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I have been bored tonight, so here is an explanation of a quite surprising fact: there exists one, and only one, contradictory hypothesis.

Entailment

The turnstile symbol '⊢' represents entailment. If we assume that A is a set of statements and B is a set of statements, then '⊢' can be defined in the following way.

1. If 'A ⊢ B', then there is no intepretation of A, where A is true and B is false.

The word 'interpretation' here means any assignment of truth-values to every statement in some set A. In other words, 'A ⊢ B' means that if A is true then B is true, or B is a logical consequence of A, or B is a valid derivation from A, where 'valid' means truth-preserving. Here is another use of the turnstile.

2. If '⊢ B', then there is no interpretation where B is false.

To say '⊢ B' is simply to say that B is a tautology i.e. B exemplifies a logical form that cannot be false. The definition is precisely the same as before, but we just delete the reference to A, since here the set of premises is empty. The turnstile can be used again to express the following.

3. If 'A ⊢', then there is no intepretation where A is true.

Here again we leave one side of the entailment empty, and this time we express a contradiction. That is, 'A ⊢' says that there is no intepretation of every statement in A, where A is true. In other words, A is an inconsistent set of premises.

Logical Content

The concept of logical content can be defined in the following way.

4. The logical content for a set of statements A is the set which contains every logical consequence of A.

The logical content of any set of statements is infinite. For example, if '⊢ B' then any tautology can be derived from the empty set of premises. The empty set is a member of every set, therefore anything which is a consequence of the empty set of premises is a consequence of every set of premises. In short, the set of every tautology, of which there are infinitely many, is a subset of every set of premises.

5. If 'A ⊢ B', then the logical content of B is a subset of the logical content of A.

If 'A ⊢ B', then there is no interpretation of A, where A is true and B is false. That is, if A is true then B must also be true. It should be clear that the logical content of A i.e. the set of every logical consequence of A, has B as one of its members, otherwise A would not entail B. However, if B is a member of the logical content of A, then every member of the logical content of B is also a member of A. Therefore, the logical content of B is a subset of the logical content of A.

6. If 'A ⊢ B' and 'B ⊢ A' then A and B have the same logical content.

Following on from the above. It should be clear that there is only one subset of the logical content of A which entails A, namely, the entire logical content of A. Therefore, if 'A ⊢ B' and 'B ⊢ A', then A is equal to B.

Hypotheses

I wish to define the notion of a hypothesis in the following way.

7. If H is a hypothesis then H is a set of statements.
8. If 'H ⊢ B', then B is a member of the logical content of hypothesis H.
9. If hypothesis H1 and hypothesis H2 have the same logical content, then H1 and H2 are the same hypothesis.

From a contradiction anything follows. This can be seen from the definition of entailment provided earlier.

10. If 'A ⊢' then 'A ⊢ B'.

That is, if there is no interpretation in which A is true, then there is surely no interpretation in which A is true and B is false. Therefore, B is a consequence of A where A is an inconsistent set of premises, whatever the logical content of B. This same point can be expressed in a formal proof, raplacing the metalogical variables of A, B, C, ... with the sentential variables of P, Q, R, ...

P → ~P, P ⊢ Q

Premise-----1. P → ~P
Premise-----2. P
Assumption--3. ~Q
MP (1, 2)----4. ~P
I& (2, 3)----5. P & ~Q
&E (2, 3)----6. P
AD (3, 6)----7. ~Q → P
MT (4, 7)----8. ~~Q
DNE (7)-----9. Q

The consequence of this argument is that anything, any statement whatever, is entailed by a contradiction.

It follows from the definition of a 'hypothesis', and the fact that anything follows from a contradiction, that there is one, and only one, contradictory hypothesis. If anything follows from a contradiction, then the logical content of an inconsistent set of premises contains every other hypothesis as a subset.

Let us suppose that there exist two contradictory hypothesis, H1 and H2. If they are indeed different hypotheses then their logical content is not equal, which means that H1 or H2 must contain a member which the other does not. If we suppose that H2 contains a member B which is not a member of H1 then we run into a problem, since we can immediately show that B does indeed follow from H1 using the proof above, since anything does because it is a contradiction.

Therefore, the logical contents of H1 and H2 are equal--they are the same hypothesis.

2. Perhaps I am thrown of my some conventions here, but it seems to me that.

Although, you can define "having the same logical content" as "being equal," but this is not the same as set "equality" in the zermelo-frankel (nor Von neumann) sense of set equality.

I am claiming:
'A ⊢ B' and 'B ⊢ A' while there exists elements of B that are not in A.

Suppose A={P,Q}, and B={P,Q,R} with Q→R.
Then 'A ⊢ B' and 'B ⊢ A', but it is not true that B=A.
IOW, The logical content of both A and B is {P,Q,R}, but it is not true that B=A.

Your use of "member" is confusing to me.
It seems like sometimes you mean "subset" instead of member.

Otherwise, it is fascinating to think about.

3. I am, perhaps, using the terminology somewhat loosely, particularly with regard to set theory, which I have not spent much time studying. In fact, I think you are right about my use of the term 'member', but it is a harmless ambiguity in this context.

If A={P, Q} and B={P, Q, R}, then there is an element of B which is not an element of A, and no element of A which is not an element of B. Therefore, B entails A but A does not entail B. However, I am confused as to why you wrote 'with Q → R', because if 'Q → R' is a premise of A or B then it should be included in the set. (though 'Q → R' is a consequence of B).

If 'A ⊢ B' is true, then the logical content of B is a subset of A. If B is a proper subset of A then 'B ⊢ A' is false, and if B is not a proper subset of A then 'B ⊢ A' is true. In other words, if B is not a proper subset of A and yet A entails B, then to write 'B ⊢ A' is equivalent to writing 'A ⊢ A', which is obviously true. [edit: In other words, B is either equal to A or a proper subset of A, and since a proper subset does not imply its set, so to speak, then 'A ⊢ B' and 'B ⊢ A', if and only if, A = B.]

(I may be using the term 'proper subset' incorrectly, tell me if I am).

4. The Cliff's Notes version:

1. The logical content of a hypothesis h is the set of every logical consequence of h.
2. If hypothesis h1 and h2 have the same logical content then h1 and h2 are the same hypothesis.
3. If a hypothesis h is contradictory then any consequence is a consequence of h.
4. If any consequence is a consequence of a contradictory hypothesis h, then its logical content contains every consequence.
5. If the logical content of h contains every consequence, then no hypothesis has a consequence which is not a consequence of h.
6. Therefore, there is one, and only one, contradictory hypothesis, because no contradictory hypothesis can have a consequence which another does not.

5. I should have suscribed to the thread. It got lost in all the new posts.

I disagree with point 2.

I think you are conflating the implication of statements with membership in a set.

I have yet to think of an interpretation that gives meaning to what you state (though that does not mean one doesn't exist)

I think you are trying to say that all sets of statements with at least one false statement have the same logical content (the set of all statements).

Unfortunately, I don't think there is a way to assign consistent semantics to "All statements are in the logical content of a set containing a false statement", without it being vacuously ture.

Consider the following statement, P:
P is not a logical consequence of {P}.

6. I am still not clear on your objection.

The logical content of the set {P, Q, Q → R} is the set of every conclusion which can be validly derived from {P, Q, Q → R}. For example,

P, Q, Q → R ⊢ P
P, Q, Q → R ⊢ Q
P, Q, Q → R ⊢ P & Q
P, Q, Q → R ⊢ R
P, Q, Q → R ⊢ Q v S
P, Q, Q → R ⊢ ~~R
P, Q, Q → R ⊢ P → P
...

Therefore, the logical content of {P, Q, Q → R} is the set {P, Q, P & Q, R, Q v S, ~~R, P → P,...}.

The 'logical content' for some set of statements X, is the set of every logical consequence of X, which we can represent by the notation C(X). The following rules hold for every valid argument (I am using the backwards turnstile '⊣' to mean 'does not entail').

1. If A ⊢ B then C(B) ⊆ C(A)
2. If A ⊢ B and B ⊢ A then C(B) = C(A)
3. If A ⊢ B and B ⊣ A then C(B) ⊂ C(A)

7. I add the following that the set {P & ~P} has a logical content which includes every statement, because anything follows from a contradition.

P & ~P ⊢ Q
P & ~P ⊢ Q → S
P & ~P ⊢ (P & ~Q) → (S & ~S)
P & ~P ⊢ ~~S
P & ~P ⊢ ~Q
P & ~P ⊢ (S → Q) → R
P & ~P ⊢ R v S
...

If A is the set {P & ~P} then C(X) ⊆ C(A), for any set X.

8. Originally Posted by ygolo
I think you are trying to say that all sets of statements with at least one false statement have the same logical content (the set of all statements).
You guys are, like, forcing me to think and stuff.

Not just a false statement, a contradiction.

If a set of statements includes a contradiction, and the members of that set are taken to be the axioms of a logical language, L, and both modus ponens and universal substitution are taken as primitive transformation rules of L, then L will contain all statements as theorems.

(You could use this language to derive A, where A has n operators, and you could use it to derive A with n+1 operators, i.e. ~A, AvA, A&A etc.)

Wait... You'd need to include certain PC valid theorems as axioms.

sigh...

9. Originally Posted by Owl
Wait... You'd need to include certain PC valid theorems as axioms.
Which is just to say that if you accept the laws of identity, non-contradiction, and excluded middle as necessary/axiomatic, (i.e. derivable from the empty set), then there is only one contradictory hypothesis.

10. Originally Posted by nocturne
I add the following that the set {P & ~P} has a logical content which includes every statement, because anything follows from a contradition.

P & ~P ⊢ Q
P & ~P ⊢ Q → S
P & ~P ⊢ (P & ~Q) → (S & ~S)
P & ~P ⊢ ~~S
P & ~P ⊢ ~Q
P & ~P ⊢ (S → Q) → R
P & ~P ⊢ R v S
...

If A is the set {P & ~P} then C(X) ⊆ C(A), for any set X.
Yes. I understand this much, and agree.

And I see you objection to my objection. You are not talking about just any sets, but any sets taken to be a complete set of premises. This does get tricky.

Now I'll try and parse Owl's statements.

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