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Boolean algebra and logic gate help

Hey guys I'm new to this boolean algebra and logic gates module and was wondering if someone could help me with this question? Any help will be appreciated.

(edited 9 years ago)

Reply 1

Algebraholic

So the first logic gate at the top is a NAND gate- for inputs it has A and B- so this is 'A NAND B'
The gate underneath it is a NOR gate- for inputs it has B and C- so this is 'B NOR C'
The output from these two gates are the inputs to a second NAND gate- so this is (A NAND B) NAND (B NOR C)
The output from this gate becomes one of the the inputs to a second NOR gate- the other input being C
So the system is...
[ (A NAND B) NAND (B NOR C) ] NOR C
This can be simplified by first of all getting rid of the last NOR I mentioned.
Recall that 'A NOR B' means ¬(A OR B) and by DeMorgans law is the same as ¬A AND ¬B
So we have ¬[ (A NAND B) NAND (B NOR C) ] AND ¬C
So what does 'A NAND B' mean? Apply DeMorgans Law to this- what is it the same as? Now apply this to the line of Boolean Algebra above.