Math Is Fun Forum

ilovealgebra

Member

Registered: 2006-10-02

Posts: 40

Logic (easy)

Hey guys.

A for “Min is at home”
B for “Min is on board”
C for “Henry is at home”, and
D for “Henry is on board”.

For...
(iii) Either both Min is on board and Henry is on board or neither is on board.

I was wondering whether (B ∧ D) ∨ (¬B ∧ ¬D) [ model answer ]  is the same as (B ∧ D) ∨ ¬(B ∧ D) [my answer]

This may be really obvious but i'm just making sure cheers for the help guys

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"...nothing physical which sense-experience sets before our eyes, or which necessary demonstrations prove to us, ought to be called into question (much less condemned) upon the testimony of biblical passages."
-Galileo Galilei

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Logic (easy)

Not quite. If you want to take a NOT out of brackets like that, you need to switch the sign inside the bracket.

So (B ∧D) ∨ ¬(B ∨ D) is the same as the model answer.

Your answer is actually always true, since it's in the form <event> OR (NOT <event>)

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Why did the vector cross the road?
It wanted to be normal.

ilovealgebra

Member

Registered: 2006-10-02

Posts: 40

Re: Logic (easy)

Thanks for the reply

Also, would that make ¬(B ∨ D) equivalent to (¬B ∧ ¬D)

Last edited by ilovealgebra (2009-11-22 10:47:44)

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"...nothing physical which sense-experience sets before our eyes, or which necessary demonstrations prove to us, ought to be called into question (much less condemned) upon the testimony of biblical passages."
-Galileo Galilei

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Logic (easy)

Yes, that is De Morgan's Law and it can be proved with venn diagrams
http://en.wikipedia.org/wiki/De_Morgan's_laws

Last edited by Identity (2009-11-22 11:31:48)