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December 25th, 2016, 05:10 PM   #1
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Boolean logic problem

Given:

x=b
y=a implies b (i.e. not a or b)

Problem: Express a in terms of x and y without using a or b.

(I don't know whether a solution exists.)
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December 25th, 2016, 05:20 PM   #2
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$y = \overline{a} + x$

There may be some relation to simplify the above. But I can't currently recall of such a thing.
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December 25th, 2016, 05:33 PM   #3
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That's the thing. It amazes me that even after expressing that relation, there seems no way to isolate 'a' on one side of the '=' operator AFAICT. Is knowledge of 'a' really lost even if you have full knowledge of 'b' and '~a.b'?

Last edited by zambod; December 25th, 2016 at 05:40 PM.
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December 25th, 2016, 05:40 PM   #4
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If you think about it in terms of inputs and outputs, it might help. You have two outputs, and only 2 inputs. Draw a circuit diagram with x and y as outputs, and a and b as inputs.
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December 25th, 2016, 06:04 PM   #5
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Yeah, from a and b, I got x and y. The next step was to treat x and y as inputs and get a as an output. This is impossible:

a b x y
0 0 0 1
0 1 1 1
1 0 0 0
1 1 1 1

The lines in the truth table represent the conditions where, for the original inputs a and b, only one is true (middle two), both are true (last) and none are true (first).

It is easy to write a formula for when a is true and b is false using x and/or y: ~y. Now all we need to do is OR '~y' with a formula for 'a&b' to get 'a'. Unfortunately, x and y do not behave differently between the cases where a and b are both true and where b is true but a is false. Both the second and fourth lines read '1 1' for columns x y in the table.

This is the piece of information I can't recover to reconstruct a from x and y alone. Did that make sense?
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December 25th, 2016, 06:37 PM   #6
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I have to ask where the problem comes from.. More specifically, what are a, b, x and y exactly. Suppose a is an input, why would you want to express it explicitly in terms of two output variables, one of which (x) is equivalent to one of the input variables (b)?

If you don't want to use the OR operator, you can express y as,

$y = \overline{(a \cdot \overline{b})} = \overline{(a \cdot \overline{x})}$
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Last edited by Joppy; December 25th, 2016 at 06:44 PM.
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January 3rd, 2017, 12:30 AM   #7
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Quote:
Originally Posted by zambod View Post
y=a implies b (i.e. not a or b)
Can someone tell me what this means?
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January 3rd, 2017, 12:35 AM   #8
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Quote:
Originally Posted by weirddave View Post
Can someone tell me what this means?
I'm no longer sure if it has any meaning.
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