Discrete Mathematics Homework #8

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# Discrete Mathematics

## Homework #8

### 1.

Let $A=\neg q\lor r$, $B=s\land p$, $C=\neg r\rightarrow B$, $D=p\rightarrow A$, so that the formula is $F=D\rightarrow\neg C$.

So that we have

### 2.

$(1)$ **Contingency.**

Let $A=\neg p\lor q$, $B=\neg r\lor\neg p$, $C=p\lor r$, $D=q\rightarrow B$, so that the formula is $F=A\land C\land D$.

So that we have

Since the result is either True or False, the formula is a **contingency**.

$(2)$ **Tautology.**

Let $A=p\rightarrow q$, $B=q\rightarrow r$, $C=p\rightarrow r$, $D=A\land B$, so that the formula is $F=D\rightarrow C$.

So that we have

Since the result is always True, the formula is a **tautology.**

$(3)$ **Tautology.**

Let $A=p\rightarrow r$, $B= q\rightarrow s$, $C=p\lor q$, $D=r\lor s$, and $E=A\land B\land C$, so that the formula is $F=E\rightarrow D$.

So that we have

Since the result is always True, the formula is a **tautology.**

### 3.

$\textbf{(1)}$ $d\leftrightarrow (c\land\neg a)$

$\textbf{(2)}$ $(\neg d\rightarrow c)\land(d\rightarrow\neg b)$

$\textbf{(3)}$ $a\rightarrow (\neg b\land\neg c\rightarrow d)$

$\textbf{(4)}$ $(a\land b\land c\leftrightarrow\neg d)\land((\neg a\land\neg b)\rightarrow(d\rightarrow c))$

### 4.

The specifications are:

$(1)$ $\neg l\rightarrow q$; $(2)$ $\neg l\leftrightarrow n$; $(3)$ $\neg q\rightarrow b$; $(4)$ $\neg l\rightarrow b$; $(5)$ $\neg b$.

To make all of the specifications $\textcolor{green}{\text{True}}$, from $(5)$ we have $b=\textcolor{red}{\text{False}}$. Then $q=l=\textcolor{green}{\text{True}}$ according to $(3)(4)$. From that we find $(1)$ also holds. So that $n=\textcolor{red}{\text{False}}$ according to $(2)$.

Now that we have a system

$$
\{l=\textcolor{green}{\text{True}},q=\textcolor{green}{\text{True}},n=\textcolor{red}{\text{False}},b=\textcolor{red}{\text{False}}\}
$$

where all the five specifications hold.

So that such system **exists.**

### 5.

$A_1=p\land q\land r$;

$A_2=p\land q\land\neg r$;

$A_3=p\land\neg q\land r$;

$A_4=p\land\neg q\land\neg r$;

$A_5=\neg p\land q\land r$;

$A_6=\neg p\land q\land\neg r$;

$A_7=\neg p\land\neg q\land r$;

$A_8=\neg p\land \neg q\land\neg r$.

From the truth table of $p,q,r,A$ we have

so that

$$
A=(\neg p\land\neg q)\lor\neg r.
$$

### 6.

Since that

|  p   |  q   | $\textbf{p}\rightarrow \textbf{q}$ | $\neg \textbf{p}\lor \textbf{q}$ |
| :--: | :--: | :--------------------------------: | :------------------------------: |
|  T   |  T   |                 T                  |                T                 |
|  T   |  F   |                 F                  |                F                 |
|  F   |  T   |                 T                  |                T                 |
|  F   |  F   |                 T                  |                T                 |

we reach the **conclusion** that $p\rightarrow q\equiv \neg p\lor q$.

So that

$$
\begin{align}
&\quad(P\land Q\land S)\lor(P\land\neg Q\land\neg R)\lor(P\land Q\land\neg S)\lor\neg(P\land R\rightarrow Q)
\\&\equiv(P\land Q)\lor(P\land\neg Q\land\neg R)\lor\neg(\neg(P\land R)\lor Q)
\\&\equiv(P\land Q)\lor(P\land\neg Q\land\neg R)\lor\neg(\neg P\lor\neg R\lor Q)
\\&\equiv(P\land Q)\lor(P\land\neg Q\land\neg R)\lor(P\land\neg Q\land R)
\\&\equiv(P\land Q)\lor(P\land\neg Q)
\\&\equiv P.
\end{align}
$$

**The proof is complete.**

### 7.

Since that $\Delta p$ is always False whatever $p$ is, we say $\Delta p\equiv F$.

$(1)$ Since that we have proven $p\rightarrow q\equiv \neg p\lor q$ in **Question 6**, we have

$$
\neg p =\neg p\lor F=p\rightarrow \Delta p.
$$

$(2)$ Using $p\rightarrow q\equiv \neg p\lor q$ and $\neg p=p\rightarrow\Delta p$, we have

$$
\begin{align}
p\land q&=\neg(\neg p\lor\neg q)=\neg(p\rightarrow\neg q)=\neg (p\rightarrow(q\rightarrow\Delta q))\\
&=(p\rightarrow(q\rightarrow\Delta q))\rightarrow\Delta(p\rightarrow(q\rightarrow\Delta q))
\\&=(p\rightarrow(q\rightarrow\Delta q))\rightarrow F
\\&=(p\rightarrow(q\rightarrow\Delta q))\rightarrow\Delta p.
\end{align}
$$

$(3)$ Similarly we have

$$
p\lor q=\neg(\neg p)\lor q=\neg p\rightarrow q=p\rightarrow\Delta p\rightarrow q.
$$

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Discrete Mathematics Homework #8

Bi_Gu • 2022 年 05 月 05 日

# Discrete Mathematics

## Homework #8

### 1.

Let $A=\neg q\lor r$, $B=s\land p$, $C=\neg r\rightarrow B$, $D=p\rightarrow A$, so that the formula is $F=D\rightarrow\neg C$.

So that we have

### 2.

$(1)$ **Contingency.**

Let $A=\neg p\lor q$, $B=\neg r\lor\neg p$, $C=p\lor r$, $D=q\rightarrow B$, so that the formula is $F=A\land C\land D$.

So that we have

Since the result is either True or False, the formula is a **contingency**.

$(2)$ **Tautology.**

Let $A=p\rightarrow q$, $B=q\rightarrow r$, $C=p\rightarrow r$, $D=A\land B$, so that the formula is $F=D\rightarrow C$.

So that we have

Since the result is always True, the formula is a **tautology.**

$(3)$ **Tautology.**

Let $A=p\rightarrow r$, $B= q\rightarrow s$, $C=p\lor q$, $D=r\lor s$, and $E=A\land B\land C$, so that the formula is $F=E\rightarrow D$.

So that we have

Since the result is always True, the formula is a **tautology.**

### 3.

$\textbf{(1)}$ $d\leftrightarrow (c\land\neg a)$

$\textbf{(2)}$ $(\neg d\rightarrow c)\land(d\rightarrow\neg b)$

$\textbf{(3)}$ $a\rightarrow (\neg b\land\neg c\rightarrow d)$

$\textbf{(4)}$ $(a\land b\land c\leftrightarrow\neg d)\land((\neg a\land\neg b)\rightarrow(d\rightarrow c))$

### 4.

The specifications are:

$(1)$ $\neg l\rightarrow q$; $(2)$ $\neg l\leftrightarrow n$; $(3)$ $\neg q\rightarrow b$; $(4)$ $\neg l\rightarrow b$; $(5)$ $\neg b$.

Now that we have a system

$$
\{l=\textcolor{green}{\text{True}},q=\textcolor{green}{\text{True}},n=\textcolor{red}{\text{False}},b=\textcolor{red}{\text{False}}\}
$$

where all the five specifications hold.

So that such system **exists.**

### 5.

$A_1=p\land q\land r$;

$A_2=p\land q\land\neg r$;

$A_3=p\land\neg q\land r$;

$A_4=p\land\neg q\land\neg r$;

$A_5=\neg p\land q\land r$;

$A_6=\neg p\land q\land\neg r$;

$A_7=\neg p\land\neg q\land r$;

$A_8=\neg p\land \neg q\land\neg r$.

From the truth table of $p,q,r,A$ we have

so that

$$
A=(\neg p\land\neg q)\lor\neg r.
$$

### 6.

Since that

we reach the **conclusion** that $p\rightarrow q\equiv \neg p\lor q$.

So that

**The proof is complete.**

### 7.

Since that $\Delta p$ is always False whatever $p$ is, we say $\Delta p\equiv F$.

$(1)$ Since that we have proven $p\rightarrow q\equiv \neg p\lor q$ in **Question 6**, we have

$$
\neg p =\neg p\lor F=p\rightarrow \Delta p.
$$

$(2)$ Using $p\rightarrow q\equiv \neg p\lor q$ and $\neg p=p\rightarrow\Delta p$, we have

$(3)$ Similarly we have

$$
p\lor q=\neg(\neg p)\lor q=\neg p\rightarrow q=p\rightarrow\Delta p\rightarrow q.
$$

Discrete Mathematics Homework #8

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