StonyBrook-Lin539-F18/main

start logic

thomas--graf committed Nov 30, 2018
1 parent 8f06ca6 commit 90a1e2e6c590eea4b671b826039968c0f95d5818
 @@ -0,0 +1,115 @@ # Propositions ## Intuition A proposition is a statement that is either true or false. This does not mean that one can necessarily tell whether a proposition is true, nor that the proposition makes any sense considering what our world is like. It just means that the statement is such that it can have a truth value. \begin{example} All of the following are propositions, even though whether they're true or false can depend greatly on context:
• Nobody likes ponies.
• He always snores like a pig.
• $2 + 2 = 4$
• We might see another stock market crash soon.
• A language is regular iff it is recognized by a finite-state automaton.
\end{example} \begin{example} The following are not propositions.
• Hell, yeah!
• A cute kitten
• John and Mary
\end{example} \begin{example} Some cases are borderline and require some more detailed analysis. Take *Either John leaves or I will*. As a literal description of how the world is, this can be either true or false, so it's a proposition. But we can also interpret it as a command that the addressee must choose between John and the speaker. Commands have no truth value and thus aren't propositions. \end{example} \begin{exercise} For each one of the following, say whether it is a proposition. If you say no, justify your answer.
1. John and Peter met yesterday.
2. The DeLorean DMC-12 is a sports car originally manufactured by John DeLorean's DeLorean Motor Company.
3. Who would say no to that.
4. That son of a gun has fooled us all.
5. Colorless green ideas sleep furiously.
6. John and Mary have only one child, who has three siblings.
\end{exercise} ## Entailment Some propositions are atomic in the sense that one cannot break them down into smaller propositions, whereas others are clearly built up from smaller propositions. \begin{example} The sentence *John rode a bike and Mary her dragon* might be true or false, so it's a proposition. But it is not an atomic proposition because it's built up from two smaller propositions. One is *John rode a bike*, the other one *Mary rode her dragon* (that English allows for the verb to be omitted in the second conjunct is irrelevant for this point). \end{example} As a result, propositions can be related to each other so that the truth value of one follows immediately from the other. This is called **entailment**. \begin{example} If *John rode a bike and Mary her dragon* is true, then it must also be true that John rode a bike, and it must be true that Mary rode her dragon. In the other direction, if *John rode a bike* is false, then so is *John rode a bike and Mary her dragon*. \end{example} Entailment can also hold between propositions that do not stand in an obvious part-whole relation to each other. \begin{example} Consider the sentences *If pigs can fly, then Sokrates is a chain-smoking dragon* and *No human is a dragon*. Assuming that both are true, the sentence *Sokrates is human* entails that pigs cannot fly. \end{example} The entailments in these two examples are slightly different. For the first, it suffices to recognize how one proposition was built up from smaller ones. The second one presents a more complex reasoning pattern that hinges on the meaning of the quantifier *no*. Logic was devised as a means to study these different kinds of entailments, but over the centuries it has evolved to a highly flexible system that can be used for all kinds of purposes. \begin{exercise} For each one of the following pairs of sentences, say whether entailment holds, and if so, in which direction. If entailment does not hold, explain why.
1. John likes his son's sisters. | John likes his daughters.
2. I am 30 years old. | My age is a multiple of 15.
3. 1 is not a prime number. | No prime numbers is less than 2.
4. John and Bill both left the party early. | Bill left the party around 9.
5. The 45th US president isn't universally beloved. | Donald Trump isn't universally beloved.
6. If $2 + 2 = 4$, then $2 + 2 = 4$. | $2 + 2 = 4$
\end{exercise}

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 @@ -0,0 +1,93 @@ # Tautologies A tautology is a formula that is always true, irrespective of what assignment function one picks. A simple example is $a \vee \neg a$. | $a$ | $\vee$ | $\neg$ | $a$ | | :-: | :-: | :-: | :-: | | 0 | 1 | 1 | 0 | | 1 | 1 | 0 | 1 | Sometimes the **verum** symbol $\top$ is used to denote an arbitrary tautology. So $\top$ is short for a proposition that is always true. Its opposite is the **falsum** $\bot$, which is always false. Tautologies are very useful for simplifying formulas. For one thing, if one knows that the formula $\phi$ is a tautology, then one can always replace it by $1$ without computing the truth values of any of its parts. But one can also use tautologies to convert $\phi$ into a fully equivalent yet distinct formula $\psi$. The rest of this section lists some important tautologies. ## Commutativity of $\wedge$ and $\vee$ From the truth tables of $\wedge$ and $\vee$, it follows immediately that - $\phi \wedge \psi \leftrightarrow \psi \wedge \phi$ - $\phi \vee \psi \leftrightarrow \psi \vee \phi$ ## Double negation It always holds that $\phi$ is equivalent to $\neg \neg \phi$, no matter what formula or propositional symbol $\phi$ corresponds to. So $\phi \leftrightarrow \neg \neg \phi$ is a tautology. \begin{exercise} Explain how this also entails that $\neg \phi \leftrightarrow \neg \neg \neg \phi$. \end{exercise} ## De Morgan's Law In set theory, De Morgan's law states that $A \cap B = \overline{\overline{A} \cup \overline{B}}$ and $A \cup B = \overline{\overline{A} \cap \overline{B}}$. A direct analogue of that holds in propositional logic. Both of the following are tautologies: - $\phi \wedge \psi \leftrightarrow \neg (\neg \phi \vee \neg \psi)$ - $\phi \vee \psi \lefrightarrow \neg (\neg \phi \wedge \neg \psi)$ This means that every instance of $\phi \wedge \psi$ can be replaced by $\neg (\neg \phi \wedge \neg \psi)$, and similarly for negation. \begin{exercise} Give the truth table for each one of the two tautologies. \end{exercise} ## Decomposition of implication Implication can be decomposed into negation and disjunction: - $\phi \rightarrow \psi \leftrightarrow \neg \phi \vee \psi$ \begin{exercise} Rewrite each one of the following formulae using only $\neg$ and $\wedge$ (yes, logical and, not logical or):
1. $p \rightarrow q$
2. $(p \vee (q \rightarrow \neg p))$
3. $p \rightarrow (q \rightarrow r)$
\end{exercise} ## Implication chains as conjunction Some implicational chains can be converted to a single implication with $\wedge$ in the antecedent. - $\phi \rightarrow (\psi \rightarrow \rho) \leftrightarrow \phi \wedge \psi \rightarrow \rho$ Bracketing is crucial. It is not the case that $(\phi \rightarrow \psi) \rightarrow \rho$ is equivalent to $\phi \wedge \psi \rightarrow \rho$. \begin{exercise} Give truth tables for the two formulas below to show that only the first is a tautology. - $\phi \rightarrow (\psi \rightarrow \rho) \leftrightarrow \phi \wedge \psi \rightarrow \rho$ - $(\phi \rightarrow \psi) \rightarrow \rho \leftrightarrow \phi \wedge \psi \rightarrow \rho$ \end{exercise} ## Distributivity of $\wedge$ and $\vee$ The operators $\wedge$ and $\vee$ distribute over each other - $\phi \wedge (\psi \vee \rho) \leftrightarrow (\phi \wedge \psi) \vee (\phi \wedge \rho)$ - $\phi \vee (\psi \wedge \rho) \leftrightarrow (\phi \vee \psi) \wedge (\phi \vee \rho)$ \begin{exercise} Show via truth tables that both formulae are indeed tautologies. \end{exercise}
 @@ -0,0 +1,66 @@ # Boolean lattices The fact that De Morgan's law holds for sets as well as propositional logic is no accident. The two are close siblings from an algebraic perspective. Let $S$ be some fixed set. We may order $\w(S)$ by the subset relation to obtain the familiar powerset lattice, with join corresponding to union and meet to intersection. In fact, this lattice satisfies two additional properties that make it a **Boolean lattice** (or **Boolean algebra**). \begin{definition} A lattice $\tuple{L, \leq}$ is
1. **distributive** iff join distributes over meet, and the other way round,
2. **bounded** iff there is a least element 0 and and a greatest element 1 such that for all $l \in L$, $0 \leq l \leq 1$,
3. **complemented** iff every eleement $a$ has a complement $\neg a$ such that $a \wedge \neg a = 0$ and $a \vee \neg a = 1$.
A lattice is **Boolean** iff it satisfies all three of the criteria. \end{definition} \begin{exercise} Show that $\tuple{\wp(S), \subseteq}$ is a Boolean lattice. More specifically:
1. Show that distributivity holds: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ and $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
2. What are the least and greatest elements of $\wp(S)$?
3. For any arbitrary $A \in \wp(S)$, what is its complement $\neg A$?
\end{exercise} As you might already expect, propositional logic also has the structure of a Boolean lattice. All we have to do is figure out the carrier set, and the rest will follow immediately. First, let $L$ be the set of all propositional formulas. We define an ordering $\leq$ such that $\phi \leq \psi$ ($\phi, \psi \in L$) iff $\phi$ always entails $\psi$. In other words, $\phi \leq \psi$ iff $\phi \rightarrow \psi$ is a tautology. We're almost done, but not quite. Note that $\tuple{L, \leq}$ is not a poset because antisymmetry is violated. \begin{example} Consider the formulas $a$ and $\neg \neg a$. Clearly $a \rightarrow \neg \neg a$ and $\neg \neg a \rightarrow a$. So we ahve $a \leq \neg \neg a$ and $\neg \neg a \leq a$, yet $a$ and $\neg \neg a$ are distinct memebers of $L$. This contradicts antisymmetry. \end{example} In order to fix this, we instead look at the quotient structure $\tuple{L_\sim, \leq}$, where $\phi \sim \psi$ iff $\phi$ and $\psi$ entail each other. So $L_\sim$ consists of equivalence classes of propositional formulas whose truth values never differ from each other. Strictly speaking, we have to define $\leq$ for this new set $L_\sim$, but the idea is simple: $[\phi] \leq [\psi]$ iff $\phi$ entails $\psi$. That is to say, we can order any two equivalence classes by looking at the entailment between their respective members. This makes $\tuple{L_\sim, \leq}$ a poset. Note that there is a unique least element, the equivalence class $[a \wedge \neg a]$, or simply $[\bot]$. And there is a unique greatest element $[a \vee \neg a]$, or simply $[\top]$. Any two elements $[\phi]$ and $[\psi]$ have a unique join $[\phi \vee \psi]$, and a unique meet $[\phi \wedge \psi]$. So the lattice operations of join ($\vee$) and meet ($\wedge$) correspond exactly to logical or ($\vee$) and logical and ($\wedge$). We already know that the latter are distributive, and it is also easy to see that for every $[\phi] \in L$ there is some complement $\neg [\phi] \in L$ such that $[\phi] \wedge \neg [\phi] = [\bot]$ and $[\phi] \vee \neg [\phi] = [\top]$. In fact, this complement $\neg [\phi]$ is simply $[\neg \phi]$, showing the close connection between logical negation and complement in this structure. Overall, then, $\tuple{L_\sim, \leq}$ is a poset where any two elements have a unique join and meet, where join and meet distribute over each, with unique least and greatest bounds, and a complement for every element of the $L_\sim$. That's just a very roundabout way of saying that $\tuple{L_\sim, \leq}$ is a Boolean lattice, just like $\tuple{\wp(S), \subseteq}$, and that's why De Morgan's Law holds for both sets and propositional logic. \begin{exercise} Instead of the propositional formulas themselves, we could also look at their truth values. Show that $\tuple{ \setof{F, T}, \leq }$ with $F \leq T$ is also a Boolean lattice. \end{exercise}
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 @@ -0,0 +1,12 @@ \documentclass[tikz]{standalone} \usepackage[linguistics]{forest} \begin{document} \begin{forest} [$\wedge$ [$p$] [$\neg$ [$q$] ] ] \end{forest} \end{document}
 @@ -0,0 +1,28 @@ \documentclass[tikz]{standalone} \usepackage[linguistics]{forest} \begin{document} \begin{forest} [$\neg$ [$\neg$ [$\neg$ [$\rightarrow$ [$\righarrow$ [$p$] [$q$] ] [$\neg$ [$\vee$ [$\neg$ [$p$] ] [$\neg$ [$q$] ] ] ] ] ] ] ] \end{forest} \end{document}