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Is the following statement equivalent: $((A \vee \neg C) \wedge ((\neg B) \leftarrow C)) \vee \neg (A \wedge B) \equiv (\neg A \vee \neg B \vee \neg C) $

Our prof gave us an exercise with this statement and said we should simplify it with truth table. Could that be the right solution? Hope somebody can help.

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    $\begingroup$ Did you try the truth table approach? With three variables, it'll be fairly simple. $\endgroup$ – Rick Decker Feb 13 '16 at 19:57
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Here is a proof by boolean calculation.

We have $¬B ← C ≡ ¬B ∨ ¬C$, so $(A ∨ ¬C) ∧ ((¬B) ← C)) ≡ (A ∨ ¬C) ∧ (¬B ∨ ¬C) ≡ (A ∧ ¬B) ∨ ¬C$. We have also $¬(A ∧ B) ≡ ¬A ∨ ¬B$, and hence the original proposition is equivalent to $¬A ∨ ¬B ∨ ¬C ∨ (A ∧ ¬B)$, which is just $¬A ∨ ¬B ∨ ¬C$ since $A ∧ ¬B$ implies $¬B$.

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I hopefully got it right... It seems to work out, but please double check.

Truth table

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  • $\begingroup$ The right hand side gets true if there is one single 1 in the rows. Had some trouble to mark the outcome of the right side in a "smart" way. Therefore the edits, I hope it's correct now. By the way, if you have to make such a truth table and you want to ensure that you got all the combinations of A, B and C, the only thing you have to do is to count from 0 to 7 in binary. $\endgroup$ – Andreas Schliebitz Feb 13 '16 at 22:12

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