Noelanijd

2022-07-15

Can someone help me prove that:

$((A\cap B)\oplus A{)}^{\prime}={A}^{\prime}\cup B$

A′ is A complement.

$((A\cap B)\oplus A{)}^{\prime}={A}^{\prime}\cup B$

A′ is A complement.

berouweek

Beginner2022-07-16Added 11 answers

Step 1

$\begin{array}{}\text{aka Xor}& p\oplus q\equiv (p\wedge \mathrm{\neg}q)\vee (\mathrm{\neg}p\wedge q)\equiv (p\vee q)\wedge (\mathrm{\neg}p\vee \mathrm{\neg}q)\end{array}$

This set notation '$\oplus $' corresponding to xor in logic.

Any $x\in ((A\cap B)\oplus A{)}^{\prime}$ if and only if:

$\mathrm{\neg}((x\in A\wedge x\in B)\oplus x\in A)$

Since $\mathrm{\neg}((p\wedge q)\oplus p)\leftrightarrow (\mathrm{\neg}p\vee q)$ is a tautology,

It's clearly equivalent to $x\notin A\vee x\in B$, hence proved…

Apply def. of Xor:

$\mathrm{\neg}(((x\in A\wedge x\in B)\wedge x\notin A)\vee (\mathrm{\neg}(x\in A\wedge x\in B)\wedge x\in A))$

Apply Commutative law & Associative law:

$\mathrm{\neg}(((x\in A\wedge x\notin A)\wedge x\in B)\vee (\mathrm{\neg}(x\in A\wedge x\in B)\wedge x\in A))$

Apply Negation law:

$\mathrm{\neg}((\mathrm{\perp}\wedge x\in B)\vee (\mathrm{\neg}(x\in A\wedge x\in B)\wedge x\in A))$

Apply Domination law:

$\mathrm{\neg}(\mathrm{\perp}\vee (\mathrm{\neg}(x\in A\wedge x\in B)\wedge x\in A))$

Apply Identity law:

$\mathrm{\neg}(\mathrm{\neg}(x\in A\wedge x\in B)\wedge x\in A)$

Apply De Morgan's law:

$\mathrm{\neg}((x\notin A\vee x\notin B)\wedge x\in A)$

Apply Distributive law:

$\mathrm{\neg}((x\notin A\wedge x\in A)\vee (x\notin B\wedge x\in A))$

Apply Negation law:

$\mathrm{\neg}(\mathrm{\perp}\vee (x\notin B\wedge x\in A))$

Apply Domination law:

$\mathrm{\neg}(x\notin B\wedge x\in A)$

Apply De Morgan's law:

$x\in B\vee x\notin A$

Apply Commutative law:

$x\notin A\vee x\in B$

This hold if and only if $x\in {A}^{\prime}\cup B$

Step 2

Hence we proved $\begin{array}{}\u25fb& ((A\cap B)\oplus A{)}^{\prime}={A}^{\prime}\cup B\end{array}$

$\begin{array}{}\text{aka Xor}& p\oplus q\equiv (p\wedge \mathrm{\neg}q)\vee (\mathrm{\neg}p\wedge q)\equiv (p\vee q)\wedge (\mathrm{\neg}p\vee \mathrm{\neg}q)\end{array}$

This set notation '$\oplus $' corresponding to xor in logic.

Any $x\in ((A\cap B)\oplus A{)}^{\prime}$ if and only if:

$\mathrm{\neg}((x\in A\wedge x\in B)\oplus x\in A)$

Since $\mathrm{\neg}((p\wedge q)\oplus p)\leftrightarrow (\mathrm{\neg}p\vee q)$ is a tautology,

It's clearly equivalent to $x\notin A\vee x\in B$, hence proved…

Apply def. of Xor:

$\mathrm{\neg}(((x\in A\wedge x\in B)\wedge x\notin A)\vee (\mathrm{\neg}(x\in A\wedge x\in B)\wedge x\in A))$

Apply Commutative law & Associative law:

$\mathrm{\neg}(((x\in A\wedge x\notin A)\wedge x\in B)\vee (\mathrm{\neg}(x\in A\wedge x\in B)\wedge x\in A))$

Apply Negation law:

$\mathrm{\neg}((\mathrm{\perp}\wedge x\in B)\vee (\mathrm{\neg}(x\in A\wedge x\in B)\wedge x\in A))$

Apply Domination law:

$\mathrm{\neg}(\mathrm{\perp}\vee (\mathrm{\neg}(x\in A\wedge x\in B)\wedge x\in A))$

Apply Identity law:

$\mathrm{\neg}(\mathrm{\neg}(x\in A\wedge x\in B)\wedge x\in A)$

Apply De Morgan's law:

$\mathrm{\neg}((x\notin A\vee x\notin B)\wedge x\in A)$

Apply Distributive law:

$\mathrm{\neg}((x\notin A\wedge x\in A)\vee (x\notin B\wedge x\in A))$

Apply Negation law:

$\mathrm{\neg}(\mathrm{\perp}\vee (x\notin B\wedge x\in A))$

Apply Domination law:

$\mathrm{\neg}(x\notin B\wedge x\in A)$

Apply De Morgan's law:

$x\in B\vee x\notin A$

Apply Commutative law:

$x\notin A\vee x\in B$

This hold if and only if $x\in {A}^{\prime}\cup B$

Step 2

Hence we proved $\begin{array}{}\u25fb& ((A\cap B)\oplus A{)}^{\prime}={A}^{\prime}\cup B\end{array}$

iarc6io

Beginner2022-07-17Added 2 answers

Step 1

Note that $\oplus $ denotes the symmetric difference between two sets. In other words,

$P\oplus Q=(P-Q)\cup (Q-P)=(P\cap {Q}^{\prime})\cup (Q\cap {P}^{\prime})$

Step 2

Using this definition as well as set identities, we have the following proof:

$(}(A\cap B)\oplus A{{\textstyle )}}^{\prime$

$={\textstyle [}{\textstyle (}(A\cap B)\cap {A}^{\prime}{\textstyle )}\cup {\textstyle (}A\cap (A\cap B{)}^{\prime}{\textstyle )}{{\textstyle ]}}^{\prime}$ ----- by definition of $\oplus $

$={\textstyle [}{\textstyle (}(B\cap A)\cap {A}^{\prime}{\textstyle )}\cup {\textstyle (}A\cap (A\cap B{)}^{\prime}{\textstyle )}{{\textstyle ]}}^{\prime}$ ----- by the commutative law

$={\textstyle [}{\textstyle (}B\cap (A\cap {A}^{\prime}){\textstyle )}\cup {\textstyle (}A\cap (A\cap B{)}^{\prime}{\textstyle )}{{\textstyle ]}}^{\prime}$ ----- by the associative law

$={\textstyle [}(B\cap \mathrm{\varnothing})\cup {\textstyle (}A\cap (A\cap B{)}^{\prime}{\textstyle )}{{\textstyle ]}}^{\prime}$ ----- by the complement law

$={\textstyle [}\mathrm{\varnothing}\cup {\textstyle (}A\cap (A\cap B{)}^{\prime}{\textstyle )}{{\textstyle ]}}^{\prime}$ ----- by the domination law

$={\textstyle [}A\cap (A\cap B{)}^{\prime}{{\textstyle ]}}^{\prime}$ ----- by the identity law

$={A}^{\prime}\cup (A\cap B{)}^{\u2033}$ ----- by DeMorgan's law

$={A}^{\prime}\cup (A\cap B)$ ----- by the double complement law

$=({A}^{\prime}\cup A)\cap ({A}^{\prime}\cup B)$ ----- by the distributive law

$=U\cap ({A}^{\prime}\cup B)$ ----- by the complement law

$={A}^{\prime}\cup B$ ----- by the identity law

Note that $\oplus $ denotes the symmetric difference between two sets. In other words,

$P\oplus Q=(P-Q)\cup (Q-P)=(P\cap {Q}^{\prime})\cup (Q\cap {P}^{\prime})$

Step 2

Using this definition as well as set identities, we have the following proof:

$(}(A\cap B)\oplus A{{\textstyle )}}^{\prime$

$={\textstyle [}{\textstyle (}(A\cap B)\cap {A}^{\prime}{\textstyle )}\cup {\textstyle (}A\cap (A\cap B{)}^{\prime}{\textstyle )}{{\textstyle ]}}^{\prime}$ ----- by definition of $\oplus $

$={\textstyle [}{\textstyle (}(B\cap A)\cap {A}^{\prime}{\textstyle )}\cup {\textstyle (}A\cap (A\cap B{)}^{\prime}{\textstyle )}{{\textstyle ]}}^{\prime}$ ----- by the commutative law

$={\textstyle [}{\textstyle (}B\cap (A\cap {A}^{\prime}){\textstyle )}\cup {\textstyle (}A\cap (A\cap B{)}^{\prime}{\textstyle )}{{\textstyle ]}}^{\prime}$ ----- by the associative law

$={\textstyle [}(B\cap \mathrm{\varnothing})\cup {\textstyle (}A\cap (A\cap B{)}^{\prime}{\textstyle )}{{\textstyle ]}}^{\prime}$ ----- by the complement law

$={\textstyle [}\mathrm{\varnothing}\cup {\textstyle (}A\cap (A\cap B{)}^{\prime}{\textstyle )}{{\textstyle ]}}^{\prime}$ ----- by the domination law

$={\textstyle [}A\cap (A\cap B{)}^{\prime}{{\textstyle ]}}^{\prime}$ ----- by the identity law

$={A}^{\prime}\cup (A\cap B{)}^{\u2033}$ ----- by DeMorgan's law

$={A}^{\prime}\cup (A\cap B)$ ----- by the double complement law

$=({A}^{\prime}\cup A)\cap ({A}^{\prime}\cup B)$ ----- by the distributive law

$=U\cap ({A}^{\prime}\cup B)$ ----- by the complement law

$={A}^{\prime}\cup B$ ----- by the identity law

Show that the sequence $a}_{n$ is an solution of the recurrence relation [a^n = a^n−1 + 2a^n−2 + 2n − 9 if a] if

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Why does a magnet always attract iron, not wood?

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In a later example, a question asked which of these functions is a bijection, the answer included $f(x)={x}^{3}$

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Showing that if n is a natural number larger than $3$, then $n!>{2}^{n}$

My try:

Base Case:

If $n=4$, then $4!>{2}^{4}$

$24>16$

So, the base case is true.

Assuming $P(k)$ is true.

$k!>{2}^{k}$

Now we need to show that $P(k+1)$ is true.

$(k+1)!={2}^{k+1}$

Proof:

$(k+1)!>(k+1)k!$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(k+1){2}^{k}$

After this I have no idea how to solve further.

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In this example, you could use:

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$\{1,3\},\{2,4\}\to 1\cdot 3+2\cdot 4=3+8=11$

$\{1,4\},\{2,3\}\to 1\cdot 4+2\cdot 3=4+6=10$

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Reminder: Suppose $T\subseteq X\times X$ is an equivalence relation over X. $\text{}\text{}A\subseteq X$ will be called a Representative set of T, if it occurs that: $\mathrm{\forall}x\in X.|[x{]}_{T}\cap A|=1$.

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