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This is what I have so far, so I just want to know if I am doing everything right or wrong. I am novice in the topic.

I have been given:

$(A \oplus B) \land \neg A \lor A$

  • XOR = $\oplus$

By order of operations regroup:

$(A \oplus B ) \land (\neg A \lor A)$ // XOR

(~A B + A ~B) * ~A + A // Distributive law

(~A ~A B + ~A A ~B) + A // Idempotent Law

~A B + A ~A ~B + A // Inverse Law

~A B + 0 ~B + A

~A B + A

Is this statement right?

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  • $\begingroup$ Where has $Z$ gone? Where is $\lnot A \lor A$ from? $\endgroup$ – hengxin Mar 3 '16 at 1:47
  • $\begingroup$ My bad, already switch that Z for A instead. So you can have a look at it $\endgroup$ – Fred Mar 3 '16 at 9:20
  • $\begingroup$ We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. $\endgroup$ – D.W. Mar 4 '16 at 0:32
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Note that $\land$ has higher precedence than $\lor$ (see order of precedence@wiki). Therefore,

$$(A \oplus B) \land \lnot A \lor A = ((A \oplus B) \land \lnot A) \lor A = (((\lnot A \land B) \lor (A \land \lnot B)) \land \lnot A) \lor A$$

Then iteratively apply the distributive law and you will get $A \lor B$.

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  • $\begingroup$ I did the correction based in that point. Is this now valid? $\endgroup$ – Fred Mar 3 '16 at 10:13
  • $\begingroup$ I got this, you finally apply Law 11. $\endgroup$ – Fred Mar 3 '16 at 10:21

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