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It is given that

  1. $SA = B \land \neg C$
  2. $SB =A \to C$
  3. $SC = \neg C \land (A \lor B)$

How do I get $(SA \land SB \land SC) \to (B \land \neg A \land \neg C)$?

I did this (below) but now I'm stuck.

$$B \land \neg C \land A \to C \land \neg C \land (A \lor B) \\ = \neg C \land (B \land (A \lor B)) \land \neg A \lor C$$

Lee David Chung Lin
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llamaro25
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  • You can simplify the last expression still more, e.g. $B \wedge (A\vee B) = B$. What about the $\neg A \vee C$? Also you have only simplified $SA\wedge SB \wedge SC$ so far. Next you can apply $X\to Y \iff \neg X \vee Y$ to the whole expression. – Ben Sep 03 '19 at 15:51
  • why is B∧(∨)=? – llamaro25 Sep 03 '19 at 15:52
  • $B \wedge (A\vee B)$ is true iff $B$ is true and $B$ or $A$ is true. Because the first assumption is already that $B$ is true, then $A\vee B$ doesn't add any extra conditions - $B$ is true so $A\vee B$ is true as well, regardless of $A$. I'm sure this rule has a name, but I don't recall it. – Ben Sep 03 '19 at 15:57
  • so like this we can conclude that ¬C∨C is true so ¬C∧ B ∧ ¬A? – llamaro25 Sep 03 '19 at 15:59
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    @Ben It's called absorption. – Vsotvep Sep 03 '19 at 16:00
  • @llamaro25 I don't quite follow your explanation but yes that's the simplification I was thinking of, $\neg C \wedge (\neg A \vee C) = \neg C \wedge \neg A$. – Ben Sep 03 '19 at 16:05
  • @llamaro25 When you have come up with the full solution I encourage you to answer your own question :) – Ben Sep 03 '19 at 16:22

1 Answers1

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So you have:

$\begin{align}&(B\wedge\neg C)\wedge(A\to C)\wedge(\neg C\wedge (A\vee B)) \\&(B\wedge\neg C)\wedge(\neg A\vee C)\wedge(\neg C\wedge(A\vee B))&&\text{by conditional equivalence}\\&\neg C\wedge\neg C\wedge B\wedge (A\vee B)\wedge (\neg A\vee C)&&\text{by association and commutivity}\\&\neg C\wedge B\wedge (A\vee B)\wedge (\neg A\vee C)&&\text{by idempotence}\end{align}$

That checks out okay. So, now use absorption: $B\wedge (A\vee B)~=~B$ $$\begin{align}&B\wedge (A\vee B)\\&(B\vee \bot)\wedge(B\vee A)\\&B\vee(\bot\wedge A)\\&B\vee\bot\\&B\end{align}$$

Next use distribution, and the path should be clear from then on.

Graham Kemp
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