In my proof the rules I used were modus ponens (for step 3) and reductio ad absurdum (for step 4). — Pfhorrest
In my proof the rules I used were modus ponens (for step 3) and reductio ad absurdum (for step 4). — Pfhorrest
1. How do I use natural deduction to show that a contradiction (p & ~p) is entailed by (~a) = a? — TheMadFool
1. ~(a = a) [Hypothesis]
2. | a = a [=Introduction] // Law of identity
3. | (a = a)∨(P∧¬P) [2, ∨Introduction] // Addition
4. | P∧~P [1, 3, ∨Elimination] // Disjunctive Syllogism
5. ∴ ~(a = a) ⊢ ⊥ [1 - 4, Deduction]
2. Is (~a) = a already a contradiction? How? — TheMadFool
3. How does (~a) = a differ from rejecting the law of identity like so: ~(a = a)? — TheMadFool
Well, (~a) = a is not a well formed formula because the negation operator has to be used before formulas, and 'a' is an individual, so you cannot use '~a' — Nicholas Ferreira
"The state of being unconditional is itself a condition". How would you translate this statement? — TheMadFool
Hmm, it's a hard sentence to translate... I thought that using definite descriptions we could have something like K((ιx)(Sx∧¬Cx)), with K(x) standing for the predicate "x is a condition" and (ιx)(Sx∧¬Cx) for the definite description "the thing x that is a state and is not conditional".
But then I realized that this formalization is wrong, because in the phrase "the state of being unconditional", "being unconditional" is not a property of this state itself, but rather of some other thing. It seems that this statement assigns a second order property, namely the property of being a condition, to the property of being unconditional, which doesn't seems to be contradictory, since it operates in diferent hierarchy levels.
I thought then that it may be simpler than that. Using the function f(x) for "the state x of being unconditional" and C(x) for "x is a condition", we can get ∀x(C(f(x))), which states that anything that is the state of being unconditional is a condition. But this formalization shows again that it is not a contradiction, and it doesn't entails one, since being unconditional isn't the negation of being a condition.
If you had the statement "The state of being unconditional is itself conditional", then it would be K((ιx)(Sx∧¬Kx)), which is obviously contradictory, since it is equivalent to Ǝx(Sx∧¬Kx∧Kx). But in the way you stated it, I think it isn't contradictory at all. — Nicholas Ferreira
You don't see the statement non-condition is a condition as contradictory? — TheMadFool
I think set theory can help us. The set of non-conditions contains no conditions. The set of conditions contains non-condition. The set of non-conditions must be empty i.e. it's the null set. The null set is a subset of every set, am I right? — TheMadFool
you are obviously more concerned with the logic than with your premise. But I do disagree with the latter.
I have loved unconditionally and have done so without me or anyone else requiring me to do so.
It just turned out that way. — Arne
Hmm, think I got it. So you are talking about a necessary condition, right? If P is a necessary condition to Q, then if you have Q, you also have P, because it is necessary for Q. And vice-versa. So P is a necessary condition for Q if and only if Q⊃P.
Now, if we say that X is unconditional, then there is no Y such that X⊃Y, because, if there was, then Y would be a condition to X. Hence any proposition which is unconditional must not imply any other proposition. But this cannot be true, because for any proposition P, if it is true, then it implies every tautology, and if it is false, it implies anything. Also, for any proposition P, P always implies P. So for any proposition P, there is a proposition Q such that P⊃Q. So there can be no such thing as a proposition that is unconditional (in the sense treated here).
It's worth remembering that I'm using the formal definition of the implication. Of course, in common sense language, it is weird to say that a false proposition implies any proposition, and that any true proposition is implied by any proposition. So I don't know how useful this reasoning will be in your discussion. — Nicholas Ferreira
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