The Mathematical Proof Thread

[Kernel Sohcahtoa]

Modification to Part one of Proof#1

If A ,B, and C are sets, then A∩(B∪C)=(A∩B)∪(A∩C)

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[secret]
Let’s first demonstrate that A∩(B∪C) ⊆ (A∩B)∪(A∩C)  IMO, the text in bold adds more clarity to the proof.


Suppose x ∈ A∩(B∪C).  Then by the definition of intersection, x ∈ A and x ∈ B∪C.  Now, by the definition of union, x ∈ B or x ∈ C. So, x ∈ A and x ∈ B or x ∈ A and x ∈ C. Thus, by the definition of intersection, x ∈ (A∩B) or x ∈ (A∩C).  Consequently, by the definition of union, x ∈ (A∩B)∪(A∩C).  Hence, we have shown that x ∈ A∩(B∪C) implies x ∈ (A∩B)∪(A∩C), so it follows that A∩(B∪C) ⊆ (A∩B)∪(A∩C).

Let’s demonstrate that (A∩B)∪(A∩C) ⊆  A∩(B∪C)

Suppose x ∈ (A∩B)∪(A∩C).  Then by the definition of union, x ∈ (A∩B) or x ∈ (A∩C). By the definition of intersection, x ∈ A and x  ∈ B or x ∈ A and x ∈ C.  Thus, x ∈ A and x ∈ B or C.  By the definition of union, x ∈ A and x ∈ (B∪C).  Also, by the definition of intersection, x ∈ A ∩(B∪C).  Hence, we’ve shown that x ∈ (A∩B)∪(A∩C) implies  x ∈ A ∩(B∪C), so it follows that (A∩B)∪(A∩C) ⊆  A∩(B∪C).

In summary, we’ve shown that A∩(B∪C) ⊆ (A∩B)∪(A∩C) and (A∩B)∪(A∩C) ⊆  A∩(B∪C).  Therefore, A∩(B∪C)=(A∩B)∪(A∩C).   Hence, the proof is complete.
 

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