx & Gx)<->(AxFx & AxGx)
(37) Ex(Fx v Gx)<->(ExFx v ExGx)
(38) (AxFx v AxGx)->Ax(Fx v Gx)
(39) Ex(Fx & Gx)->(ExFx & ExGx)
(40) (Ax(Fx->Gx)->Ex(Fx & Gx))<->ExFx
(41) ~ExFx->Ax(Fx->Gx) (42) AxGx->Ax(Fx->Gx)

(43) (Ex(Fx & Gx) & Ex(Fx & ~Gx))->AxFx is invalid
(44) (Ex(Fx & ~Gx) & Ex(~Fx & Gx))->Ax(Fx v Gx) is invalid
(45) (Ex(Fx & Gx) & Ex(Fx & ~Gx) & Ex(~Fx & Gx))->Ax(Fx v Gx)
is invalid.
etc.

Proofs of validity and invalidity of monadic predicate logic
expressions with three predicates requires truth tables that
have 64 function values.
-----------------------


To simplify these calculations a small 10 line program of QBasic is sufficient.

PS: for a correctly aligned copy of this post go to,

http://groups.google.ca/groups?hl=e...h1.37380@news01.bloor.is.net.cable.rogers.com
 
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  • #14
honestrosewater
Gold Member
2,105
5
Owen,
Interesting. I don't really understand it all yet, but I'll play around with it eventually and let you know what I think, if you'd like. I got my books today! :biggrin: So I'll be spending the next week or so working through them.
 
  • #15
517
0
I don't think this argument can be interpreted easily in terms of categorical syllogisms.

Informally, the argument is easy to understand. Take the set of thoughts and memories that are observable, and call it set T. T is non-empty by D1. D2 means that if there exists evidence for the elements of T, it is observable. D3 says that there does exist some evidence for the elements of T. Therefore all of this evidence is observable, and also by D3 some of this evidence is qualia. So some qualia--those qualia that provide such evidence--are observable.

Symbolically,

Thoughts and memories: set M
Observables: set O
Qualia: set Q
Define a function "evidence" V(x) which maps set M union O to set D, the set of all evidence. So if x is an element of M union O, then V(x) is the evidence for x.

D1: Ex(xeO & xeM) (using e for element of)
D2: Ax((xeO & xeM) --> V(x)eO)
D3: I'm going to use some license here and say that "provides evidence" is the same as "is evidence." The argument is not valid if "provides evidence" is not the same as "is evidence." So anyway:
D3: Ex(xeQ & xeV(OnM)) (n for intersection)

Conclusion to find: Ex (xeQ & xeO)

1: yeQ & yeV(OnM) (D3, existential instantiation)
2. yeV(OnM) (1, Simp)
3. Ex(xe(OnM) & V(x)=y) (this is what line 2 means)
4. ze(OnM) & V(z)=y (3,existential instantiation)
5. zeO & zeM & V(z)=y (4, substituting zeO & zeM for ze(OnM))
6. zeO & zeM --> V(z)eO (D2, universal instantiation)
7. zeO & zeM (5, Simp)
8. V(z)eO (6, 7, MP)
9. V(z)=y (5, Simp)
10. yeO (8, 9, substitution)
11. yeQ (1, Simp)
12. yeO & yeQ (10, 11, Conj)
13. Ex(xeO & xeQ) (12, EI)

It turns out you don't even need premise D1.

D1. Some thoughts and memories are observables.
D2. If thoughts and memories are observables, then the evidence for them is observable.
D3. Phenomenal contents (i.e. qualia) provide evidence for observable kinds of thoughts and memories.
D4. Therefore, qualia are observable.


Edit: I had some minor mistakes in this proof mainly because I edited it and forgot to change all the explanations for lines.
 
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  • #16
honestrosewater
Gold Member
2,105
5
Bartholomew said:
D1: Ex(xeO & xeM) (using e for element of)
D2: Ax((xeO & xeM) --> V(x)eO)
D3: I'm going to use some license here and say that "provides evidence" is the same as "is evidence." The argument is not valid if "provides evidence" is not the same as "is evidence." So anyway:
D3: Ex(xeQ & xeV(OnM)) (n for intersection)

Conclusion to find: Ex (xeQ & xeO)
Thanks, but the conculsion to find is Ex[xeO & ~(xeQ)].
 
  • #17
517
0
The conclusion is "qualia are observable" (which is best interpreted as "some qualia are observable"). Ex(xeO & ~(xeQ)) means "some things that are observable are not qualia," which is not what you want (in fact, it's nearly the opposite).
 
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