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      CommentAuthorMathdude314 (Advanced Member)
    • CommentTimeFeb 11th 2017 edited
     
    A long time ago, I was trying to come up with puzzles based on an infinite list of sentences. Then I started coming up with variations, and soon there were over 100 ways to modify the puzzle.

    BASE PUZZLE:
    There is an infinite list of sentences, as follows:
    1. Exactly 1 sentence on this list is false.
    2. Exactly 2 sentences on this list are false.
    3. Exactly 3 sentences on this list are false.
    ...
    And so on, one sentence for each natural number. (I was inspired by puzzles that were finite, usually going to whatever the year was, because you could not brute force 2017 sentences. Here, they go on forever.)
    Your goal is to identify the true/false value of each statement (if there is 1 solution), or give all solutions (may need a schema), or prove no solution.

    So here are the variations (you of course have to change the grammar if appropriate):
    Quantifier (3-5): Change the first word to "exactly" (1), "at least" (2), or "at most" (3). (I am not sure if strict "fewer than" and "more than" are equivalent to existing problems, so if not add them.)
    Boolean (2): Change the last word to "false" (1) or "true" (2).
    Zero (2): This adds sentence 0 which does the obvious. (So for the base, it is "Exactly 0 sentences on this list are false", which is of course false.)
    Omega Sentence (4+): There are three varieties of the omega sentence. You can try to see what happens when you add more than one, but there are enough as it is.
    Ω1. (quantifier) all numbered sentences are (boolean).
    Ω2. (quantifier) all sentences are (boolean). (Ω2 includes itself, Ω1 does not.)
    Ω3. (quantifier) an infinite number of sentences are (boolean). (I thought of Ω4 which includes itself but they are equivalent.)


    Well, with that I end up with 5 · 2 · 2 · 4 = 80 puzzles. (Q4, Q5, and Ω3 were invented as I wrote this, so I must be missing something else as I had not 36 but 108 some time ago...)-----------------
    If less people are active, that will only make for even less activity. Start making those designs!
    Currently working on a one-round no-elimination quick RP game...
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      CommentAuthorMathdude314 (Advanced Member)
    • CommentTimeFeb 11th 2017
     
    Just to start things off, I will solve six of them.
    Base: No two sentences can be true, which means an infinite number of sentences are false and thus no sentence can be true. So they are all false.
    Q2: Any sentence n being true implies sentence m for all m ≤ n. However, we run into a problem: if some sentence is false, then everything above it is false, but that contradicts all the statements being false. If all statements are true, then that is a contradiction too. Hence, no solution
    Q3: Any sentence being true implies all the sentences above it. So either everything is false, everything is true, or all sentences above some n are true (and the others are false).
    B2: There are two solutions: either all sentences are false, or sentence 1 is true and no others are. (No two distinct sentences can be true, because they would contradict each other.)
    Q2,B2: Any sentence n being true implies sentence m for all m ≤ n, so the answers are as follows:
    • No sentences are true
    • All sentences are true
    • There exists n such that if m ≤ n then sentence m is true, and if m > n then sentence m is false.
    Q3,B2: Any sentence being true implies all the ones above it. But that contradicts them (they all imply a finite number of true statements). Therefore no sentence is true. However, at most 1 sentence is true, so sentence 1 is true and we have a contradiction. No solution-----------------
    If less people are active, that will only make for even less activity. Start making those designs!
    Currently working on a one-round no-elimination quick RP game...
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      CommentAuthorMathdude314 (Advanced Member)
    • CommentTimeFeb 22nd 2017
     
    OK, so if you add the "zero sentence" idea, then:
    Base + Zero: This changes nothing. They must still be all false.
    True + Zero: Statement zero must be false, because it being true implies everything is false. Here the only solution is that sentence 1 is true and nothing else is, similar to the above except that now something has to be true.

    I might work on these more later.-----------------
    If less people are active, that will only make for even less activity. Start making those designs!
    Currently working on a one-round no-elimination quick RP game...
    • CommentAuthorXyuzhg (Moderator)
    • CommentTimeFeb 26th 2017
     
    I'm not sure how the quantifier modification works with Ω1 and Ω2, so I'm leaving that out. I am including strictly inequality. The zero statement is interesting as it allows for transfinite induction in some cases. So there are a total of 5 · 2 · 2 · 2 = 40 puzzles here.

    Here I'm using the following numbering:

    Q[Quantifier =/≤/≥/</>]|[Boolean T/F]|[optional 0][optional Ω(3)]
    e.g. Q<|F|0Ω is 'Strictly fewer than n are false' including 0 and the statement 'Finitely many are false'.


    • Q=|F, Q=|F|0
    No two are true, so infinitely many are false, so all are false.

    • Q=|F|Ω, Q=|F|0Ω
    No two are true, so infinitely many are false, so only (Ω) is true.

    • Q=|T, Q=|T|Ω
    No two are true, so either all are false or only (1) is true.

    • Q=|T|0, Q=|T|0Ω
    No two are true, and none true is in contradiction to (0) being false, so only (1) is true.


    • Q≤|F, Q≤|F|0, Q≤|F|Ω, Q≤|F|0Ω
    If (α) is true, (β) is true for all β ≥ α. (Ω) is true if included. We have infinitely many solutions of the form (β) is true iff β ≥ α where α = 0, 1, 2, ..., ω.

    • Q≤|T, Q≤|T|0, Q≤|T|Ω, Q≤|T|0Ω
    If any finite (α) is true, (β) is true for all β ≥ α, so infinitely many are true, a contradiction to (α) being true. If all finite (α) are false, this contradicts (1) being false. Contradiction.


    • Q≥|F, Q≥|F|0, Q≥|F|Ω, Q≥|F|0Ω
    If any finite (α) is false, (β) is false for all β ≥ α, so infinitely many are false, a contradiction to (α) being false. If all finite (α) are true, this contradicts (2) being true. Contradiction.

    • Q≥|T, Q≥|T|Ω
    If (α) is false, (β) is false for all β ≥ α. We have infinitely many solutions of the form (β) is false iff β > α where α = 0, 1, 2, ..., ω.

    • Q≥|T|0, Q≥|T|0Ω
    Let α ≤ ω. If (β) is true for all β < α, clearly (α) is true. By transfinite induction, all are true.


    • Q<|F, Q<|F|Ω
    If (α) is true, (β) is true for all β ≥ α. We have infinitely many solutions of the form (β) is true iff β > α where α = 0, 1, 2, ..., ω.

    • Q<|F|0, Q<|F|0Ω
    Let α ≤ ω. If (β) is false for all β < α, clearly (α) is false. By transfinite induction, all are false.

    • Q<|T, Q<|T|0, Q<|T|Ω, Q<|T|0Ω
    If any finite (α) is true, (β) is true for all β ≥ α, so infinitely many are true, a contradiction to (α) being true. If all finite (α) are false, this contradicts (2) being false. Contradiction.


    • Q>|F, Q>|F|0, Q>|F|Ω, Q>|F|0Ω
    If any finite (α) is false, (β) is false for all β ≥ α, so infinitely many are false, a contradiction to (α) being false. If all finite (α) are true, this contradicts (1) being true. Contradiction.

    • Q>|T, Q>|T|0, Q>|T|Ω, Q>|T|0Ω
    If (α) is false, (β) is false for all β ≥ α. (Ω) is false if included. We have infinitely many solutions of the form (β) is false iff β ≥ α where α = 0, 1, 2, ..., ω.
    -----------------
    Hopefully PA is inconsistent.