- #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