Math / Discrete

Question(3) لأى ثلاث مجموعات C, B , A أثبت صحة القو انين التالية : (i) $A \Delta(B \Delta C)=(A \Delta B) \Delta C$ (ii) $A \cap(B \cup C)^{c}=(A-B) \cup(A-C)$ (iii) $(A \cap B) \cup\left(A^{c} \cap C\right) \cup(B \cap C)=(A \cap B) \cup\left(A^{c} \cap C\right)$

Studdy Solution

STEP 1

What is this asking? We need to prove three set theory statements: the associativity of symmetric difference, a set intersection with a complement, and a simplification of a union of intersections. Watch out! Don't get lost in the symbols!
Keep track of what each operation means as we go.

STEP 2

1. Associativity of Symmetric Difference
2. Intersection with a Complement
3. Simplification of a Union of Intersections

STEP 3

Let's prove $A \Delta (B \Delta C) = (A \Delta B) \Delta C$.
Remember, the symmetric difference $X \Delta Y$ means elements that are in $X$ or $Y$, but not in both.

STEP 4

An element $x$ is in $A \Delta (B \Delta C)$ if it's in an odd number of the sets $A$, $B$, and $C$.
Think about it!
If $x$ is in $A$ and not in $B \Delta C$, it's in one set.
If $x$ is in $B \Delta C$ and not in $A$, and $B \Delta C$ means $x$ is in one of $B$ or $C$ but not both, then $x$ is in one set.
If $x$ is in $A$ and in $B \Delta C$, then $x$ is in either $B$ or $C$ but not both, so $x$ is in $A$, and either $B$ or $C$.
If $x$ is in $A$ and $B$ but not $C$, then $x$ is in two sets.
If $x$ is in $A$ and $C$ but not $B$, then $x$ is in two sets.
If $x$ is in all three sets, then $x$ is not in $B \Delta C$ so $x$ is in $A$ only, which is one set.

STEP 5

Similarly, an element $x$ is in $(A \Delta B) \Delta C$ if it's in an odd number of the sets $A$, $B$, and $C$.

STEP 6

Since both sides describe elements being in an odd number of the sets, they are equal!

STEP 7

We want to show that $A \cap (B \cup C)^c = (A - B) \cup (A - C)$.

STEP 8

$A \cap (B \cup C)^c$ means elements in $A$ and not in $B \cup C$.

STEP 9

Not being in $B \cup C$ means not being in $B$ and not being in $C$.
So, we have elements in $A$, not in $B$, and not in $C$.

STEP 10

$(A - B) \cup (A - C)$ means elements in $A$ but not $B$, or elements in $A$ but not $C$.

STEP 11

If an element is in $A$ and not in $B$ and not in $C$, it's certainly in $A$ and not in $B$ or in $A$ and not in $C$.

STEP 12

Therefore, both sides are equivalent!

STEP 13

Let's prove $(A \cap B) \cup (A^c \cap C) \cup (B \cap C) = (A \cap B) \cup (A^c \cap C)$.

STEP 14

Consider the case where an element $x$ is in $B \cap C$.
This means $x$ is in both $B$ and $C$.

STEP 15

Now, either $x$ is in $A$ or it's not.

STEP 16

If $x$ is in $A$, then since it's also in $B$, it's in $A \cap B$.

STEP 17

If $x$ is not in $A$ (meaning it's in $A^c$), then since it's in $C$, it's in $A^c \cap C$.

STEP 18

So, if $x$ is in $B \cap C$, it must be in either $A \cap B$ or $A^c \cap C$.

STEP 19

Thus, adding $B \cap C$ to the union doesn't add any new elements, and both sides are equal!

STEP 20

We have proven all three set equalities:
1. $A \Delta (B \Delta C) = (A \Delta B) \Delta C$
2. $A \cap (B \cup C)^c = (A - B) \cup (A - C)$
3. $(A \cap B) \cup (A^c \cap C) \cup (B \cap C) = (A \cap B) \cup (A^c \cap C)$

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Studdy Solution

STEP 1

What is this asking? We need to prove three set theory statements: the *associativity of symmetric difference*, a *set intersection with a complement*, and a *simplification of a union of intersections*. Watch out! Don't get lost in the symbols!Keep track of what each operation *means* as we go.

STEP 2

1. Associativity of Symmetric Difference2. Intersection with a Complement3. Simplification of a Union of Intersections

STEP 3

Let's **prove** \(A \Delta (B \Delta C) = (A \Delta B) \Delta C\).Remember, the *symmetric difference* \(X \Delta Y\) means elements that are in \(X\) or \(Y\), but *not* in both.

STEP 4

STEP 5

Similarly, an element \(x\) is in \((A \Delta B) \Delta C\) if it's in an *odd* number of the sets \(A\), \(B\), and \(C\).

STEP 6

Since both sides describe elements being in an odd number of the sets, they are **equal**!

STEP 7

We want to **show** that \(A \cap (B \cup C)^c = (A - B) \cup (A - C)\).

STEP 8

\(A \cap (B \cup C)^c\) means elements in \(A\) and *not* in \(B \cup C\).

STEP 9

Not being in \(B \cup C\) means not being in \(B\) *and* not being in \(C\).So, we have elements in \(A\), not in \(B\), and not in \(C\).

STEP 10

\((A - B) \cup (A - C)\) means elements in \(A\) but not \(B\), or elements in \(A\) but not \(C\).

STEP 11

If an element is in \(A\) and not in \(B\) and not in \(C\), it's certainly in \(A\) and not in \(B\) *or* in \(A\) and not in \(C\).

STEP 12

Therefore, both sides are **equivalent**!

STEP 13

Let's **prove** \((A \cap B) \cup (A^c \cap C) \cup (B \cap C) = (A \cap B) \cup (A^c \cap C)\).

STEP 14

Consider the case where an element \(x\) is in \(B \cap C\).This means \(x\) is in both \(B\) and \(C\).

STEP 15

Now, either \(x\) is in \(A\) or it's not.

STEP 16

If \(x\) is in \(A\), then since it's also in \(B\), it's in \(A \cap B\).

STEP 17

If \(x\) is *not* in \(A\) (meaning it's in \(A^c\)), then since it's in \(C\), it's in \(A^c \cap C\).

STEP 18

So, if \(x\) is in \(B \cap C\), it *must* be in either \(A \cap B\) or \(A^c \cap C\).

STEP 19

Thus, adding \(B \cap C\) to the union doesn't add any new elements, and both sides are **equal**!

STEP 20

We have **proven** all three set equalities:1. \(A \Delta (B \Delta C) = (A \Delta B) \Delta C\)2. \(A \cap (B \cup C)^c = (A - B) \cup (A - C)\)3. \((A \cap B) \cup (A^c \cap C) \cup (B \cap C) = (A \cap B) \cup (A^c \cap C)\)

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What is this asking? We need to prove three set theory statements: the associativity of symmetric difference, a set intersection with a complement, and a simplification of a union of intersections. Watch out! Don't get lost in the symbols!
Keep track of what each operation means as we go.

1. Associativity of Symmetric Difference
2. Intersection with a Complement
3. Simplification of a Union of Intersections

Let's prove \(A \Delta (B \Delta C) = (A \Delta B) \Delta C\).
Remember, the symmetric difference \(X \Delta Y\) means elements that are in \(X\) or \(Y\), but not in both.

Similarly, an element \(x\) is in \((A \Delta B) \Delta C\) if it's in an odd number of the sets \(A\), \(B\), and \(C\).

Since both sides describe elements being in an odd number of the sets, they are equal!

We want to show that \(A \cap (B \cup C)^c = (A - B) \cup (A - C)\).

\(A \cap (B \cup C)^c\) means elements in \(A\) and not in \(B \cup C\).

Not being in \(B \cup C\) means not being in \(B\) and not being in \(C\).
So, we have elements in \(A\), not in \(B\), and not in \(C\).

If an element is in \(A\) and not in \(B\) and not in \(C\), it's certainly in \(A\) and not in \(B\) or in \(A\) and not in \(C\).

Therefore, both sides are equivalent!

Let's prove \((A \cap B) \cup (A^c \cap C) \cup (B \cap C) = (A \cap B) \cup (A^c \cap C)\).

Consider the case where an element \(x\) is in \(B \cap C\).
This means \(x\) is in both \(B\) and \(C\).

If \(x\) is not in \(A\) (meaning it's in \(A^c\)), then since it's in \(C\), it's in \(A^c \cap C\).

So, if \(x\) is in \(B \cap C\), it must be in either \(A \cap B\) or \(A^c \cap C\).

Thus, adding \(B \cap C\) to the union doesn't add any new elements, and both sides are equal!

We have proven all three set equalities:
1. \(A \Delta (B \Delta C) = (A \Delta B) \Delta C\)
2. \(A \cap (B \cup C)^c = (A - B) \cup (A - C)\)
3. \((A \cap B) \cup (A^c \cap C) \cup (B \cap C) = (A \cap B) \cup (A^c \cap C)\)