Posted by: dynamite October 16, 2009
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assuming bnot = not b
a.b.c + a.notb.a + a.notb.c (distributive)
a.b.c + a.notb + a.notbc ( a.a = a)
a.c.b + a.c.notb +a.notb (re-arranging)
ac(b+ notb) + a.notb ( distributive)
b+ notb = true
therefore ac + a.notb
a.b.c + a.notb.a + a.notb.c (distributive)
a.b.c + a.notb + a.notbc ( a.a = a)
a.c.b + a.c.notb +a.notb (re-arranging)
ac(b+ notb) + a.notb ( distributive)
b+ notb = true
therefore ac + a.notb
Last edited: 16-Oct-09 11:36 AM