Prove-that-If-A-B-and-C-are-subset-of-the-same-universal-set-then-A-B-A-C-A-B-C-

Question Number 7440 by Tawakalitu. last updated on 29/Aug/16

P r o v e t h a t

I f A, B a n d C a r e s u b s e t o f t h e s a m e u n i v e r s a l s e t

t h e n (A - B) \cap (A - C) = A - (B - C)

Commented by Yozzia last updated on 29/Aug/16

L e t A = {1, 2, 3, 4}, B = {2, 3, 4, 5}, C = {3, 4, 5, 6} .

\Rightarrow A - B = {1}, A - C = {1, 2}

\Rightarrow (A - B) \cap (A - C) = {1} .

B - C = {2}

\Rightarrow A - (B - C) = {1, 3, 4} \neq {1} = (A - B) \cap (A - C) .

T h e c o r r e c t i d e n t i t y i s (A - B) \cap (A - C) = A - (B \cup C) .

Commented by Rasheed Soomro last updated on 29/Aug/16

C a n w e s a y (A - B) \cap (A - C) = A - (B \cup C) a s

o n e o f t h e t w o d e {m a r g a n}^{'} s l a w s ? O t h e r l a w m a y b e

(A - B) \cup (A - C) = A - (B \cap C) ?

Commented by Yozzia last updated on 29/Aug/16

I o n l y k n o w (\sim a) \land (\sim b) =\sim (a \lor b) a n d (\sim a) \lor (\sim b) =\sim (a \land b)

a s D e {M o r g a n}^{'} s l a w s i n l o g i c . T h e r e s u l t s y o u

s p e a k o f m a y h a v e n o {o n e}^{'} s n a m e a t t a c h e d

t o t h e m, a s t h e y s t e m f r o m d e f i n i t i o n s

a n d m a y n o t b e v e r y f r e q u e n t l y u s e d .

M y t e x t o n a b s t r a c t a l g e b r a d o e s n o t

a s s o c i a t e {a n y o n e}^{'} s n a m e t o t h e s e r e s u l t s

i n v o l v i n g t h e s e t d i f f e r e n c e .

Commented by Tawakalitu. last updated on 29/Aug/16

T h a n k y o u s i r, i u n d e r s t a n d .

Commented by Rasheed Soomro last updated on 30/Aug/16

(A - B) \cap (A - C) = A - (B \cup C)

(A - B) \cup (A - C) = A - (B \cap C)

A n y w a y t h e a b o v e l a w s l a w s h a v e c l o s e

c o n n e c t i o n w i t h d e - m a r g a n l a w s .

I f w e l e t A b e u n i v e r s a l s e t U t h e n

(A - B) \cap (A - C) = A - (B \cup C) \Rightarrow (U - B) \cap (U - C) = U - (B \cup C)

\Rightarrow B^{'} \cap C^{'} = {(B \cup C)}^{'} \Rightarrow {(B \cup C)}^{'} = B^{'} \cap C^{'}

S i m i l a r l y (A - B) \cup (A - C) = A - (B \cap C) \Rightarrow {(B \cap C)}^{'} = B^{'} \cup C^{'}

I t s e e m s t h a t t h e a b o v e l a w s a r e g e n e r a l a n d

d e - m a r g a n l a w s a r e s p e c i a l c a s e s! W h e n B, C \subseteq A a n d

a n d A i s r e c o g n i z e d a s U (U n i v e r s a l S e t) .

I t h i n k I r e a d a n y w h e r e i n S e t t h e o r y t h a t t h e a b o v e l a w s

a r e “ D e m a r g a n l a w s^{″} . A l t h o u g h I c a n n o t c o n f i r m i t

n o w .

Commented by Yozzia last updated on 30/Aug/16

H m m \dots i n t e r e s t i n g . T h a n k s f o r s h a r i n g!

Commented by Tawakalitu. last updated on 31/Aug/16

I r e a l l y a p p r e c i a t e y o u r e f f o r t s i r s