Truth table and proving a tautology

[arnold28]

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

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 can't see it

 

[D H]

Are you sure you have the parentheses right in the expression that is to be proven?

 

[arnold28]

yes, it is just like that in the exercise paper

 

[arnold28]

so do you think that I have done it right and the exercise is badly worded?