Page 1 of 3 123 LastLast
Results 1 to 10 of 23

Thread: Contradictions imply everything (a proof)

  1. Top | #1
    Junior Member
    Join Date
    Jun 2019
    Location
    England
    Posts
    76
    Rep Power
    1

    Contradictions imply everything (a proof)

    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).

  2. Top | #2
    Contributor
    Join Date
    Nov 2004
    Location
    South Carolina
    Posts
    5,239
    Archived
    14,025
    Total Posts
    19,264
    Rep Power
    60
    Are P and Q collectively exhaustive? Like P and not P are

  3. Top | #3
    Junior Member
    Join Date
    Jun 2019
    Location
    England
    Posts
    76
    Rep Power
    1
    Quote Originally Posted by fast View Post
    Are P and Q collectively exhaustive? Like P and not P are
    P and Q cannot both be false. They may both be true.

  4. Top | #4
    Contributor
    Join Date
    Nov 2004
    Location
    South Carolina
    Posts
    5,239
    Archived
    14,025
    Total Posts
    19,264
    Rep Power
    60
    Quote Originally Posted by A Toy Windmill View Post
    Quote Originally Posted by fast View Post
    Are P and Q collectively exhaustive? Like P and not P are
    P and Q cannot both be false. They may both be true.
    If P and Q cannot both be false, then one must be true rendering Q equivalent to not P, but if they may both be true, Q isn’t equivalent to not P.

  5. Top | #5
    Veteran Member
    Join Date
    Nov 2017
    Location
    seattle
    Posts
    4,752
    Rep Power
    11
    I am used to using trtuth tables and symbolic logic.

    OR function. A || B = C

    A B C
    F F F
    F T T
    T F T
    T T T

    A || !A = TRUE From the truth table for OR any true in gives a true out. If A is false !A is true if A is false !A is true.
    Last edited by steve_bank; 06-29-2019 at 04:49 AM.

  6. Top | #6
    Junior Member
    Join Date
    Jun 2019
    Location
    England
    Posts
    76
    Rep Power
    1
    Quote Originally Posted by fast View Post
    Quote Originally Posted by A Toy Windmill View Post
    Quote Originally Posted by fast View Post
    Are P and Q collectively exhaustive? Like P and not P are
    P and Q cannot both be false. They may both be true.
    If P and Q cannot both be false, then one must be true rendering Q equivalent to not P
    That doesn't follow. Not sure if it's just a silly mistake, but you might like to check its invalidity using a truth-table as steve suggests.

  7. Top | #7
    Contributor
    Join Date
    Nov 2004
    Location
    South Carolina
    Posts
    5,239
    Archived
    14,025
    Total Posts
    19,264
    Rep Power
    60
    Quote Originally Posted by A Toy Windmill View Post
    Quote Originally Posted by fast View Post
    If P and Q cannot both be false, then one must be true rendering Q equivalent to not P
    That doesn't follow. Not sure if it's just a silly mistake, but you might like to check its invalidity using a truth-table as steve suggests.
    Is there any part of what I said you agree with?

  8. Top | #8
    Junior Member
    Join Date
    Jun 2019
    Location
    England
    Posts
    76
    Rep Power
    1
    Quote Originally Posted by fast View Post
    Quote Originally Posted by A Toy Windmill View Post
    Quote Originally Posted by fast View Post
    If P and Q cannot both be false, then one must be true rendering Q equivalent to not P
    That doesn't follow. Not sure if it's just a silly mistake, but you might like to check its invalidity using a truth-table as steve suggests.
    Is there any part of what I said you agree with?
    No. Everything was wrong.

    I'm happy to help with basic logic, but given the OP, this isn't the appropriate thread.

  9. Top | #9
    Veteran Member
    Join Date
    Jan 2006
    Location
    Buenos Aires
    Posts
    2,039
    Archived
    7,588
    Total Posts
    9,627
    Rep Power
    52
    Thanks for trying to do that.

    One would need to restrict the propositions, though, otherwise a universal set of propositions runs into trouble (proper classes won't help, either). Do you have some restrictions in mind?

    In any case, (P&¬P)->Q is more basic than the axiom of choice. But I feel like every attempt at proving it will fail to persuade those who don't see it.

  10. Top | #10
    Junior Member
    Join Date
    Jun 2019
    Location
    England
    Posts
    76
    Rep Power
    1
    Quote Originally Posted by Angra Mainyu View Post
    Thanks for trying to do that.

    One would need to restrict the propositions, though, otherwise a universal set of propositions runs into trouble (proper classes won't help, either). Do you have some restrictions in mind?
    Assume that P and Q are carved out of subsets of X = {T, all propositions hold}. Line 5 then changes to

    5) Then the sets considered above are the same set: X.

    In any case, (P&¬P)->Q is more basic than the axiom of choice. But I feel like every attempt at proving it will fail to persuade those who don't see it.
    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.

    So even if mathematicians are using an intuitive logic without 1 and 2 (it wouldn't surprise me), they are still forced to conclude that inconsistencies imply everything by axioms of logic to which they explicitly hold, specifically choice.

    Obviously, I don't think this way, and I take it as basic that falsehoods imply everything. My favourite axiom system for classical propositional logic is this one:

    1) P → (Q → P)
    2) (P → Q) → (Q → R) → (P → R)
    3) ((P → Q) → P) → P
    4) ⊥ → P

    Here, negation isn't even a basic connective. Contradictions are captured directly by a special proposition ⊥ whose single defining characteristic is that it implies everything. Negation is then defined as P → ⊥.

    I like to justify this in a few ways. First, you can read ⊥ as an arbitrary absurdity like "I'm the Queen of Sheba", and so negation is just everywhere the colloquialism "if that's true, then I'm the Queen of Sheba!" Secondly, it understands negation as a kind of bet. To declare a proposition false, I must put up a forfeit should I be mistaken and the proposition turn out to be true. And that forfeit is ⊥, a proposition designed to render the entire enterprise pointless and boring by being the ultimate cheat code for all other problems, the master key to the dungeon.

    As Hardy declared of proof by contradiction, the mathematician is willing to sacrifice the game (⊥).

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •