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Thread: Contradictions imply everything (a proof)

  1. Top | #11
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    I think I'm probably not getting the notation, but what I was trying to get at is something like this: We are using AC (if we don't have AC, one can make other arguments), so every set has a cardinality. Suppose that Y is a set of propositions with cardinality a, and let b be the next cardinal. For every cardinal c<=b such that c=\=a, let P(c) be the proposition "It is not the case that the cardinality of Y is c", and let P(a) be the proposition "The cardinality of Y is a". Then, for all c<=b, P(c) is a true proposition, and at least one of those true propositions is not an element of Y (else, we would have a>=c, contradicting the hypothesis). Thus, there is no set of all true propositions.


    Quote Originally Posted by A Toy Windmill
    The features of the proof I was going for are:

    1) The absence of clear irrelevancies, cases where a Q comes out of nowhere. Instead, I consider "everything holds" (∀p. p).
    2) No reliance on the definition of material implication. I don't assume that everything implies true, or that false implies everything.
    I agree, those sound like good ideas for a proof.

  2. Top | #12
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    Quote Originally Posted by Angra Mainyu View Post
    I think I'm probably not getting the notation, but what I was trying to get at is something like this: We are using AC (if we don't have AC, one can make other arguments), so every set has a cardinality. Suppose that Y is a set of propositions with cardinality a, and let b be the next cardinal. For every cardinal c<=b such that c=\=a, let P(c) be the proposition "It is not the case that the cardinality of Y is c", and let P(a) be the proposition "The cardinality of Y is a". Then, for all c<=b, P(c) is a true proposition, and at least one of those true propositions is not an element of Y (else, we would have a>=c, contradicting the hypothesis). Thus, there is no set of all true propositions.
    Yes. Sets of propositions would either need to be suitably stratified or an impredicative class. But we can sidestep this and only consider subsets of the finite set X = {T, everything holds}. So long as X can be admitted to exist, we go with

    Code:
    1) Consider the subset of X consisting of propositions Q such that P or Q. This is non-empty since it at least contains T.
    2) Consider the subset of X consisting of propositions R such that either P or else (R implies everything). This is non-empty since it at least contains the proposition "everything holds".

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    Quote Originally Posted by A Toy Windmill View Post
    First, we'll prove that either a proposition P holds, or else it implies everything.

    To do so, we will admit that we can quantify over and form sets of propositions and that we have the axiom of choice. We'll let T stand for a true proposition, whichever you like.

    Code:
    1) Consider the set of propositions Q such that P or Q. This is non-empty since it at least contains T.
    2) Consider the set of propositions R such that either P or else (R implies everything). This is non-empty since it at least contains the proposition "everything holds".
    3) Pair the above sets and use the axiom of choice to pick out a Q and R respectively.
      4) Assume P
      5) Then the sets considered above are the same set: the universal set of propositions.
      6) Thus choice picked out only one proposition above. In other words, Q = R
        7) Assume Q
        8) Assume R implies everything
        9) Hence Q implies everything (from 6 and 8)
        10) Everything (from 7 and 9)
      11) If Q and R implies everything, then everything (from 7-10)
    12) If P, Q and R implies everything, then everything (from 4-11).
    13) (if Q and R implies everything) then P implies everything (reordering the antecedents of 12)
    14) (P or Q) and (P or R implies everything) (from 1, 2 and 3)
    15) P or (Q and R implies everything) (from 14 by distributivity)
    16) P or P implies everything (from 13 and 15)
    As a corollary:

    Code:
    1) Assume P.
    2) Assume not-P.
    3) Have P or P implies everything (from the proof above)
    4) P implies everything (disjunctive syllogism from 2 and 3).
    5) Anything. (from 1 and 4).
    Non-Aristotelian logic. Not human.
    EB

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    Quote Originally Posted by A Toy Windmill View Post
    Quote Originally Posted by Angra Mainyu View Post
    I think I'm probably not getting the notation, but what I was trying to get at is something like this: We are using AC (if we don't have AC, one can make other arguments), so every set has a cardinality. Suppose that Y is a set of propositions with cardinality a, and let b be the next cardinal. For every cardinal c<=b such that c=\=a, let P(c) be the proposition "It is not the case that the cardinality of Y is c", and let P(a) be the proposition "The cardinality of Y is a". Then, for all c<=b, P(c) is a true proposition, and at least one of those true propositions is not an element of Y (else, we would have a>=c, contradicting the hypothesis). Thus, there is no set of all true propositions.
    Yes. Sets of propositions would either need to be suitably stratified or an impredicative class. But we can sidestep this and only consider subsets of the finite set X = {T, everything holds}. So long as X can be admitted to exist, we go with

    Code:
    1) Consider the subset of X consisting of propositions Q such that P or Q. This is non-empty since it at least contains T.
    2) Consider the subset of X consisting of propositions R such that either P or else (R implies everything). This is non-empty since it at least contains the proposition "everything holds".

    Okay, now I see what I misunderstood in the notation.

    I take it that "everything holds" the proposition 'all propositions are true', or something like that?

    By the way, for finite sets, the axiom of choice follows from ZF, so it's a theorem of choice, so one does not need to add it to the proof.

    Still, I think I'm not reading this right, for the following reason:

    Assuming P, the subset of X consisting of propositions Q such that P or Q is X. The subset of X consisting of propositions R such that either P or else (R implies everything) is also X. Okay, so you're saying choice was used only to pick one proposition because you're picking from the singleton {X}. I see.

    We could also give a proof independent of the ZF axioms. But Speakpigeon will not accept any proofs.
    Last edited by Angra Mainyu; 07-04-2019 at 01:16 AM.

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    Could someone please translate “or else open parenthesis the letter “R” implies everything close parenthesis” in plain English? I’m especially confused with the “(“ following the word, “else.”

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    Quote Originally Posted by Angra Mainyu View Post
    Okay, now I see what I misunderstood in the notation.

    I take it that "everything holds" the proposition 'all propositions are true', or something like that?
    Yes. In second order propositional calculus, it's just (∀p. p).

    By the way, for finite sets, the axiom of choice follows from ZF, so it's a theorem of choice, so one does not need to add it to the proof.
    Ah, of course!

    The proof in the OP is adapted from another theorem, but when you unfold it using finite choice, it simplifies to the classic!

    1) Assume P
    2) Then P or R
    3) Assume ~P
    4) Therefore R (by 2 and 3)

  7. Top | #17
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    Quote Originally Posted by fast View Post
    Could someone please translate “or else open parenthesis the letter “R” implies everything close parenthesis” in plain English? I’m especially confused with the “(“ following the word, “else.”
    I want to say that there are two possible cases:

    Either:
    1) P is true.
    2) R implies everything.

    Getting that into one line can create ambiguity. I could write "either P or R implies everything", but that could be read as

    If it is the case that P or it is the case that R, then we have everything.

    Brackets disambiguate. The intended reading is "P or (R implies everything)". The other reading is "(P or R) implies everything."

    There should be even more brackets in the OP, but they're ugly to write. It would have been easier to do everything symbolically.

  8. Top | #18
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    Thank you. That helps.

  9. Top | #19
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    Quote Originally Posted by A Toy Windmill View Post
    First, we'll prove that either a proposition P holds, or else it implies everything.

    To do so, we will admit that we can quantify over and form sets of propositions and that we have the axiom of choice. We'll let T stand for a true proposition, whichever you like.

    Code:
    1) Consider the set of propositions Q such that P or Q. This is non-empty since it at least contains T.
    2) Consider the set of propositions R such that either P or else (R implies everything). This is non-empty since it at least contains the proposition "everything holds".
    3) Pair the above sets and use the axiom of choice to pick out a Q and R respectively.
      4) Assume P
      5) Then the sets considered above are the same set: the universal set of propositions.
      6) Thus choice picked out only one proposition above. In other words, Q = R
        7) Assume Q
        8) Assume R implies everything
        9) Hence Q implies everything (from 6 and 8)
        10) Everything (from 7 and 9)
      11) If Q and R implies everything, then everything (from 7-10)
    12) If P, Q and R implies everything, then everything (from 4-11).
    13) (if Q and R implies everything) then P implies everything (reordering the antecedents of 12)
    14) (P or Q) and (P or R implies everything) (from 1, 2 and 3)
    15) P or (Q and R implies everything) (from 14 by distributivity)
    16) P or P implies everything (from 13 and 15)
    As a corollary:

    Code:
    1) Assume P.
    2) Assume not-P.
    3) Have P or P implies everything (from the proof above)
    4) P implies everything (disjunctive syllogism from 2 and 3).
    5) Anything. (from 1 and 4).
    We would need first to prove empirically that your method of logical proof here is correct.

    Me, I'm 100% confident it's wrong and I'm also not aware of any empirical proof that it would be correct.

    Short of that, this is just a mathematical pie in the sky meaning nothing.

    You should refrain from making it look as if your proof had any authoritative value. You are misleading posters here.

    You know, a disclaimer would be appropriate here. Someone might decide to sue you and I think he would win.
    EB

  10. Top | #20
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    I finally figured out why I have trouble with those logic threads. People shift between informal and attempts a t formal proofs with no constancy or structure.

    When I read (P&¬P)->Q I see Q as always false. I am always trying to convert text to symbolic logic.

    I see the OP trying to prove a tautology, or the definition of OR and AND.I am a little slow so I like pictures.

    There is a box of apples and a box of peaches. I say put one or the other in the car. That is an exclusive or XOR. One or the other but not both.

    If I say take one or the other or both that is inclusive or.

    If I say take the apples and peaches t means take both.

    See the link on Boolean Algebra on the tread I started for logic functions and the truth tables.

    .

    This is where truth tables come in. To precisely define what you mean post a truth tale. It is unambiguous and not open to interpretation.

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