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!

Printable View

- Sep 10th 2008, 05:11 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, 08:21 AMPlato