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$(¬a∨b)∧(a∨b)∧¬a$

So I have been looking at this question all day and and i have no idea how to start.

Can someone please help with me proving this algebra logic and what laws I would need to use?

I have also looked through already answered questions, and could not find anything similar.

Simplify the following statement using the laws and axioms of logic

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closed as off-topic by Graham Kemp, DMcMor, Yanior Weg, Leucippus, GNUSupporter 8964民主女神 地下教會 Apr 22 at 23:06

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    $\begingroup$ Hello. That's not an equation, it's a formula of propositional logic, and it's not tautological, therefore it can't be proven. Also: I'm not sure what you mean by "simplify"; do you mean having a logically equivalent formula with only one logical operator? $\endgroup$ – Simone Apr 21 at 11:53
  • $\begingroup$ Use the rules of commutation, absorption, and redundancy. $\endgroup$ – Graham Kemp Apr 21 at 12:08
  • $\begingroup$ Thank you for helping me. but for me to understand this can you please show me the step thanks $\endgroup$ – Daisoh Apr 22 at 5:52
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This expression is a "formula" because it has a well-defined mathematical value (for each allowable set of values for the variables in it). It is a "statement" because that mathematical value is either "true" or "false". But an "equation" is a statement claiming that two expressions are equal to each other (equal $\equiv$ equation), and there is no "=" sign in this statement. Therefore it is not an "equation".

Examine when your statement will be true. It has three parts joined together by ands: $$(\lnot a\lor b)\\(a\lor b)\\\lnot a$$

For the whole expression to be true, all three must be true. But the last one requires $a$ to be false. And if $a$ is false, the first will be true, regardless of $b$. Finally $a\lor b$ requires either $a$ or $b$ to be true. But we already know $a$ has to be false, so $b$ has to be true.

Therefore, this will be true only if $a$ is false and $b$ is true.

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  • $\begingroup$ hi can you please help me this question clarify and simplify. what law i need to use to simplify. $\endgroup$ – Daisoh Apr 22 at 5:51
  • $\begingroup$ No. I abhor it when people would prefer to mindlessly apply mechanical methods rather than think about what the mathematics actually means. I walked you the simplification and explained exactly why it is the case. That is all I intend to do. $\endgroup$ – Paul Sinclair Apr 23 at 0:00

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