December 25th, 2016, 04:10 PM  #1 
Newbie Joined: Mar 2016 From: Alabama Posts: 18 Thanks: 0  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.) 
December 25th, 2016, 04:20 PM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,293 Thanks: 442 Math Focus: Yet to find out. 
$y = \overline{a} + x$ There may be some relation to simplify the above. But I can't currently recall of such a thing. 
December 25th, 2016, 04:33 PM  #3 
Newbie Joined: Mar 2016 From: Alabama Posts: 18 Thanks: 0 
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 04:40 PM. 
December 25th, 2016, 04:40 PM  #4 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,293 Thanks: 442 Math Focus: Yet to find out. 
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.

December 25th, 2016, 05:04 PM  #5 
Newbie Joined: Mar 2016 From: Alabama Posts: 18 Thanks: 0 
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? 
December 25th, 2016, 05:37 PM  #6 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,293 Thanks: 442 Math Focus: Yet to find out. 
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, $y = \overline{(a \cdot \overline{b})} = \overline{(a \cdot \overline{x})}$ Last edited by Joppy; December 25th, 2016 at 05:44 PM. 
January 2nd, 2017, 11:30 PM  #7 
Senior Member Joined: Apr 2014 From: UK Posts: 769 Thanks: 290  
January 2nd, 2017, 11:35 PM  #8 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,293 Thanks: 442 Math Focus: Yet to find out.  

Tags 
boolean, logic, problem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Boolean algebra simplification using boolean rules  JustForFun2000  Computer Science  2  September 26th, 2016 09:11 AM 
Derive a Boolean expression from logic circuit and simplify  unwisetome3  Applied Math  4  February 17th, 2014 05:00 AM 
Boolean logic combining properties rule  unwisetome3  Applied Math  3  February 2nd, 2014 01:09 PM 
Help please with a logic problem  FSU_91  Applied Math  1  May 25th, 2013 02:51 PM 
2 simple boolean algebra questions (logic)  OriaG  Applied Math  2  April 20th, 2013 05:35 PM 