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I know I am making a mistake somewhere but consider the following truth table:

p0   p1   p2  |
0    0    0   | 1  
0    0    1   | 0  
0    1    0   | 0  
0    1    1   | 1  
1    0    0   | 1  
1    0    1   | 1  
1    1    0   | 1  
1    1    1   | 1

This coresponds to the boolean formula in DNF:

p0 v (~p1 ^ ~p2) v (p1 ^ p2)

or in CNF:

(p0 v ~p1 v p2) ^ (p0 v p1 v ~p2)

Using the last two clauses and applying the resolution inference rule on p1:

(p0 v ~p1 v p2) ^ (p0 v p1 v ~p2) |- (p0 v p2 v ~p2) = (p0 v T) = p0

So from the two clauses I can infer p0. This does not seem right since p0 does not always have to be true according to the truth table.

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up vote 6 down vote accepted

Your mistake is that $(p_0 \vee T) = T$ and not $p_0$ as you wrote.

In fact it is provable from the empty set that $(p_0 \vee p_2\vee\neg p_2)$.

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You are right. What a newbie mistake. My head was spinning for the last hour :) – attractor Mar 7 '11 at 21:30

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