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JackRyan
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Discrete Math: prove B intersection A = A, given A-B = null set
1. Problem Statement:
Prove B [itex]\cap[/itex] A = A, given A-B = ∅ (empty set)
xε(B[itex]\cap[/itex]A) => xεB and xεA => Logic given A-B = ∅ => xεA
I tried using A-B = A[itex]\cap[/itex]!B for xε(A[itex]\cap[/itex]!B)=∅ => xεA and x not in !B or x not in A and Xε!B
I am unsure how to fill in that logic section and prove that B[itex]\cap[/itex]A=A
1. Problem Statement:
Prove B [itex]\cap[/itex] A = A, given A-B = ∅ (empty set)
The Attempt at a Solution
xε(B[itex]\cap[/itex]A) => xεB and xεA => Logic given A-B = ∅ => xεA
I tried using A-B = A[itex]\cap[/itex]!B for xε(A[itex]\cap[/itex]!B)=∅ => xεA and x not in !B or x not in A and Xε!B
I am unsure how to fill in that logic section and prove that B[itex]\cap[/itex]A=A
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