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Offline superjames

Senior Newbie





« Posted 2006-07-12 11:57:44 »

If A is B, but B is A and C, then assuming that A is not C is it possible that B = C, or is it certain that B = C (assuming obviously that A = A, B = B etc)

Offline Riven
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« Reply #1 - Posted 2006-07-12 12:04:59 »

Option 1:

A = B
B = A & C

thus: B = A   &   B = C
thus: A  = B = C
so that B == C



Option 2:
A = B
B = A + C

thus: B = B + C
thus: C = 0
so that B != C (if A and B are not 0)



So... it depends.

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Offline superjames

Senior Newbie





« Reply #2 - Posted 2006-07-12 12:10:54 »

Option 1:

A = B
B = A & C

thus: B = A   &   B = C
thus: A  = B = C
so that B == C



Option 2:
A = B
B = A + C

thus: B = B + C
thus: C = 0
so that B != C (if A and B are not 0)



So... it depends.

nice try but if A is equal to C then A cannot not be equal to C
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Offline endolf

JGO Coder


Medals: 7
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« Reply #3 - Posted 2006-07-12 12:22:52 »

A = B
B = A & C

thus

A = A & C

if A = 101 and C = 111 then A & C = 101, but A != C

QED Smiley

Endolf

Offline Riven
« League of Dukes »

JGO Overlord


Medals: 816
Projects: 4
Exp: 16 years


Hand over your head.


« Reply #4 - Posted 2006-07-12 14:00:27 »

Quote
nice try but if A is equal to C then A cannot not be equal to C

it can, if A, B and C are all 0.



1  
2  
If A is B, but B is A and C, then assuming that A is not C is it possible that B = C, or is it certain that B = C (assuming obviously that A = A, B = B etc)
If 0 is 0, but 0 is 0 and 0, then assuming that 0 is not 0 is it possible that 0 = 0, or is it certain that 0 = 0 (assuming obviously that 0 = 0, 0 = 0 etc)


clearly the assumption is wrong, because 0 is always 0

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Offline superjames

Senior Newbie





« Reply #5 - Posted 2006-07-13 00:52:20 »

Quote
nice try but if A is equal to C then A cannot not be equal to C

it can, if A, B and C are all 0.



1  
2  
If A is B, but B is A and C, then assuming that A is not C is it possible that B = C, or is it certain that B = C (assuming obviously that A = A, B = B etc)
If 0 is 0, but 0 is 0 and 0, then assuming that 0 is not 0 is it possible that 0 = 0, or is it certain that 0 = 0 (assuming obviously that 0 = 0, 0 = 0 etc)


clearly the assumption is wrong, because 0 is always 0

You need to revise your axoims
Offline kevglass

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« Reply #6 - Posted 2006-07-13 01:00:31 »

I really hate questions where the "clever" bit is to interpret the english language of the questioner - which obviously can be typoed or just poor use

More to the point this response:

Quote
nice try but if A is equal to C then A cannot not be equal to C

doesn't seem to negate this answer:

Quote
Option 2:
A = B
B = A + C

thus: B = B + C
thus: C = 0
so that B != C (if A and B are not 0)

Where A and B can be the same non-zero value as long as C is zero. i.e. C and A are not the same value.

Kev

Offline superjames

Senior Newbie





« Reply #7 - Posted 2006-07-13 03:43:27 »

I really hate questions where the "clever" bit is to interpret the english language of the questioner - which obviously can be typoed or just poor use

More to the point this response:

Quote
nice try but if A is equal to C then A cannot not be equal to C

doesn't seem to negate this answer:

Quote
Option 2:
A = B
B = A + C

thus: B = B + C
thus: C = 0
so that B != C (if A and B are not 0)

Where A and B can be the same non-zero value as long as C is zero. i.e. C and A are not the same value.

Kev


existence is not a property, and referring to it as a property confuses the distinction between a concept of something and the thing itself

Offline Riven
« League of Dukes »

JGO Overlord


Medals: 816
Projects: 4
Exp: 16 years


Hand over your head.


« Reply #8 - Posted 2006-07-13 06:32:40 »

to be or not to be, that's the question

Let me rephrase that for you: get a life



This is a Java forum, not some place for semi-elevated enlightened wannabe philosophers. Kiss


And Kev, Option 2 is not 'an option' as he didn't seem to mean + but =.
In that case the set of conditions is just wrong, instead of a paradox.
It's like saying, if 13 = 14, and we assume they are equal, they cannot be equal.

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Offline superjames

Senior Newbie





« Reply #9 - Posted 2006-07-13 07:37:11 »

to be or not to be, that's the question

Let me rephrase that for you: get a life



This is a Java forum, not some place for semi-elevated enlightened wannabe philosophers. Kiss


Bitterness, my friend, is the enemy of all...
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Offline purpleguitar

Junior Duke





« Reply #10 - Posted 2006-07-13 12:42:56 »

If A is B, but B is A and C, then assuming that A is not C  is it possible that [snip]

That's all you need to show that your antecedent is a contradiction.  Convert this into first-order logic, and it's simply a conjunction of equalities.  I'll use a plus for conjunction since this board doesn't seem to support LaTeX's \wedge. 

(A=B) + (B=A) + (B=C) + (A!=C) -> [anytthing]

The left hand side is clearly a contradiction:
A=B + B=C <=> A=C
which yields A=C + A!=C in the antecedent.

Now, once you have a contradiction as the antecedent of a material impliciation, you can imply anything at all.  Remember the truth table for material implication:

A  B  A->B
T  T  T
T  F  F
F  T  T
F  F  T

Hence, the consequent is true.  The general scientific rule here is that if your assumptions contain a contradiction, you can prove absolutely anything.  Insert discussion of evolution vs. creationism here.

The author is probably trying to be clever by inserting an ambiguous "or" in the consequent.  If this is inclusive, then we're done.  If it is exclusive, then only one of the components of the disjunction can be true.  However, once we've shown a contradiction in the antecedent, pragmatic philosophers wouldn't care.

I wouldn't call it armchair philosophy:  I would call it a mean homework assignment from a discrete maths course.
Offline superjames

Senior Newbie





« Reply #11 - Posted 2006-07-13 13:35:32 »

If A is B, but B is A and C, then assuming that A is not C  is it possible that [snip]

That's all you need to show that your antecedent is a contradiction.  Convert this into first-order logic, and it's simply a conjunction of equalities.  I'll use a plus for conjunction since this board doesn't seem to support LaTeX's \wedge. 

(A=B) + (B=A) + (B=C) + (A!=C) -> [anytthing]

The left hand side is clearly a contradiction:
A=B + B=C <=> A=C
which yields A=C + A!=C in the antecedent.

Now, once you have a contradiction as the antecedent of a material impliciation, you can imply anything at all.  Remember the truth table for material implication:

A  B  A->B
T  T  T
T  F  F
F  T  T
F  F  T

Hence, the consequent is true.  The general scientific rule here is that if your assumptions contain a contradiction, you can prove absolutely anything.  Insert discussion of evolution vs. creationism here.


Incorrect
Offline Riven
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JGO Overlord


Medals: 816
Projects: 4
Exp: 16 years


Hand over your head.


« Reply #12 - Posted 2006-07-13 13:50:20 »


Hi, appreciate more people! Σ ♥ = ¾
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