Truth table and proving a tautology

  • Thread starter arnold28
  • Start date
  • #1
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Homework Statement


I'm supposed to prove that this statement is a tautology (with truth table):

(p -> ((p OR q) AND ~q)) = (~p OR ~q) AND p


Homework Equations





The Attempt at a Solution



Code:
p | q | ~p | ~q | p OR q | (p OR q) AND ~q | p -> ((p OR q) AND ~q) | (~p OR ~q) | (~p OR ~q) AND p
----------------------------------------------------------------------------------------------------
1   1    0    0      1            0                      0                 0               0
1   0    0    1      1            1                      1                 1               1
0   1    1    0      1            0                      1                 1               0
0   0    1    1      0            0                      1                 1               0

But here cleary is that it is not tautology, because left side of equation gets 0 1 1 1 and right side 0 1 0 0

So where did i make mistake, because i cant see it
 

Answers and Replies

  • #2
D H
Staff Emeritus
Science Advisor
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Are you sure you have the parentheses right in the expression that is to be proven?
 
  • #3
14
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yes, it is just like that in the excercise paper
 
  • #4
14
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so do you think that I have done it right and the excercise is badly worded?
 

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