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solakis
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May 18th, 2018,
20:29
solakis
replied to a thread
Propositional calculus
in
Challenge Questions and Puzzles
Can you give an example of a system with a binary opperation where by using associativity you can prove that parenthesis are redundant
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9 replies | 190 view(s)
May 17th, 2018,
07:04
solakis
replied to a thread
Propositional calculus
in
Challenge Questions and Puzzles
By using associativity you can get them equal. But i am asking ,how wiki gets, a*(b*a)*c from (a*b)*(a*c) by ‹using associativity Parenthesis...
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9 replies | 190 view(s)
May 16th, 2018,
21:00
solakis
replied to a thread
Propositional calculus
in
Challenge Questions and Puzzles
If the system is associative ,then (A*B)*C =A*(B*C).........(1) ,Note i use * instead the circle,where A,B,C are elements of the structure And...
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9 replies | 190 view(s)
May 13th, 2018,
20:04
solakis
replied to a thread
Challenge problem #3 [Olinguito]
in
Challenge Questions and Puzzles
Differentiate both sides w.r.t. y and we get: $$\frac{dy}{dy}= \frac{dx}{dy}+\sqrt2\,e^{\frac{x+y}{\sqrt2}}\frac{1}{\sqrt...
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4 replies | 208 view(s)
May 12th, 2018,
21:10
solakis
replied to a thread
Propositional calculus
in
Challenge Questions and Puzzles
Sorry,but i am trying to find out how you got (p\wedge\neg q)\wedge (p\wedge\neg r) from p\wedge(\neg q\wedge\neg r)
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9 replies | 190 view(s)
May 11th, 2018,
20:52
solakis
started a thread
Propositional calculus
in
Challenge Questions and Puzzles
Given : p\wedge\neg (q\vee r) then prove whether this formula implies: !) (p\wedge\neg q)\vee (p\wedge\neg r) OR 2) (p\wedge\neg q)\wedge...
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9 replies | 190 view(s)
May 9th, 2018,
12:59
solakis
replied to a thread
axiomatics
in
Challenge Questions and Puzzles
So you thing this kind of proof strictly based on the axioms ,theorems, and definitions of an axiomatic system it should not be used in ordinary...
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16 replies | 436 view(s)
May 9th, 2018,
12:14
solakis
replied to a thread
axiomatics
in
Challenge Questions and Puzzles
To take your very good example. Can we prove : if, a+b=7 and b=2,then a+2=7 without using the substitution rule ?? proof: 1)...
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16 replies | 436 view(s)
May 8th, 2018,
15:18
solakis
replied to a thread
axiomatics
in
Challenge Questions and Puzzles
No No you do not get what i mean We have the formula 0A+(-0A)=...
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16 replies | 436 view(s)
May 8th, 2018,
13:07
solakis
replied to a thread
set theory
in
Challenge Questions and Puzzles
{a,2,b}={a,2}U{b}
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3 replies | 142 view(s)
May 8th, 2018,
13:03
solakis
replied to a thread
axiomatics
in
Challenge Questions and Puzzles
Again by 4 we have : 0A+(-0A)= (0A+0A)+(-0A) and not 0A+(-0A)= 0A+0A+(-0A) (put A=0A,B=0A+0A,C=-0A) ,Then by using the axiom 4 we have : 0A+(-0A)=...
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16 replies | 436 view(s)
May 8th, 2018,
10:01
solakis
replied to a thread
axiomatics
in
Challenge Questions and Puzzles
=> 0A=0A+0A 3) 0A+(-0A) = (0A+0A)+(-0A) By using 4 .This is correct 4) 0 =0A +0 = 0A By 4 and 10 ..............................How??
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16 replies | 436 view(s)
May 7th, 2018,
20:40
solakis
started a thread
set theory
in
Challenge Questions and Puzzles
Prove in any axiomatic set theory that: 2ε{a,2,b} , where a,b are letters
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3 replies | 142 view(s)
May 3rd, 2018,
19:51
solakis
replied to a thread
axiomatics
in
Challenge Questions and Puzzles
In my list of axioms there is no axiom like the one you mention : 0 = A-A However there is an axiom : A+(-A)=0. As for (2) that is what we want...
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16 replies | 436 view(s)
May 3rd, 2018,
12:13
solakis
started a thread
axiomatics
in
Challenge Questions and Puzzles
Given the following axioms: For all A,B,C.........we have: 1) A=A 2) A=B <=> B=A 3) A=B & B=C => A=C
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16 replies | 436 view(s)
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About solakis
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About solakis
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May 18th, 2018
11:09 -
greg1313
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Propositional calculus
by
solakis
Given : p\wedge\neg...
May 9th, 2018
16:02 -
Country Boy
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Re: axiomatics
by
solakis
I see what you mean, and...
May 8th, 2018
17:48 -
solakis
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Re: axiomatics
by
steenis
I see what you mean, and...
13:57 -
Country Boy
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Re: set theory
by
solakis
What is the definition...
10:02 -
solakis
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Re: axiomatics
by
steenis
0 = 0 + 0 (by 10 and 2)...