Part 9: Symbolic Logic (more easy proofs & some common errors)

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Комментарии
Автор

I'm confused on the last one. You have Q and ~Q; isn't that a sign that something is wrong with the premises? Basically it seems like if your premises lead you to a contradiction like that, you can prove anything you want to.

BrainsterPC
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Hi, love the series, I’m really enjoying it! I have a question about your example at 2:10. You say it’s invalid to use addition on part of a conditional. I looked at this indirectly and then wrote out the whole truth table, and I can’t find an example where A->B is true and A->(BVJ) is false. Also, addition says that B->(BVJ), so if A->B, then A->(BVJ). So I don’t see how the inference is invalid. However, A->B does NOT imply that (AVJ)->B. It’s invalid when A & B are false and J is true. So it seems that you can apply addition to the conclusion of a conditional but not the premise.

parrotkoi
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hi, so you have "not s or t". To get t, you need not not s, which is same as 's'. Where is that? I'm not sure I follow, you could write in steps here and I'll take a look. :) show every step

teachphilosophy
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hey i was doing the example at 11 mins, i did on 5. -s, 6. G and used simplificaiton on both, then i did addition on 5 to get -svt and then used modus ponens on 5, 7 to get t, is this a correct way to get the final answer?

Pitafajita
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I don't think you did the second example correctly. You can't do DS with 2, 3.

minapham