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I'm having some difficulty with a problem I think.

If X = true and Y = true

((X AND Y)' AND (X' OR Y') ' ) '

I get back true. When I put it in Wolfram Alpha it gave me back false. But I think it may have had a ' by it as well? So I'm not really sure. I'm new to this and hoping for some clarification. My thinking was that:

((TRUE AND TRUE) ' AND (TRUE' OR TRUE') ' ) '

((FALSE AND FALSE) AND (FALSE OR FALSE) ' ) '

((FALSE) AND (FALSE) ' ) '

((FALSE) AND (TRUE)) '

((FALSE)) '

((TRUE))

Can someone please tell me if this is correct?

pawello2222
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asp1124
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  • Please provide a link or show us the input for Wolfram Alpha which was evaluated to false. – Hero Wanders Jan 21 '21 at 00:28
  • I have entered your expression into Wolfram Alpha and it shows me T as the result for X = Y = T: https://www.wolframalpha.com/input/?i=NOT+%28NOT+%28X+AND+Y%29+AND+NOT+%28NOT+X+OR+NOT+Y%29+%29+ – Hero Wanders Jan 21 '21 at 00:32

1 Answers1

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The ultimate result TRUE is correct, your calculation is wrong though.

If ' is the boolean negation which turns true to false and vice versa, you have made a mistake: you did not apply De Morgan's laws.

(A AND B)' = A' OR B'
(A OR B)' = A' AND B'

In particular

((FALSE) AND (TRUE))' = FALSE' OR TRUE'

which is still TRUE.

Complete simplification of the expression is possible without making use of that law though, just by knowing how AND and OR are defined for two given booleans:

((TRUE AND TRUE)' AND (TRUE' OR TRUE')')' =
(TRUE' AND (FALSE OR FALSE)')' =
(FALSE AND FALSE')' =
(FALSE AND TRUE)' =
FALSE' =
TRUE

We can generalize even further: for any X and Y it is (now using the aforementioned law):

((X AND Y)' AND (X' OR Y')')' =
((X AND Y)' AND (X AND Y))' =
(Z' AND Z)' =
FALSE' =
TRUE

(with Z = X AND Y)

So independent of how you choose X and Y, the result is TRUE.

Hero Wanders
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  • Ok thanks for the reply. And yes, the apostrophe is supposed to be negation in the system I was using. I think I'm a little confused, please excuse me. It looks like what I wrote out ultimately was the same as the simplification that you wrote out at the end. Here's what I wrote in the calculator: ((X AND Y)' AND (X' OR Y')')' The result I got was False ' (X). Does that mean it's actually True and I read it wrong? – asp1124 Jan 21 '21 at 01:32
  • `False ' (X)` seems to be equal to `True (X)` which might have the meaning "true, independent of what X is". If you have problems interpreting your calculator's output, you should refer to its manual or ask a separate question for that. Wolfram Alpha's answer is formatted differently but it says the same: https://www.wolframalpha.com/input/?i=NOT+%28NOT+%28X+AND+Y%29+AND+NOT+%28NOT+X+OR+NOT+Y%29+%29+ – Hero Wanders Jan 21 '21 at 06:17