sep-2-17_00
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@@ -13,3 +13,10 @@ AuB = {1, 2, 3, 5, 7, 9}
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A↓uB = {3, 7}
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A\B = {1, 9}
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A"delt"B = {1, 2, 5, 9}
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Beviska:
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1) ¨A↓u(AuB) = B\A
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2) (AuB)↓u¨(A↓uB) = (A↓u¨B)u(B↓u¨A)
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@@ -19,3 +19,7 @@ Räkne regler:
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x € ¨(AuB) => x !€ AuB => {x !€ A, x !€ B => x !€ {x € ¨A, x € ¨B => x € ¨(AuB). Visar att ¨(AuB) …c A↓uB.
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x € ¨A↓u¨B => {x € ¨A, x € ¨B => {x !€ A, x !€ B => x !€ AuB => x € ¨(AuB). Visar att ¨A↓uB …c ¨(AuB).
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Resultat: ¨(AuB) = ¨Au¨B
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- Distributiv: (AuB)↓uC <=> (A↓uC)u(B↓uC), (A↓uB)uC <=> (AuC)↓u(BuC)
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- Absorbering: A↓u(AuB) = A, Au(A↓uB) = A
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- Differens: A\B <=> A↓u¨B
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