• TheMadFool
    13.8k
    Firstly some context.

    There's a thread on the forum discussing unconditional love: the claim being that unconditional itself qualifies as a condition and so...there's no such thing as unconditional love.

    I bet this has something to do with self-reference and one particular statement seems to be part of the paradox viz. a non-condition is a condition which leads to unconditional love is conditional.

    If we take c = condition, then the statement, non-condition is a condition becomes (~c) = c. If we were to instantiate the form of (~c) = c then one would be, non-dogs are dogs. It's clearly a contradiction to say that something is not that thing but I despite many hours of trying I just couldn't put it into the familiar contradiction form (p & ~p) using natural deduction.

    Then I thought it violated the law of identity which is a = a but to deny the law of identity we need to say ~(a = a) but what I have is (~a) = a.

    My questions are as follows:

    1. How do I use natural deduction to show that a contradiction (p & ~p) is entailed by (~a) = a?

    2. Is (~a) = a already a contradiction? How?

    3. How does (~a) = a differ from rejecting the law of identity like so: ~(a = a)?

    Thanks
  • Pfhorrest
    4.6k
    1. a = ~a
    2. *a
    3. .: ~a
    4. .: a ^ ~a
  • TheMadFool
    13.8k
    Could you explain the symbols a bit. Thanks.
  • Pfhorrest
    4.6k
    Given a equals not a, if we suppose a, we can conclude not a, and therefore conclude both a (which we supposed as a premise) and not a, a contradiction.

    I’m treating equality as equivalent to if-and-only-if for these purposes.
  • TheMadFool
    13.8k
    I think I got it.

    1. John is not John = it is not the case that John is John
    j = ~j is equivalent to ~(j = j)

    Similarly...

    2. non-condition is a condition = it is not the case that condition is condition
    (~c) = c is equivalent to ~(c = c)

    It seems there's no rule as such to use here. It's simply a process of translating it in the right way.
  • Pfhorrest
    4.6k
    In my proof the rules I used were modus ponens (for step 3) and reductio ad absurdum (for step 4).
  • TheMadFool
    13.8k


    c = condition
    non-condition is condition = condition is non-condition = condition is not condition = It is not the case that condition is condition = ~(c = c)

    1. ~(c = c)....premise
    2. c = c..........I'd
    3. (c = c) & (~(c = c))...1, 2 conj (contradiction)
    4. ~~(c = c)..............1 to 3 reductio ad absurdum
    5. c = c
  • TheMadFool
    13.8k
    In my proof the rules I used were modus ponens (for step 3) and reductio ad absurdum (for step 4).Pfhorrest

    :up: :up:
  • Nicholas Ferreira
    78
    In my proof the rules I used were modus ponens (for step 3) and reductio ad absurdum (for step 4).Pfhorrest

    Since 'a' is an individual, your use of negation is not the same as the negation of natural deduction, since the latter is a truth-functional operator and individuals are not things that can be truth-evaluated. Also, modus ponens is to be used with propositions of the form "P" and "P⊃Q". You can't use modus ponens in "a" and "a=~a" because neither 'a' is a proposition nor '=' is a a truth-functional operator (and therefore it isn't the implication operator). You cannot use the conjunction to assert a∧~a for the same reason, since 'a' is not a wff (it's a term, not a formula), and you are supposed to use wff as arguments to the conjunction operator. Also, you can't use reductio ad absurdum for step 4 since step 3 doesn't express a proposition, and reductio is used only in propositions.

    1. How do I use natural deduction to show that a contradiction (p & ~p) is entailed by (~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'. But if with this you mean "~(a = a)", then it is a wff, since "a=a" is a wff. And is easy to derive p∧~p from that. Actually, you can derive anything from that, because it violates the rule of introduction of identiy. The proof goes as follows:

    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]

    Which is similar to your proof. But from this proof you can derive any contradiction P∧¬P, as you asked.

    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

    The answer is exactly as before. If by "(~a) = a" you mean "it is not the case that a equals a", i.e., ~(a = a), then it is not exactly a contradiction since contradictions have the form φ∧¬φ. But a contradiction is easily derived from this as I showed. Also, in this case, there is no difference between "(~a) = a" and "~(a = a)". But if you have another use for the negation in '~a' (which as I said cannot be the same as in P∧~P), then let us know.
  • TheMadFool
    13.8k
    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

    :chin:

    "The state of being unconditional is itself a condition". How would you translate this statement?
  • Nicholas Ferreira
    78
    "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.
  • TheMadFool
    13.8k
    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?

    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?
  • Nicholas Ferreira
    78
    You don't see the statement non-condition is a condition as contradictory?TheMadFool

    But you didn't say that non-condition is a condition. Your statement was "The state of being unconditional is itself a condition". Here, "condition" applies to "the state of being unconditional", not to "non-condition". Also, there are no problems with this. The property of being a cat is not itself a cat, and there is no contradiction here. As I said, if your statement was "The state of being unconditional is itself conditional", then it would be contradictory.

    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

    Well, the set of non-conditions is the complement of the set of conditions. I don't know exactly what do you mean by "condition", but the set of non-conditions contains everything which is not a condition. So a knife or the number 42, for example, if it's not a condition, is contained in this set. So it cannot be empty. If it was empty, then since the set of conditions is it's complement, it must be the universal set, which means that everything would be a condition, which is kinda weird (but I really don't know what a condition is here).
  • TheMadFool
    13.8k
    Condition for x = prerequisite (necessary for x)

    Being unconditional to me means that there are no conditions and since everything can be a condition, the set of non-conditions = { }. The set of non-conditions represents the state of being unconditional.

    An empty set is a subset of every set. So the set of non-conditions, since it's the null set, is a subset of the set of conditions. If I recall correctly, the null set is a subset of all sets because it doesn't contain any member that isn't a member of any other set.

    It follows therefore that the state of being unconditional, expressed as the set of non-conditions, is also a condition.

    However, when people use the words "conditional" and "unconditional", they mean a finite set of conditions relevant to a particular issue. For example, conditional love will have a finite list of conditions e.g. beautiful, charming, good, etc. The notion of unconditional love would simply mean that the conditions so listed don't matter.

    The problem arises when the set of conditions for a particular fact or event also includes the state of being unconditional but doing so, although logically permissible, would not be an accurate representation of the usage of the words "conditional" and "unconditional".
  • Arne
    796
    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.
  • Nicholas Ferreira
    78
    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.
  • TheMadFool
    13.8k
    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

    IF one looks at unconditional love logically, we get a paradox but the conventional meaning of it is without any such problems.
  • TheMadFool
    13.8k
    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

    I'm talking about unconditional love and by extension anything stated to be unconditional. I guess you're right in bringing up a discussion of necessary conditions but I'd also like to include sufficient conditions as they seem to form part of the definition of conditions.
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