The simplest way to solve the jester’s puzzle is to make a table of the four cases … then determine for each case whether the inscriptions are in fact true or false as required for that case. The conclusion is that the first box has the frog and the true inscription.
If you do this, the case where the second inscription is true and the first box contains a frog is also consistent.
Interestingly enough, I just mapped this whole problem out carefully in a spreadsheet, and it appears to agree with zzz2. I’ll have to check it now that I’ve seen your comment.
The simplest way to solve the jester’s puzzle is to make a table of the four cases … then determine for each case whether the inscriptions are in fact true or false as required for that case. The conclusion is that the first box has the frog and the true inscription.
If you do this, the case where the second inscription is true and the first box contains a frog is also consistent.
If you do this, the case where the second inscription is true and the first box contains a frog is also consistent.
No, because in that case the first inscription would also be true. Both inscriptions cannot be true.
Interestingly enough, I just mapped this whole problem out carefully in a spreadsheet, and it appears to agree with zzz2. I’ll have to check it now that I’ve seen your comment.
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