Is 0/10 a reasonable fraction
2 answers:
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Answer:
hello your question is incomplete below is the missing parts
(a) A\ (A\B) = B\(B\A)
(b) A\ (BA) = B\(A\B)
answer : A\ (A\B) = B\(B\A) = always true
A\ (BA) = B\(A\B) = sometimes true and sometimes false
Step-by-step explanation:
(a) A\ (A\B) = B\(B\A). = ALWAYS TRUE
using de Morgan's law to prove this
A\ (A\B) = A\ ( A ∩ B^c )
= A ∩ ( A^C ∪ B )
= ( A ∩ A^C ) ∪ ( A ∩ B )
= Ф ∪ ( A ∩ B )
= ( A ∩ B )
ALSO : B\(B\A) = attached below is the remaining parts of the solution
B) A\ (BA) = B\(A\B) = Sometimes true and sometimes false
attached below is the prove using De Morgan's law
