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):

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...)]]>

• 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.

I might work on these more later.]]>

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

• Q=|F|Ω, Q=|F|0Ω

No two are true, so infinitely many are false, so

• Q=|T, Q=|T|Ω

No two are true, so either

• Q=|T|0, Q=|T|0Ω

No two are true, and none true is in contradiction to (0) being false, so

• 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

• 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.

• 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.

• Q≥|T, Q≥|T|Ω

If (α) is false, (β) is false for all β ≥ α. We have infinitely many solutions of the form

• Q≥|T|0, Q≥|T|0Ω

Let α ≤ ω. If (β) is true for all β < α, clearly (α) is true. By transfinite induction,

• Q<|F, Q<|F|Ω

If (α) is true, (β) is true for all β ≥ α. We have infinitely many solutions of the form

• Q<|F|0, Q<|F|0Ω

Let α ≤ ω. If (β) is false for all β < α, clearly (α) is false. By transfinite induction,

• 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.

• 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.

• 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