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Post by Deleted on Jan 11, 2018 16:25:43 GMT
Given:
(1) If #A, #A<>B, then #B
Use a theorem: if #A, #A->B, then #B; and #~A->(A->B),
to prove given formula (1).
(# - valid, truth; <> - equivalence; -> - implication)
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Post by Deleted on Jan 11, 2018 17:02:50 GMT
Ahh Boolean algebra... Brings me back in 1st year of high school  |
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Post by Deleted on Jan 12, 2018 19:29:50 GMT
Ahh Boolean algebra... Brings me back in 1st year of high school Yes, it's kind of it. This exercise is from S. C. Kleene's 'Mathematical logic'. So, can you solve it? Did you have good marks/grades in Mathematics?  ) |
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Post by Deleted on Jan 12, 2018 19:37:01 GMT
Ahh Boolean algebra... Brings me back in 1st year of high school Yes, it's kind of it. This exercise is from S. C. Kleene's 'Mathematical logic'. So, can you solve it? Did you have good marks/grades in Mathematics? ) I guess I was fine mathematician. Can't say I was nearly the best because I had some people in my class that were winning math olympiads during high school. In comparison to them I was bad. Maybe I would be able to solve it if I started to recall these stuff, because I am more into other areas of interest right now, and honestly not a lot in mathematics more in software development. |
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Post by Deleted on Jan 22, 2018 21:14:52 GMT
By the way what is the meaning of this statement:
#~A->(A->B)
Also if I rephrase all the statements, would it be like this:
Prove:
A <=> !B
using:
1) A => !B
2) A => (A => B)
(=> - implication, <=> - equivalence, ! - negation)
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Post by Elizabeth on Jan 22, 2018 23:09:11 GMT
I love math but it might be easier for me to solve it if you don't write it in English. XD
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Post by Deleted on Jan 22, 2018 23:37:18 GMT
I love math but it might be easier for me to solve it if you don't write it in English. XD I think he has one typo |
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Post by Deleted on Feb 5, 2018 15:59:09 GMT
By the way what is the meaning of this statement: #~A->(A->B) Also if I rephrase all the statements, would it be like this: Prove: A <=> !B using: 1) A => !B 2) A => (A => B) (=> - implication, <=> - equivalence, ! - negation) The statement #~A->(A->B) - it's Duns Scott's formula. It means if you claim something contradictory it will imply everything. We need to prove A <=> B just simple equivalence. It is bad I can't find English version of Kleane. To prove it, I think, I need to show there's a contradiction (?), and then imply #B. ? This Scott's formula is tearing me apart. I think your version is also possible, but Kleane doesn't answer. |
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Post by Deleted on Feb 5, 2018 17:47:35 GMT
By the way what is the meaning of this statement: #~A->(A->B) Also if I rephrase all the statements, would it be like this: Prove: A <=> !B using: 1) A => !B 2) A => (A => B) (=> - implication, <=> - equivalence, ! - negation) The statement #~A->(A->B) - it's Duns Scott's formula. It means if you claim something contradictory it will imply everything. We need to prove A <=> B just simple equivalence. It is bad I can't find English version of Kleane. To prove it, I think, I need to show there's a contradiction (?), and then imply #B. ? This Scott's formula is tearing me apart. I think your version is also possible, but Kleane doesn't answer. I am just unsure the meaning of tilde sign. You didn't state what it means  |
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Post by Deleted on Feb 5, 2018 18:49:30 GMT
The statement #~A->(A->B) - it's Duns Scott's formula. It means if you claim something contradictory it will imply everything. We need to prove A <=> B just simple equivalence. It is bad I can't find English version of Kleane. To prove it, I think, I need to show there's a contradiction (?), and then imply #B. ? This Scott's formula is tearing me apart. I think your version is also possible, but Kleane doesn't answer. I am just unsure the meaning of tilde sign. You didn't state what it means Aw, it's negation! I'm sorry that I didn't explain it. I've met this sign in many math books, so I thought it very clear to understand. I was wrong. |
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Post by Deleted on Feb 5, 2018 19:07:22 GMT
Well, I proved (or so I think) the problem I stated, but now I realized your initial problem is different. Your problem would be something like: Prove A <=> B using tautologies:
1) A => B
2) !A => (A => B)if I finally got it right |
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Post by Deleted on Feb 5, 2018 20:29:57 GMT
Almost.
According to Kleane, we need to prove tautology of "A equivalent B", but in such interpretation:
"If A and (A equivalent B), then B." Besides, A and B - are truth formulas (tautologies).
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Post by Deleted on Feb 5, 2018 20:47:24 GMT
So: (A and (A <=> B)) => B
????
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Post by Deleted on Feb 6, 2018 13:20:13 GMT
So: (A and (A <=> B)) => B ???? Probably, it is. And I suspect 'A' and 'A<=>B' might appear contradiction: !A<=>(A<=>B) or A<=>!(A<=>B) ? Then, if 'A' and 'A<=>B' contradicts to one another, using !A->(A->B) let us imply B. |
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