Demonstrate that p ->q is equivalent to (p^q')'.

This is what i got so far: T = true, F= false

p q p -> q q'

T T T F

T F F T

F T T F

F F T T

now what?? grrr

any help is appreciated thanks!

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- Sep 10th 2008, 06:11 AMNeedHelp18Logic -- have no idea how to finish this!
Demonstrate that p ->q is equivalent to (p^q')'.

This is what i got so far: T = true, F= false

p q p -> q q'

T T T F

T F F T

F T T F

F F T T

now what?? grrr

any help is appreciated thanks! - Sep 10th 2008, 09:21 AMPlato