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# 数学代写|离散数学代写Discrete Mathematics代考|MATH1061 De Morgan Laws and Other Rules of Classical Logic

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## 数学代写|离散数学代写Discrete Mathematics代考|De Morgan Laws and Other Rules of Classical Logic

In Section $1.7$ we discussed the de Morgan laws. Now that we also know about intuitionistic logic we revisit these laws.

Proposition 11.4. The following equivalences (de Morgan laws) are provable in classical logic.
\begin{aligned} &\neg(P \wedge Q) \equiv \neg P \vee \neg Q \ &\neg(P \vee Q) \equiv \neg P \wedge \neg Q . \end{aligned}
In fact, $\neg(P \vee Q) \equiv \neg P \wedge \neg Q$ and $(\neg P \vee \neg Q) \Rightarrow \neg(P \wedge Q)$ are provable in intuitionistic logic. The proposition $(P \wedge \neg Q) \Rightarrow \neg(P \Rightarrow Q)$ is provable in intuitionistic logic and $\neg(P \Rightarrow Q) \Rightarrow(P \wedge \neg Q)$ is provable in classical logic. Therefore, $\neg(P \Rightarrow Q)$ and $P \wedge \neg Q$ are equivalent in classical logic. Furthermore, $P \Rightarrow Q$ and $\neg P \vee Q$ are equivalent in classical logic and $(\neg P \vee Q) \Rightarrow(P \Rightarrow Q)$ is provable in intuitionistic logic.

Proof. We only prove the very last part of Proposition $11.4$ leaving the other parts as a series of exercises. Here is an intuitionistic proof of $(\neg P \vee Q) \Rightarrow(P \Rightarrow Q)$ :

$$\frac{\frac{\neg P^z P^x}{\frac{\perp}{Q}} \quad \frac{P^y}{\frac{Q}{P \Rightarrow Q}}{ }^y}{\frac{Q^t}{P \Rightarrow Q}{ }^x} \frac{}{\left.\frac{P}{P \Rightarrow Q}\right)^w}$$

## 数学代写|离散数学代写Discrete Mathematics代考|Formal Versus Informal Proofs

As we said before, it is practically impossible to write formal proofs (i.e., proofs written as proof trees using the rules of one of the systems presented earlier) of “real” statements that are not “toy propositions.” This is because it would be extremely tedious and time-consuming to write such proofs and these proofs would be huge and thus very hard to read.

What we do instead is to construct “informal” proofs in which we still make use of the logical rules that we have presented but we take shortcuts and sometimes we even omit proof steps (some elimination rules, such as $\wedge$-elimination and some introduction rules, such as $\vee$-introduction) and we use a natural language (here, presumably, English) rather than formal symbols (we say “and” for $\wedge$, “or” for $\vee$, etc.). We refer the readetr to Section $1.8$ for a discussion of these issues. We also urge our readers to read Chapter 3 of Gowers $[11]$ which contains very illuminating remarks about the notion of proof in mathematics.

Here is a concrete example illustrating the usefulnes of auxiliary lemmas in constructing informal proofs.
Say we wish to prove the implication

$$\neg(P \wedge Q) \Rightarrow((\neg P \wedge \neg Q) \vee(\neg P \wedge Q) \vee(P \wedge \neg Q))$$
It can be shown that the above proposition is not provable intuitionistically, so we have to use the proof-by-contradiction method in our proof. One quickly realizes that any proof ends up re-proving basic properties of $\wedge$ and $\vee$, such as associativity, commutativity, idempotence, distributivity, and so on, some of the de Morgan laws, and that the complete proof is very large. However, if we allow ourselves to use the de Morgan laws as well as various basic properties of $\wedge$ and $\vee$, such as distributivity,
$$(A \wedge B) \vee C \equiv(A \wedge C) \vee(B \wedge C),$$
commutativity of $\wedge$ and $\vee(A \wedge B \equiv B \wedge A, A \vee B \equiv B \vee A)$, associativity of $\wedge$ and $\vee$ $(A \wedge(B \wedge C) \equiv(A \wedge B) \wedge C, A \vee(B \vee C) \equiv(A \vee B) \vee C)$, and the idempotence of $\wedge$ and $\vee(A \wedge A \equiv A, A \vee A \equiv A)$, then we get
\begin{aligned} (\neg P \wedge \neg Q) \vee(\neg P \wedge Q) \vee(P \wedge \neg Q) \equiv &(\neg P \wedge \neg Q) \vee(\neg P \wedge \neg Q) \ & \vee(\neg P \wedge Q) \vee(P \wedge \neg Q) \ \equiv &(\neg P \wedge \neg Q) \vee(\neg P \wedge Q) \ & \vee(\neg P \wedge \neg Q) \vee(P \wedge \neg Q) \ & \equiv(\neg P \wedge(\neg Q \vee Q)) \vee(\neg P \wedge \neg Q) \vee(P \wedge \neg Q) \ & \equiv \neg P \vee(\neg P \wedge \neg Q) \vee(P \wedge \neg Q) \ & \equiv \neg P \vee((\neg P \vee P) \wedge \neg Q) \ & \equiv \neg P \vee \neg Q, \end{aligned}
where we make implicit uses of commutativity and associativity, and the fact that $R \wedge(P \vee \neg P) \equiv R$, and by de Morgan,
$$\neg(P \wedge Q) \equiv \neg P \vee \neg Q,$$
using auxiliary lemmas, we end up proving $(*)$ without too much pain.

## 数学代写|离散数学代写离散数学代考|德摩根定律和其他经典逻辑的规则

\begin{aligned} &\neg(P \wedge Q) \equiv \neg P \vee \neg Q \ &\neg(P \vee Q) \equiv \neg P \wedge \neg Q . \end{aligned}

$$\frac{\frac{\neg P^z P^x}{\frac{\perp}{Q}} \quad \frac{P^y}{\frac{Q}{P \Rightarrow Q}}{ }^y}{\frac{Q^t}{P \Rightarrow Q}{ }^x} \frac{}{\left.\frac{P}{P \Rightarrow Q}\right)^w}$$

## 数学代写|离散数学代写离散数学代考|正式证明与非正式证明

$$\neg(P \wedge Q) \Rightarrow((\neg P \wedge \neg Q) \vee(\neg P \wedge Q) \vee(P \wedge \neg Q))$$

$$(A \wedge B) \vee C \equiv(A \wedge C) \vee(B \wedge C),$$

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