I am trying to self study Probability Theory. Below is an equation that came up from the chapter about Plausible Reasoning.
The goal is to prove that below equation can be deduced to the following: C = ( B $ \Rightarrow \neg A$)
Equation: Original Equation directly from the book
$C \equiv (A \vee \neg B)(\neg A \vee A \wedge \neg B) \vee \neg A \wedge B ( A \vee B) $
Any idea helps!
Thanks in advanced!
Edit: Thank you for pointing out that + is not a logical operation. That only shows that I am really unfamiliar with such kind of language.
1. The book that I am reading is Probability Theory: The Logic of Science by dwin Thompson Jaynes.
- My take is embarrassing to show to the internet. But since you suggested, here is what I have:
$(A \vee \neg B)(\neg A \vee A \wedge \neg B) \vee \neg A \wedge B ( A \vee B)$
$= (A \neg A) \vee (AA \neg B ) \vee (\neg B \neg A) \vee (\neg B A \neg B) \vee (\neg A B A ) \vee ( \neg A BB) $
$= 0 \vee (A \neg B) \vee (\neg B \neg A ) \vee (\neg B) \vee 0 \vee (\neg A B) $
= no clue from there..
Edit 2: I have included actual equation from the book because a lot of professionals in this thread seems to be confused.
Thank you for your help!
Assuming $ + = \vee$
Let $$\begin{align} X = B \Rightarrow \neg A & = \neg B \vee \neg A \\ \end{align}$$
and $$\begin{align} Y & = (A \vee \neg B)\wedge ( \neg A \vee A \wedge \neg B) +\neg A\wedge B \wedge(A \vee B)\\ & = A \wedge \neg B \vee \neg A \wedge\neg B \vee A \wedge \neg B \vee \neg A \wedge B \\ &= \neg A \wedge B \vee A \wedge B \vee A \wedge B \\ & = B \vee A\wedge B \\ & = (\neg B \vee \neg A) \wedge (\neg B \vee B) \\ & = \neg A \vee \neg B \\ & = X \end{align}$$
Note: For the sake of simplicity, I have omitted the step like $ X \wedge \neg X = 0 $