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De Morgan's Law - Propositional Calculus

Can someone tell me what's the correct way to solve this proposition with De Morgan's Law?

Proposition: $q \vee ¬[ (p \wedge q) \vee ¬q ]$

[Answers]

Option 1:

$q \vee ¬(p \wedge q) \wedge q$

Option 2:

$q \vee [¬(p \wedge q) \wedge q ]$

As you can see, the difference between both is that one of them don't have square brackets "[ ]". So, what's the correct one?

md2perpe
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3 Answers3

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If you consider $\land$ as having higher precedence that $\lor$ (just like $\cdot $ has higher precedence than $+$), then $$ q \vee (¬(p \wedge q) \wedge q )\equiv q \vee ¬(p \wedge q) \wedge q $$ in the same way as $$ q + \left((p \cdot q)^2 \cdot q \right)=q + (p \cdot q)^2 \cdot q.$$ If you do not define such precedence, then the extra brackets are needed (as would be differently placed brackets for any alternative interpretation).

Hagen von Eitzen
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The second version is correct because you are just applying DeMorgan to the second part. You can simplify it further, but I don't see that as part of the question. If the first $q$ were $r$ throughout, it would matter how you grouped the three terms at the end.

Ross Millikan
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  • @Arthur: you are right. I had noticed that all the clauses were joined by $\vee$ at the start, but missed that it changed at the end, as it should because of DeMorgan – Ross Millikan Jul 11 '19 at 19:23
  • Ross Millikan, thanks for the observation, I edited it. Thanks again and sorry. –  Jul 11 '19 at 19:26
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Because of how the original expression you can use replacement of variables and rewrite it as $q \vee \neg A$ where A is the expression in braces, because of this idea option 2 is the correct one

Sam
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