(A △B)∩C = (A ∩C)△(B ∩C) proof | Mathematics | Page 1

idle totem Oct 22, 2023, 9:25 AM

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help me prove that (A △B)∩C = (A ∩C)△(B ∩C) using definitions of the operators,

graceful solarBOT Oct 22, 2023, 9:25 AM

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distant ginkgo Oct 22, 2023, 10:21 AM

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$(A \Delta B)={x:((x \in A ∧ x \notin B) ∨ (x \notin A ∧ x \in B))}$

median windBOT Oct 22, 2023, 10:23 AM

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distant ginkgo Oct 22, 2023, 10:25 AM

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$A \cap B = {x : (x \in A ∧ x \in B)}$

median windBOT Oct 22, 2023, 10:26 AM

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distant ginkgo Oct 22, 2023, 10:28 AM

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$(A \Delta B) \cap C = {x:(((x \in A ∧ x \notin B) ∨ (x \notin A ∧ x \in B)) ∧ x \in C)}$

median windBOT Oct 22, 2023, 10:28 AM

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distant ginkgo Oct 22, 2023, 10:38 AM

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$(A \cap C) \Delta (B \cap C) = {x : (((x \in A ∧ x \in C) ∧ ¬(x \in B ∧ x \in C)) ∨ (¬(x \in A ∧ x \in C) ∧ (x \in B ∧ x \in C)))}$

median windBOT Oct 22, 2023, 10:38 AM

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distant ginkgo Oct 22, 2023, 10:45 AM

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De Morgan’s Law: $(A \cap C) \Delta (B \cap C) = {x : (((x \in A ∧ x \in C) ∧ (x \notin B ∨ x \notin C)) ∨ ((x \notin A ∨ x \notin C) ∧ (x \in B ∧ x \in C)))}$

median windBOT Oct 22, 2023, 10:46 AM

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distant ginkgo Oct 22, 2023, 10:48 AM

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The conjunctions show us that $x \in C$

median windBOT Oct 22, 2023, 10:48 AM

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distant ginkgo Oct 22, 2023, 10:49 AM

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$(A \cap C) \Delta (B \cap C) = {x : (((x \in A ∧ x \in C) ∧ (x \notin B ∧ x \in C)) ∨ ((x \notin A ∧ x \in C) ∧ (x \in B ∧ x \in C)))}$

median windBOT Oct 22, 2023, 10:49 AM

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distant ginkgo Oct 22, 2023, 10:51 AM

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Now we just factor out the $x \in C$ and get: $(A \cap C) \Delta (B \cap C) = {x:(((x \in A ∧ x \notin B) ∨ (x \notin A ∧ x \in B)) ∧ x \in C)}$

median windBOT Oct 22, 2023, 10:52 AM

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distant ginkgo Oct 22, 2023, 11:09 AM

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Rules of inference (not including the most obvious ones):

De Morgan’s Law

((¬P ∨ ¬R) ∧ (Q ∧ R)) ⇔ ((¬P ∧ R) ∧ (Q ∧ R))

((P ∧ R) ∨ (Q ∧ R)) ⇔ ((P ∨ Q) ∧ R)

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@idle totem have fun

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Definitions:

$A \Delta B={x:((x \in A ∧ x \notin B) ∨ (x \notin A ∧ x \in B))}$

$A \cap B = {x : (x \in A ∧ x \in B)}$

Proof:

(1) $(A \Delta B) \cap C = {x:(((x \in A ∧ x \notin B) ∨ (x \notin A ∧ x \in B)) ∧ x \in C)}$

(2) $(A \cap C) \Delta (B \cap C) = {x : (((x \in A ∧ x \in C) ∧ ¬(x \in B ∧ x \in C)) ∨ (¬(x \in A ∧ x \in C) ∧ (x \in B ∧ x \in C)))}$

(3) $(A \cap C) \Delta (B \cap C) = {x : (((x \in A ∧ x \in C) ∧ (x \notin B ∨ x \notin C)) ∨ ((x \notin A ∨ x \notin C) ∧ (x \in B ∧ x \in C)))}$

(4) $(A \cap C) \Delta (B \cap C) = {x : (((x \in A ∧ x \in C) ∧ (x \notin B ∧ x \in C)) ∨ ((x \notin A ∧ x \in C) ∧ (x \in B ∧ x \in C)))}$

(5) $(A \cap C) \Delta (B \cap C) = {x:(((x \in A ∧ x \notin B) ∨ (x \notin A ∧ x \in B)) ∧ x \in C)}$

Rules of inference (not including the most obvious ones):

(2) ⇒ (3): De Morgan’s Law

(3) ⇒ (4): ((¬P ∨ ¬R) ∧ (Q ∧ R)) ⇔ ((¬P ∧ R) ∧ (Q ∧ R))

(4) ⇒ (5): ((P ∧ R) ∨ (Q ∧ R)) ⇔ ((P ∨ Q) ∧ R)

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Each of those inference rules are easy to prove using truth tables

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I hope formal logic is allowed in your assignment @idle totem

median windBOT Oct 22, 2023, 11:21 AM

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distant ginkgo Oct 22, 2023, 1:00 PM

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@idle totem if you have any questions feel free to ask

idle totem Oct 22, 2023, 1:00 PM

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Thank you so much 😮

distant ginkgo Oct 22, 2023, 5:31 PM

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I was motivated enough to invest time into formatting this so I might as well send it in here

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I would expand on it a bit if I needed to send this in, but I think most of the implied steps here are obvious enough as to not necessitate being explicitly stated in a less formal setting

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Prove that
[(A \Delta B) \cap C = (A \cap C) \Delta (B \cap C)]
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Definitions:
[A \Delta B={x:((x \in A ∧ x \notin B) ∨ (x \notin A ∧ x \in B))}]
[A \cap B = {x : (x \in A ∧ x \in B)}]
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Proof:

(1) $(A \Delta B) \cap C = \ {x:(((x \in A ∧ x \notin B) ∨ (x \notin A ∧ x \in B)) ∧ x \in C)}$
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(2) $(A \cap C) \Delta (B \cap C) = \ {x : (((x \in A ∧ x \in C) ∧ ¬(x \in B ∧ x \in C)) \ ∨ (¬(x \in A ∧ x \in C) ∧ (x \in B ∧ x \in C)))}$
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(3) $(A \cap C) \Delta (B \cap C) = \ {x : (((x \in A ∧ x \in C) ∧ (x \notin B ∨ x \notin C)) \ ∨ ((x \notin A ∨ x \notin C) ∧ (x \in B ∧ x \in C)))}$
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(4) $(A \cap C) \Delta (B \cap C) = \ {x : (((x \in A ∧ x \in C) ∧ (x \notin B ∧ x \in C)) \ ∨ ((x \notin A ∧ x \in C) ∧ (x \in B ∧ x \in C)))}$
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(5) $(A \cap C) \Delta (B \cap C) = \ {x:(((x \in A ∧ x \notin B) ∨ (x \notin A ∧ x \in B)) ∧ x \in C)}$
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From (1) and (5), we can see that $(A \Delta B) \cap C = (A \cap C) \Delta (B \cap C)$
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Rules of inference:

(2) ⇒ (3): De Morgan’s Law

(3) ⇒ (4): ((¬P ∨ ¬R) ∧ (Q ∧ R)) ⇔ ((¬P ∧ R) ∧ (Q ∧ R))

(4) ⇒ (5): ((P ∧ R) ∨ (Q ∧ R)) ⇔ ((P ∨ Q) ∧ R)\

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In case you want to use my text as a draft

#(A △B)∩C = (A ∩C)△(B ∩C) proof