3 ms·

If (1) must be verified, how can you know that the blue card has a number on the other side? To verify (1) you must prove that each card has a color on one sid

by greggyb 11y ago

If (1) must be verified, how can you know that the blue card has a number on the other side? To verify (1) you must prove that each card has a color on one side and a number on the other. The only way to do this is to see both sides of each card.

ikeboy 11y ago
Oh, you mean actually verifying 1. I wasn't doing that. I was just doubting 1 being true, but interpreting the problem as asking you to verify 2, in which case you only need to turn over 5,8, and green.
- greggyb 11y ago
  Okay, now I'm really confused about what you meant with your original post in this thread. If you doubt (1), how is it that you need to verify that the 5 has a color, but not that the blue has a number on the other side? To verify (2) you only need to prove that every even card has a blue back. An odd may be blue or green, but an even must be blue. Hence the 5 is fine, it can have any color. The blue is fine, because it can have any number. But that 8 must have a blue back, and that green must have an odd number.
  - ikeboy 11y ago
    >The problem with this is that the 5 card also needs to be turned; it might have an even number on the back. If you can't be sure that the cards have a color and a numbers, there's nothing outlawing a 5-6 card, which breaks (2).
    - greggyb 11y ago
      So if I understand correctly: You trust that if a card has a color on one side, it has a number on the other side. You do not trust that if a card has a number on one side, it has a color on the other side. Is this correct?
      - ikeboy 11y ago
        No. I don't make any assumptions as to anything I can't see. However, regardless of what's on the back of blue, (2) will still be correct; the consequent is true, therefore the conditional is true.
        greggyb 11y ago
        Okay, I see where you're coming from now. (1) may or may not be true. Understanding that we cannot trust (1), but that we don't care to verify it, then the 5 is necessary to prove (2). That is much more subtle than I originally interpreted from you original post. Thanks for your patient explanation.

by greggyb 11y ago

ikeboy 11y ago

Oh, you mean actually verifying 1. I wasn't doing that. I was just doubting 1 being true, but interpreting the problem as asking you to verify 2, in which case you only need to turn over 5,8, and green.

greggyb 11y ago
Okay, now I'm really confused about what you meant with your original post in this thread. If you doubt (1), how is it that you need to verify that the 5 has a color, but not that the blue has a number on the other side? To verify (2) you only need to prove that every even card has a blue back. An odd may be blue or green, but an even must be blue. Hence the 5 is fine, it can have any color. The blue is fine, because it can have any number. But that 8 must have a blue back, and that green must have an odd number.
- ikeboy 11y ago
  >The problem with this is that the 5 card also needs to be turned; it might have an even number on the back. If you can't be sure that the cards have a color and a numbers, there's nothing outlawing a 5-6 card, which breaks (2).
  - greggyb 11y ago
    So if I understand correctly: You trust that if a card has a color on one side, it has a number on the other side. You do not trust that if a card has a number on one side, it has a color on the other side. Is this correct?
    - ikeboy 11y ago
      No. I don't make any assumptions as to anything I can't see. However, regardless of what's on the back of blue, (2) will still be correct; the consequent is true, therefore the conditional is true.
      - greggyb 11y ago
        Okay, I see where you're coming from now. (1) may or may not be true. Understanding that we cannot trust (1), but that we don't care to verify it, then the 5 is necessary to prove (2). That is much more subtle than I originally interpreted from you original post. Thanks for your patient explanation.