Sample Homework Problems

Numbers 9, 11, 12 , and 14 A\(B\C) = (A\B)∪(A∩C)

Proof: we show that x ∈ LHS <=> x ∈ RHS x ∈ LHS <=> x ∈ A\(B\C) <=> (x ∈ A) ^ ¬(x ∈ B\C) <=> (x ∈ A) ^ ¬(x ∈ B ^ X ∈ C) <=> (X ∈ A \ B)∪(X ∈ A)∩(X ∈ C) <=> x ∈ (A\B)∪(A ∩ C) <=> x ∈ RHS

Proof: We show LHS ⊂ RHS and RHS ⊂ LHS LHS ⊂ RHS : Let x ∈ LHS = (A ∩ B)\C. Then x ∈ A ∩ and x ∈ C. Since A ∈ A ∩ B, then X ∈ A and x ∈B. Hence X ∈ A, X ∈ B, and X ∈ X. X ∈ A\C and X ∈ B\C, so X ∈ (A\C)∩(B\C) = RHS

RHS ⊂ LHS : Let X ∈ RHS = (A\C)∩(B\C). Then X ∈ A\C and x ∈ B\C, so X ∈ A, X∉ and x ∈ B, x ∉ C. So X ∈ A and X ∈ B and X∈B ∩ B. Since also uses C, X ∈(A ∩ B)\C = LHS. Since LHS ⊂ RHS and RHS ⊂ LHS, we conclude that LHS = RHS.

Proof By Drawing

![[scrnli_10_11_2023_7-42-02 PM.png]]

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Problem #2

(A∩B)\C = (A\C)∩(B\C) Proof: We show that LHS ⊂ RHS and RHS ⊂ LHS. LHS ⊂ RHS : Let X ∈ LHS = (A∩B)\C. Then X ∈ A∩B and X ∈ C. Since X ∈ A ∩ B, X ∈ A and X ∈ B. Hence X∈A,X∈B, and X∉C, X ∈ A\C abd X ∈ B\C, So x∈(A\C)∩(B\C) = RHS.

RHS ⊂ LHS: Let ∈ RHS = (A\C)∩(B\C). Then X∈A\C and X∈ B\C, so x∈A,X∉C and X∈B, X∉ C. So X∈A and X∈B and X∈A∩B. Since also X∉C, X∈(A∩B)\C = RHS. Since also X∉C, X∈(A∩B)\C = RHS.