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Puzzle:

Prove that $1 + 1$ is $1$ using formal logic.

Hint:

$1 + 0$ is $0$

Hint 2 (edit):

No new definitions are required. The solution does have to do with the symbol + other actual (perhaps less formal) definitions.

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closed as too broad by Bass, GOTO 0, Oray, Beastly Gerbil, Peregrine Rook Feb 17 '18 at 19:54

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Howw can yu explain that with "formal logic" but tagged as lateral thinking? I just thought thays ironic, but you probably have your own explanation. $\endgroup$ – North Feb 17 '18 at 15:18
  • $\begingroup$ @North There is definitely a reason. It's what makes this puzzle work. $\endgroup$ – user72528 Feb 17 '18 at 15:19
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    $\begingroup$ Given that you have to go outside normal mathematics, there are endlessly many different solutions, each one just as good as the other, and no way to tell which answer is the correct one. $\endgroup$ – Bass Feb 17 '18 at 18:14
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    $\begingroup$ I’m thinking this is more a coding question where + is more of an ‘and’ so 1 and 1 is true (1) and 1 and 0 is false (0) $\endgroup$ – Aggie Kidd Feb 19 '18 at 20:19
  • $\begingroup$ @AggieKidd That's what I was thinking. And I used "formal logic" because and is an operation in formal logic. $\endgroup$ – user72528 Feb 21 '18 at 1:40
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I suspect there will be multiple answers to this, but here's one:

Define the "+" operator to mean "the following number."

Then:

1 + 0 is 0
1 + 1 is 1

In fact,

x + 0 is 0, for any x, and x + 1 is 1, for any x, etc.

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    $\begingroup$ Similar to my first thought. Formally define the + symbol as a multiplication operator. $\endgroup$ – Matt Stevens Feb 18 '18 at 3:48
  • $\begingroup$ @MattStevens that would also work - and it's perhaps simpler than mine :) $\endgroup$ – puzzledPig Feb 18 '18 at 6:41
  • $\begingroup$ That's not exactly what I was thinking. No new definitions are required. The solution does have to do with the symbol + other actual (perhaps less formal) definitions. $\endgroup$ – user72528 Feb 18 '18 at 18:30

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