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Example 2.7.7. Since, for all \( x > 0 \) | \[ \frac{d}{dx}\log \left( x\right) = \frac{1}{x} > 0 \] and \[ \frac{{d}^{2}}{d{x}^{2}}\log \left( x\right) = - \frac{1}{{x}^{2}} < 0 \] the function \( y = \log \left( x\right) \) is increasing on \( \left( {0,\infty }\right) \) and its graph is concave downward on \( \left( {0,\infty }\right) \) . See Figure 2.7.2. | Yes |
Example 2.7.9. We may now complete the example, discussed in Example 2.5.8 and continued in Example 2.6.19, of finding the length \( L \) of the graph of \( y = {x}^{2} \) over the interval \( \left\lbrack {0,1}\right\rbrack \) . | In those examples we found that\n\n\[ L = {\int }_{0}^{1}\sqrt{1 + 4{x}^{2}}{dx} = \frac{\sqrt{5}}{2} + \frac{1}{4}{\int }_{1}^{2 + \sqrt{5}}\frac{1}{w}{dw}. \]\n\nNow we see that\n\n\[ {\int }_{1}^{2 + \sqrt{5}}\frac{1}{w}{dw} = {\left. \log \left( w\right) \right| }_{1}^{2 + \sqrt{5}} = \log \left( {2 + \sqrt{5}}\rig... | Yes |
To evaluate\n\n\[ \n{\int }_{0}^{1}\frac{x}{1 + {x}^{2}}{dx} \n\] | we first make the change of variable\n\n\[ \nu = 1 + {x}^{2} \]\n\n\[ {du} = {2xdx}. \]\n\nThen\n\n\[ \n{\int }_{0}^{1}\frac{x}{1 + {x}^{2}}{dx} = {\left. \frac{1}{2}{\int }_{1}^{2}\frac{1}{u}du = \frac{1}{2}\log \left( u\right) \right| }_{1}^{2} = \frac{\log \left( 2\right) }{2}. \n\] | Yes |
Evaluating\n\n\[ \n{\int }_{1}^{10}\log \left( x\right) {dx} \n\] | provides an interesting application of integration by parts. If we let\n\n\[ \nu = \log \left( x\right) \]\n\n\[ {dv} = {dx} \]\n\n\[ {du} = \frac{1}{x}{dx} \]\n\n\[ v = x, \]\n\nthen\n\n\[ \n{\int }_{1}^{10}\log \left( x\right) {dx} = {\left. x\log \left( x\right) \right| }_{1}^{10} - {\int }_{1}^{10}{dx} = {10}\log \... | Yes |
Example 2.7.12. We could use the change of variable \( u = {3x} + 2 \) to evaluate\n\n\[ \n{\int }_{0}^{5}\frac{4}{{3x} + 2}{dx} \n\] | \[ \n{\int }_{0}^{5}\frac{4}{{3x} + 2}{dx} = {\left. \frac{4}{3}\log \left( 3x + 2\right) \right| }_{0}^{5} = \frac{4}{3}\left( {\log \left( {17}\right) - \log \left( 2\right) }\right) = \frac{4}{3}\log \left( \frac{17}{2}\right) . \n\] | Yes |
Example 2.7.14. Carbon-14 is a naturally occurring radioactive isotope of carbon with a half-life of 5730 years. A living organism will maintain constant levels of carbon-14, which will begin to decay once the organism dies and is buried. Because of this, the amount of carbon-14 in the remains of an organism may be use... | \[ \n{0.14}{y}_{0} = {y}_{0}{e}^{kT} \n\]\n\nwhere, from (2.7.51),\n\n\[ \nk = - \frac{\log \left( 2\right) }{5730} \n\]\n\nIt follows that \( {0.14} = {e}^{kT} \), and so\n\n\[ \n{kT} = \log \left( {0.14}\right) \text{.} \n\]\n\nThus\n\[ \nT = \frac{\log \left( {0.14}\right) }{k} = - \frac{{5730}\log \left( {0.14}\rig... | Yes |
Example 2.7.15. Suppose that a certain habitat initially holds a population of 1000 deer, and that five years later the population has grown to 1200 deer. If we let \( y \) be the size of the population after \( t \) years, and assuming no constraints on the growth of the population, we would have\n\n\[ y = {1000}{e}^{... | after five more years. | Yes |
Example 2.7.16. In our previous example, where \( y \) represented the number of deer in a certain habitat after \( t \) years, we found\n\n\[ k = \frac{\log \left( {1.2}\right) }{5}. \]\n\nNow suppose the habitat can support no more than 20,000 deer. Then the logistic model would give us\n\n\[ z = \frac{\left( {1000}\... | After ten years, this model would predict a population of\n\n\[ z\left( {10}\right) = \frac{20000}{1 + {19}{e}^{-{10k}}} \approx {1409}\text{ deer,}\] | Yes |
To evaluate\n\n\\[ \n{\\int }_{0}^{1}\\frac{x}{{x}^{2} - 4}{dx} \n\\]\n\nwe first note that, since \\( {x}^{2} - 4 = \\left( {x - 2}\\right) \\left( {x + 2}\\right) \\), there exist constants \\( A \\) and \\( B \\)\n\nfor which\n\\[ \n\\frac{x}{{x}^{2} - 4} = \\frac{A}{x - 2} + \\frac{B}{x + 2}. \n\\] | It follows that\n\\[ \n\\frac{x}{{x}^{2} - 4} = \\frac{A\\left( {x + 2}\\right) + B\\left( {x - 2}\\right) }{{x}^{2} - 4}, \n\\]\n\nand so\n\n\\[ \nx = A\\left( {x + 2}\\right) + B\\left( {x - 2}\\right) \n\\]\n\nfor all values of \\( x \\) . In particular, when \\( x = 2 \\) we have \\( 2 = {4A} \\), and when \\( x = ... | Yes |
A number is called perfect iff it is equal to the sum of its proper divisors (i.e., numbers that evenly divide it but aren't identical to the number). For instance, 6 is perfect because its proper divisors are 1,2, and 3, and 6 = \( 1 + 2 + 3 \) . In fact,6 is the only positive integer less than 10 that is perfect. So,... | \[ \{ 6\} = \{ x : x\text{ is perfect and }0 \leq x \leq {10}\} \] We read the notation on the right as \ | Yes |
Every set is a subset of itself, and \( \varnothing \) is a subset of every set. | The set of even numbers is a subset of the set of natural numbers. Also, \( \{ a, b\} \subseteq \) \( \{ a, b, c\} \) . But \( \{ a, b, e\} \) is not a subset of \( \{ a, b, c\} \) . | No |
Extensionality gives a criterion of identity for sets: \( A = B \) iff every element of \( A \) is also an element of \( B \) and vice versa. | The definition of \ | No |
What are all the possible subsets of \( \{ a, b, c\} \) ? | They are: \( \varnothing \) , \( \{ a\} ,\{ b\} ,\{ c\} ,\{ a, b\} ,\{ a, c\} ,\{ b, c\} ,\{ a, b, c\} \) . The set of all these subsets is \( \wp \left( {\{ a, b, c\} }\right) \) :\n\n\[ \wp \left( {\{ a, b, c\} }\right) = \{ \varnothing ,\{ a\} ,\{ b\} ,\{ c\} ,\{ a, b\} ,\{ b, c\} ,\{ a, c\} ,\{ a, b, c\} \} | Yes |
Example 1.16. Since the multiplicity of elements doesn't matter, the union of two sets which have an element in common contains that element only once, e.g., \( \{ a, b, c\} \cup \{ a,0,1\} = \{ a, b, c,0,1\} \) . | The union of a set and one of its subsets is just the bigger set: \( \{ a, b, c\} \cup \) \( \{ a\} = \{ a, b, c\} . \n\nThe union of a set with the empty set is identical to the set: \( \{ a, b, c\} \cup \varnothing = \) \( \{ a, b, c\} \) . | No |
Example 1.25. If \( A = \{ 0,1\} \), and \( B = \{ 1, a, b\} \), then their product is | \[ A \times B = \{ \langle 0,1\rangle ,\langle 0, a\rangle ,\langle 0, b\rangle ,\langle 1,1\rangle ,\langle 1, a\rangle ,\langle 1, b\rangle \} . \] | Yes |
If \( A \) is a set, the product of \( A \) with itself, \( A \times A \), is also written \( {A}^{2} \). It is the set of all pairs \( \langle x, y\rangle \) with \( x, y \in A \). The set of all triples \( \langle x, y, z\rangle \) is \( {A}^{3} \), and so on. We can give a recursive definition: | \[ {A}^{1} = A \] \[ {A}^{k + 1} = {A}^{k} \times A \] | Yes |
Proposition 1.27. If \( A \) has \( n \) elements and \( B \) has \( m \) elements, then \( A \times B \) has \( n \cdot m \) elements. | Proof. For every element \( x \) in \( A \), there are \( m \) elements of the form \( \langle x, y\rangle \in \) \( A \times B \) . Let \( {B}_{x} = \{ \langle x, y\rangle : y \in B\} \) . Since whenever \( {x}_{1} \neq {x}_{2},\left\langle {{x}_{1}, y}\right\rangle \neq \left\langle {{x}_{2}, y}\right\rangle \) , \( ... | Yes |
Theorem 1.29 (Russell’s Paradox). There is no set \( R = \{ x : x \notin x\} \). | Proof. For reductio, suppose that \( R = \{ x : x \notin x\} \) exists. Then \( R \in R \) iff \( R \notin R \), since sets are extensional. But this is a contradicion. | Yes |
The set \( {\mathbb{N}}^{2} \) of pairs of natural numbers can be listed in a 2-dimensional matrix like this: | \[ \langle \mathbf{0},\mathbf{0}\rangle \;\langle 0,1\rangle \;\langle 0,2\rangle \;\langle 0,3\rangle \;\ldots \] \[ \langle 1,0\rangle \;\langle \mathbf{1},\mathbf{1}\rangle \;\langle 1,2\rangle \;\langle 1,3\rangle \;\ldots \] \[ \langle 2,0\rangle \;\langle 2,1\rangle \;\langle \mathbf{2},\mathbf{2}\rangle \;\langl... | Yes |
Proposition 2.12. If \( R \subseteq {A}^{2} \) is an equivalence relation, then \( {Rxy} \) iff \( {\left\lbrack x\right\rbrack }_{R} = {\left\lbrack y\right\rbrack }_{R} \) . | Proof. For the left-to-right direction, suppose \( {Rxy} \), and let \( z \in {\left\lbrack x\right\rbrack }_{R} \) . By definition, then, \( {Rxz} \) . Since \( R \) is an equivalence relation, \( {Ryz} \) . (Spelling this out: as \( {Rxy} \) and \( R \) is symmetric we have \( {Ryx} \), and as \( {Rxz} \) and \( R \)... | Yes |
A nice example of equivalence relations comes from modular arithmetic. For any \( a, b \), and \( n \in \mathbb{N} \), say that \( a{ \equiv }_{n}b \) iff dividing \( a \) by \( n \) gives remainder \( b \) . (Somewhat more symbolically: \( a{ \equiv }_{n}b \) iff \( \left( {\exists k \in \mathbb{N}}\right) a - b = {kn... | And there are exactly \( n \) distinct equivalence classes generated by \( { \equiv }_{n} \) ; that is, \( \mathbb{N}/{ \equiv }_{n} \) has \( n \) elements. These are: the set of numbers divisible by \( n \) without remainder, i.e., \( {\left\lbrack 0\right\rbrack }_{{ \equiv }_{n}} \) ; the set of numbers divisible b... | Yes |
Consider the no longer than relation \( \preccurlyeq \) on \( {\mathbb{B}}^{ * } : x \preccurlyeq y \) iff \( \operatorname{len}\left( x\right) \leq \) len \( \left( y\right) \) . | This is a preorder (reflexive and transitive), and even connected, but not a partial order, since it is not anti-symmetric. For instance, \( {01} \preccurlyeq {10} \) and \( {10} \preccurlyeq {01} \), but \( {01} \neq {10} \). | Yes |
Proposition 2.25. If \( R \) is a strict order on \( A \), then \( {R}^{ + } = R \cup {\operatorname{Id}}_{A} \) is a partial order. Moreover, if \( R \) is total, then \( {R}^{ + } \) is a linear order. | Proof. Suppose \( R \) is a strict order, i.e., \( R \subseteq {A}^{2} \) and \( R \) is irreflexive, asymmetric, and transitive. Let \( {R}^{ + } = R \cup {\operatorname{Id}}_{A} \) . We have to show that \( {R}^{ + } \) is reflexive, antisymmetric, and transitive.\n\n\( {R}^{ + } \) is clearly reflexive, since \( \la... | Yes |
Proposition 2.26. If \( R \) is a partial order on \( X \), then \( {R}^{ - } = R \smallsetminus {\operatorname{Id}}_{X} \) is a strict order. Moreover, if \( R \) is linear, then \( {R}^{ - } \) is total. | Proof. This is left as an exercise. | No |
Proposition 2.28. If \( < \) totally orders \( A \), then:\n\n\[ \left( {\forall a, b \in A}\right) \left( {\left( {\forall x \in A}\right) \left( {x < a \leftrightarrow x < b}\right) \rightarrow a = b}\right) \] | Proof. Suppose \( \left( {\forall x \in A}\right) \left( {x < a \leftrightarrow x < b}\right) \) . If \( a < b \), then \( a < a \), contradicting the fact that \( < \) is irreflexive; so \( a \nless b \) . Exactly similarly, \( b \nless a \) . So \( a = b \), as \( < \) is connected. | Yes |
Example 3.3. Multiplication is a function because it pairs each input-each pair of natural numbers-with a single output: \( \times : {\mathbb{N}}^{2} \rightarrow \mathbb{N} \) . | By contrast, the square root operation applied to the domain \( \mathbb{N} \) is not functional, since each positive integer \( n \) has two square roots: \( \sqrt{n} \) and \( - \sqrt{n} \) . We can make it functional by only returning the positive square root: \( \sqrt{} : \mathbb{N} \rightarrow \mathbb{R} \) . | Yes |
Example 3.5. Let \( f : \mathbb{N} \rightarrow \mathbb{N} \) be defined such that \( f\left( x\right) = x + 1 \). This is a definition that specifies \( f \) as a function which takes in natural numbers and outputs natural numbers. It tells us that, given a natural number \( x, f \) will output its successor \( x + 1 \... | In this case, the codomain \( \mathbb{N} \) is not the range of \( f \) , since the natural number 0 is not the successor of any natural number. The range of \( f \) is the set of all positive integers, \( {\mathbb{Z}}^{ + } \). | Yes |
Let \( g : \mathbb{N} \rightarrow \mathbb{N} \) be defined such that \( g\left( x\right) = x + 2 - 1 \) . This tells us that \( g \) is a function which takes in natural numbers and outputs natural numbers. Given a natural number \( n, g \) will output the predecessor of the successor of the successor of \( x \), i.e.,... | We just considered two functions, \( f \) and \( g \), with different definitions. However, these are the same function. After all, for any natural number \( n \), we have that \( f\left( n\right) = n + 1 = n + 2 - 1 = g\left( n\right) \) . Otherwise put: our definitions for \( f \) and \( g \) specify the same mapping... | Yes |
We can also define functions by cases. For instance, we could define \( h : \mathbb{N} \rightarrow \mathbb{N} \) by\n\n\[ h\left( x\right) = \left\{ \begin{array}{ll} \frac{x}{2} & \text{ if }x\text{ is even } \\ \frac{x + 1}{2} & \text{ if }x\text{ is odd. } \end{array}\right. \] | Since every natural number is either even or odd, the output of this function will always be a natural number. Just remember that if you define a function by cases, every possible input must fall into exactly one case. | Yes |
Proposition 3.13. Let \( R \subseteq A \times B \) be such that:\n\n1. If \( {Rxy} \) and \( {Rxz} \) then \( y = z \) ; and\n\n2. for every \( x \in A \) there is some \( y \in B \) such that \( \langle x, y\rangle \in R \) .\n\nThen \( R \) is the graph of the function \( f : A \rightarrow B \) defined by \( f\left( ... | Proof. Suppose there is a \( y \) such that \( {Rxy} \) . If there were another \( z \neq y \) such that \( {Rxz} \), the condition on \( R \) would be violated. Hence, if there is a \( y \) such that \( {Rxy} \), this \( y \) is unique, and so \( f \) is well-defined. Obviously, \( {R}_{f} = R \) . | Yes |
Proposition 3.16. Every bijection has a unique inverse. | Proof. Exercise. | No |
Proposition 3.17. Every function \( f \) has at most one inverse. | Proof. Exercise. | No |
Consider the functions \( f\left( x\right) = x + 1 \), and \( g\left( x\right) = {2x} \). Since \( \left( {g \circ f}\right) \left( x\right) = g\left( {f\left( x\right) }\right) \), for each input \( x \) you must first take its successor, then multiply the result by two. So their composition is given by \( \left( {g \... | Since \( \left( {g \circ f}\right) \left( x\right) = g\left( {f\left( x\right) }\right) \), for each input \( x \) you must first take its successor, then multiply the result by two. So their composition is given by \( \left( {g \circ f}\right) \left( x\right) = \) \( 2\left( {x + 1}\right) \). | Yes |
Proposition 3.24. Suppose \( R \subseteq A \times B \) has the property that whenever \( {Rxy} \) and \( {Rx}{y}^{\prime } \) then \( y = {y}^{\prime } \) . Then \( R \) is the graph of the partial function \( f : X \rightarrow Y \) defined by: if there is a \( y \) such that \( {Rxy} \), then \( f\left( x\right) = y \... | Proof. Suppose there is a \( y \) such that \( {Rxy} \) . If there were another \( {y}^{\prime } \neq y \) such that \( {Rx}{y}^{\prime } \), the condition on \( R \) would be violated. Hence, if there is a \( y \) such that \( {Rxy} \), that \( y \) is unique, and so \( f \) is well-defined. Obviously, \( {R}_{f} = R ... | Yes |
Proposition 4.2. If \( A \) has an enumeration, it has an enumeration without repetitions. | Proof. Suppose \( A \) has an enumeration \( {x}_{1},{x}_{2},\ldots \) in which each \( {x}_{i} \) is an element of \( A \) . We can remove repetitions from an enumeration by removing repeated elements. For instance, we can turn the enumeration into a new one in which we list \( {x}_{i} \) if it is an element of \( A \... | Yes |
The function \( f\left( n\right) = {\left( -1\right) }^{n}\left\lceil \frac{\left( n - 1\right) }{2}\right\rceil \) (where \( \lceil x\rceil \) denotes the ceiling function, which rounds \( x \) up to the nearest integer) enumerates the set of integers \( \mathbb{Z} \). | Notice how \( f \) generates the values of \( \mathbb{Z} \) by \ | No |
Proposition 4.8. There is a surjection \( f : {\mathbb{Z}}^{ + } \rightarrow A \) iff there is a surjection \( g : \mathbb{N} \rightarrow \) A. | Proof. Given a surjection \( f : {\mathbb{Z}}^{ + } \rightarrow A \), we can define \( g\left( n\right) = f\left( {n + 1}\right) \) for all \( n \in \mathbb{N} \) . It is easy to see that \( g : \mathbb{N} \rightarrow A \) is surjective. Conversely, given a surjection \( g : \mathbb{N} \rightarrow A \), define \( f\lef... | Yes |
Proposition 4.10. If \( f : {\mathbb{Z}}^{ + } \rightarrow A \) is surjective (i.e., an enumeration of \( A \) ), there is a bijection \( g : Z \rightarrow A \) where \( Z \) is either \( {\mathbb{Z}}^{ + } \) or \( \{ 1,\ldots, n\} \) for some \( n \in {\mathbb{Z}}^{ + } \) . | Proof. We define the function \( g \) recursively: Let \( g\left( 1\right) = f\left( 1\right) \) . If \( g\left( i\right) \) has already been defined, let \( g\left( {i + 1}\right) \) be the first value of \( f\left( 1\right), f\left( 2\right) ,\ldots \) not already among \( g\left( 1\right) ,\ldots, g\left( i\right) \... | Yes |
Corollary 4.11. A set \( A \) is enumerable iff it is empty or there is a bijection \( f : N \rightarrow \) \( A \) where either \( N = \mathbb{N} \) or \( N = \{ 0,\ldots, n\} \) for some \( n \in \mathbb{N} \) . | Proof. \( A \) is enumerable iff \( A \) is empty or there is a surjective \( f : {\mathbb{Z}}^{ + } \rightarrow A \) . By Proposition 4.10, the latter holds iff there is a bijective function \( f : Z \rightarrow A \) where \( Z = {\mathbb{Z}}^{ + } \) or \( Z = \{ 1,\ldots, n\} \) for some \( n \in {\mathbb{Z}}^{ + } ... | Yes |
Proposition 4.12. \( \mathbb{N} \times \mathbb{N} \) is enumerable. | Proof. Let \( f : \mathbb{N} \rightarrow \mathbb{N} \times \mathbb{N} \) take each \( k \in \mathbb{N} \) to the tuple \( \langle n, m\rangle \in \mathbb{N} \times \mathbb{N} \) such that \( k \) is the value of the \( n \) th row and \( m \) th column in Cantor’s zig-zag array. \( ▱ \) | No |
Proposition 4.13. \( {\mathbb{N}}^{n} \) is enumerable, for every \( n \in \mathbb{N} \) . | Cantor’s zig-zag method makes the enumerability of \( {\mathbb{N}}^{n} \) visually evident. But let us focus on our array depicting \( {\mathbb{N}}^{2} \) . Following the zig-zag line in the array and counting the places, we can check that \( \langle 1,2\rangle \) is associated with the number 7. However, it would be n... | Yes |
The function \( h : {\mathbb{N}}^{2} \rightarrow \mathbb{N} \) given by\n\n\[ h\left( {n, m}\right) = {2}^{n}\left( {{2m} + 1}\right) - 1 \]\n\nis a pairing function for the set of pairs of natural numbers \( {\mathbb{N}}^{2} \) . | Accordingly, in our second enumeration of \( {\mathbb{N}}^{2} \), the pair \( \langle 0,0\rangle \) has code \( h\left( {0,0}\right) = {2}^{0}\left( {2 \cdot 0 + 1}\right) - 1 = 0;\langle 1,2\rangle \) has code \( {2}^{1} \cdot \left( {2 \cdot 2 + 1}\right) - 1 = 2 \cdot 5 - 1 = 9 \) ; \( \langle 2,6\rangle \) has code... | No |
Theorem 4.18. \( \wp \left( {\mathbb{Z}}^{ + }\right) \) is not enumerable. | Proof. We proceed in the same way, by showing that for every list of subsets of \( {\mathbb{Z}}^{ + } \) there is a subset of \( {\mathbb{Z}}^{ + } \) which cannot be on the list. Suppose the following is a given list of subsets of \( {\mathbb{Z}}^{ + } \) :\n\n\[ \n{Z}_{1},{Z}_{2},{Z}_{3},\ldots \n\] \n\nWe now define... | Yes |
Proposition 4.20. Equinumerosity is an equivalence relation. | Proof. We must show that equinumerosity is reflexive, symmetric, and transitive. Let \( A, B \), and \( C \) be sets.\n\nReflexivity. The identity map \( {\operatorname{Id}}_{A} : A \rightarrow A \), where \( {\operatorname{Id}}_{A}\left( x\right) = x \) for all \( x \in A \) , is a bijection. So \( A \approx A \) .\n\... | Yes |
Proposition 4.21. If \( A \approx B \), then \( A \) is enumerable if and only if \( B \) is. | Proof. Suppose \( A \approx B \), so there is some bijection \( f : A \rightarrow B \), and suppose that \( A \) is enumerable. Then either \( A = \varnothing \) or there is a surjective function \( g : {\mathbb{Z}}^{ + } \rightarrow \) \( A \) . If \( A = \varnothing \), then \( B = \varnothing \) also (otherwise ther... | Yes |
Theorem 4.24 (Cantor). \( A \prec \wp \left( A\right) \), for any set \( A \) . | Proof. The map \( f\left( x\right) = \{ x\} \) is an injection \( f : A \rightarrow \wp \left( A\right) \), since if \( x \neq y \) , then also \( \{ x\} \neq \{ y\} \) by extensionality, and so \( f\left( x\right) \neq f\left( y\right) \) . So we have that \( A \preccurlyeq \wp \left( A\right) \) . We show that there ... | Yes |
Let \( \lceil x\rceil \) be the ceiling function, which rounds \( x \) up to the nearest integer. Then the function \( f : \mathbb{N} \rightarrow \mathbb{Z} \) given by:\n\n\[ f\left( n\right) = {\left( -1\right) }^{n}\left\lceil \frac{n}{2}\right\rceil \]\n\nenumerates the set of integers \( \mathbb{Z} \) as follows:\... | Notice how \( f \) generates the values of \( \mathbb{Z} \) by \ | Yes |
Theorem 4.31. \( {\mathbb{B}}^{\omega } \) is non-enumerable. | Proof. Consider any enumeration of a subset of \( {\mathbb{B}}^{\omega } \) . So we have some list \( {s}_{0} \) , \( {s}_{1},{s}_{2},\ldots \) where every \( {s}_{n} \) is an infinite string of \( {0}^{\prime }\mathrm{s} \) and \( {1}^{\prime }\mathrm{s} \) . Let \( {s}_{n}\left( m\right) \) be the \( n \) th digit of... | Yes |
Theorem 4.32. \( \wp \left( \mathbb{N}\right) \) is not enumerable. | Proof. We proceed in the same way, by showing that every list of subsets of \( \mathbb{N} \) omits some subset of \( \mathbb{N} \) . So, suppose that we have some list \( {N}_{0},{N}_{1},{N}_{2},\ldots \) of subsets of \( \mathbb{N} \) . We define a set \( D \) as follows: \( n \in D \) iff \( n \notin {N}_{n} \) :\n\n... | Yes |
Proposition 5.1. \( \sim \) is an equivalence relation. | Proof. We must show that \( \sim \) is reflexive, symmetric, and transitive.\n\nReflexivity: Evidently \( \langle a, b\rangle \sim \langle a, b\rangle \), since \( a + b = b + a \).\n\nSymmetry: Suppose \( \langle a, b\rangle \sim \langle c, d\rangle \), so \( a + d = c + b \). Then \( c + b = a + d \), so that \( \lan... | Yes |
Theorem 5.4. \( \sqrt{2} \) is not rational, i.e., \( \sqrt{2} \notin \mathbb{Q} \) | Proof. Suppose, for reductio, that \( \sqrt{2} \) is rational. So \( \sqrt{2} = m/n \) for some natural numbers \( m \) and \( n \) . Indeed, we can choose \( m \) and \( n \) so that the fraction cannot be reduced any further. Re-organising, \( {m}^{2} = 2{n}^{2} \) . From here, we can complete the proof in two ways:\... | Yes |
Theorem 5.6. The set of cuts has the Completeness Property. | Proof. Let \( S \) be any non-empty set of cuts with an upper bound. Let \( \lambda = \bigcup S \) . We first claim that \( \lambda \) is a cut: 1. Since \( S \) has an upper bound, at least one cut is in \( S \), so \( \varnothing \neq \alpha \) . Since \( S \) is a set of cuts, \( \lambda \subseteq \mathbb{Q} \) . Si... | Yes |
Theorem 5.11. The Cauchy sequences constitute an ordered field. | Proof. Exercise. | No |
Theorem 5.12. Every non-empty set of Cauchy sequences with an upper bound has a least upper bound. | Proof sketch. Let \( S \) be any non-empty set of Cauchy sequences with an upper bound. So there is some \( p \in \mathbb{Q} \) such that \( {p}_{\mathbb{R}} \) is an upper bound for \( S \) . Let \( r \in S \) ; then there is some \( q \in \mathbb{Q} \) such that \( {q}_{\mathbb{R}} < r \) . So if a least upper bound ... | Yes |
Lemma 6.3. For any function \( f \) and any \( o \in A \): 1. \( o \in {\operatorname{clo}}_{f}\left( o\right) \) ; and 2. \( {\operatorname{clo}}_{f}\left( o\right) \) is \( f \) -closed; and 3. if \( X \) is \( f \) -closed and \( o \in X \), then \( {\operatorname{clo}}_{f}\left( o\right) \subseteq X \) | Proof. Note that there is at least one \( f \) -closed set, namely \( \operatorname{ran}\left( f\right) \cup \{ o\} \) . So \( {\operatorname{clo}}_{f}\left( o\right) \), the intersection of all such sets, exists. We must now check (1)-(3). (1). \( o \in {\operatorname{clo}}_{f}\left( o\right) \) as it is an intersecti... | Yes |
Theorem 6.5. If there is a Dedekind infinite set, then there is a Dedekind algebra. | Proof. Let \( D \) be Dedekind infinite. So there is an injection \( g : D \rightarrow D \) and an element \( o \in D \smallsetminus \operatorname{ran}\left( g\right) \) . Now let \( A = {\operatorname{clo}}_{g}\left( o\right) \), and note that \( o \in A \) . Let \( f = g{ \upharpoonright }_{A} \) . We will show that ... | Yes |
Theorem 6.6 (Arithmetical induction). Let \( N, s \), o yield a Dedekind algebra. Then for any set \( X \) :\n\n\[ \text{if}o \in X\text{and}\left( {\forall n \in N \cap X}\right) s\left( n\right) \in X\text{, then}N \subseteq X\text{.} \] | Proof. By the definition of a Dedekind algebra, \( N = {\operatorname{clo}}_{s}\left( o\right) \) . Now if both \( o \in X \) and \( \left( {\forall n \in N}\right) \left( {n \in X \rightarrow s\left( n\right) \in X}\right) \), then \( N = {\operatorname{clo}}_{s}\left( o\right) \subseteq X \) . | Yes |
Corollary 6.7. Let \( N, s, o \) yield a Dedekind algebra. Then for any formula \( \varphi \left( x\right) \) , which may have parameters:\n\n\[ \text{if}\varphi \left( o\right) \text{and}\left( {\forall n \in N}\right) \left( {\varphi \left( n\right) \rightarrow \varphi \left( {s\left( n\right) }\right) }\right) \text... | Proof. Let \( X = \{ n \in N : \varphi \left( n\right) \} \), and now use Theorem 6.6 | No |
Proposition 6.8. For any function \( f \), and any \( B \): 1. \( B \subseteq {\operatorname{Clo}}_{f}\left( B\right) \) ; and 2. \( {\operatorname{Clo}}_{f}\left( B\right) \) is \( f \) -closed; and 3. if \( X \) is \( f \) -closed and \( B \subseteq X \), then \( {\operatorname{Clo}}_{f}\left( B\right) \subseteq X \)... | Proof. Exactly as in Lemma 6.3. | No |
Proposition 6.9. If \( A \subseteq B \subseteq C \) and \( A \approx C \), then \( A \approx B \approx C \). | Proof. Given a bijection \( f : C \rightarrow A \), let \( F = {\operatorname{Clo}}_{f}\left( {C \smallsetminus B}\right) \) and define a function \( g \) with domain \( C \) as follows:\n\n\[ g\left( x\right) = \left\{ \begin{array}{ll} f\left( x\right) & \text{ if }x \in F \\ x & \text{ otherwise } \end{array}\right.... | Yes |
Theorem 7.4 (Principle of induction on formulas). If some property \( P \) holds for all the atomic formulas and is such that\n\n1. it holds for \( \neg \varphi \) whenever it holds for \( \varphi \) ;\n\n2. it holds for \( \left( {\varphi \land \psi }\right) \) whenever it holds for \( \varphi \) and \( \psi \) ;\n\n3... | Proof. Let \( S \) be the collection of all formulas with property \( P \) . Clearly \( S \subseteq \) \( \operatorname{Frm}\left( {\mathcal{L}}_{0}\right) \) . \( S \) satisfies all the conditions of Definition 7.1: it contains all atomic formulas and is closed under the logical operators. \( \operatorname{Frm}\left( ... | Yes |
Proposition 7.7 (Unique Readability). Any formula \( \varphi \) in \( \operatorname{Frm}\left( {\mathcal{L}}_{0}\right) \) has exactly one parsing as one of the following\n\n1. \( \bot \) .\n\n2. \( {p}_{n} \) for some \( {p}_{n} \in {\mathrm{{At}}}_{0} \) .\n\n3. \( \neg \psi \) for some formula \( \psi \) .\n\n4. \( ... | Proof. By induction on \( \varphi \) . For instance, suppose that \( \varphi \) has two distinct readings as \( \left( {\psi \rightarrow \chi }\right) \) and \( \left( {{\psi }^{\prime } \rightarrow {\chi }^{\prime }}\right) \) . Then \( \psi \) and \( {\psi }^{\prime } \) must be the same (or else one would be a prope... | Yes |
Theorem 7.11 (Local Determination). Suppose that \( {\mathfrak{v}}_{1} \) and \( {\mathfrak{v}}_{2} \) are valuations that agree on the propositional letters occurring in \( \varphi \), i.e., \( {\mathfrak{v}}_{1}\left( {p}_{n}\right) = {\mathfrak{v}}_{2}\left( {p}_{n}\right) \) whenever \( {p}_{n} \) occurs in some fo... | Proof. By induction on \( \varphi \) . | No |
Proposition 7.13. \( \mathfrak{v} \vDash \varphi \) iff \( \overline{\mathfrak{v}}\left( \varphi \right) = \mathbb{T} \) . | Proof. By induction on \( \varphi \) . | No |
Proposition 7.15. 1. \( \varphi \) is a tautology if and only if \( \varnothing \vDash \varphi \) ; | Proof. Exercise. | No |
Proposition 7.16. \( \Gamma \vDash \varphi \) if and only if \( \Gamma \cup \{ \neg \varphi \} \) is unsatisfiable. | Proof. Exercise. | No |
Theorem 7.17 (Semantic Deduction Theorem). \( \Gamma \vDash \varphi \rightarrow \psi \) if and only if \( \Gamma \cup \) \( \{ \varphi \} \vDash \psi \) . | Proof. Exercise. | No |
Every initial sequent, e.g., \( \chi \Rightarrow \chi \) is a derivation. | We can obtain a new derivation from this by applying, say, the WL rule,\n\n\[ \frac{\Gamma \Rightarrow \Delta }{\varphi ,\Gamma \Rightarrow \Delta }\mathrm{{WL}} \]\n\nThe rule, however, is meant to be general: we can replace the \( \varphi \) in the rule with any sentence, e.g., also with \( \theta \) . If the premise... | Yes |
Give an LK-derivation for the sequent \( \varphi \land \psi \Rightarrow \varphi \) . | We begin by writing the desired end-sequent at the bottom of the derivation.\n\n\[ \varphi \land \psi \Rightarrow \varphi \]\n\nNext, we need to figure out what kind of inference could have a lower sequent of this form. This could be a structural rule, but it is a good idea to start by looking for a logical rule. The o... | Yes |
Suppose we want to prove \( \Rightarrow \varphi \vee \neg \varphi \) . | Applying VR backwards would give us one of these two derivations:\n\n\[ \begin{array}{l} \Rightarrow \varphi \\ \Rightarrow \varphi \vee \neg \varphi \vee \mathrm{R} \\ \end{array} \]\n\nNeither of these of course ends in an initial sequent. The trick is to realize that the contraction rule allows us to combine two cop... | Yes |
Proposition 9.12 (Reflexivity). If \( \varphi \in \Gamma \), then \( \Gamma \vdash \varphi \) . | Proof. The initial sequent \( \varphi \Rightarrow \varphi \) is derivable, and \( \{ \varphi \} \subseteq \Gamma \) . | Yes |
Proposition 9.13 (Monotony). If \( \Gamma \subseteq \Delta \) and \( \Gamma \vdash \varphi \), then \( \Delta \vdash \varphi \) . | Proof. Suppose \( \Gamma \vdash \varphi \), i.e., there is a finite \( {\Gamma }_{0} \subseteq \Gamma \) such that \( {\Gamma }_{0} \Rightarrow \varphi \) is derivable. Since \( \Gamma \subseteq \Delta \), then \( {\Gamma }_{0} \) is also a finite subset of \( \Delta \) . The derivation of \( {\Gamma }_{0} \Rightarrow ... | Yes |
Proposition 9.14 (Transitivity). If \( \Gamma \vdash \varphi \) and \( \{ \varphi \} \cup \Delta \vdash \psi \), then \( \Gamma \cup \Delta \vdash \psi \) . | Proof. If \( \Gamma \vdash \varphi \), there is a finite \( {\Gamma }_{0} \subseteq \Gamma \) and a derivation \( {\pi }_{0} \) of \( {\Gamma }_{0} \Rightarrow \varphi \) . If \( \{ \varphi \} \cup \Delta \vdash \psi \), then for some finite subset \( {\Delta }_{0} \subseteq \Delta \), there is a derivation \( {\pi }_{... | Yes |
Proposition 9.15. \( \Gamma \) is inconsistent iff \( \Gamma \vdash \varphi \) for every sentence \( \varphi \) . | Proof. Exercise. | No |
Proposition 9.16 (Compactness). 1. If \( \Gamma \vdash \varphi \) then there is a finite subset \( {\Gamma }_{0} \subseteq \Gamma \) such that \( {\Gamma }_{0} \vdash \varphi \) . 2. If every finite subset of \( \Gamma \) is consistent, then \( \Gamma \) is consistent. | Proof. 1. If \( \Gamma \vdash \varphi \), then there is a finite subset \( {\Gamma }_{0} \subseteq \Gamma \) such that the sequent \( {\Gamma }_{0} \Rightarrow \varphi \) has a derivation. Consequently, \( {\Gamma }_{0} \vdash \varphi \) . 2. If \( \Gamma \) is inconsistent, there is a finite subset \( {\Gamma }_{0} \s... | Yes |
Proposition 9.17. If \( \Gamma \vdash \varphi \) and \( \Gamma \cup \{ \varphi \} \) is inconsistent, then \( \Gamma \) is inconsistent. | Proof. There are finite \( {\Gamma }_{0} \) and \( {\Gamma }_{1} \subseteq \Gamma \) such that \( \mathbf{{LK}} \) derives \( {\Gamma }_{0} \Rightarrow \varphi \) and \( \varphi ,{\Gamma }_{1} \Rightarrow \) . Let the LK-derivation of \( {\Gamma }_{0} \Rightarrow \varphi \) be \( {\pi }_{0} \) and the LK-derivation of ... | Yes |
Proposition 9.18. \( \Gamma \vdash \varphi \) iff \( \Gamma \cup \{ \neg \varphi \} \) is inconsistent. | Proof. First suppose \( \Gamma \vdash \varphi \), i.e., there is a derivation \( {\pi }_{0} \) of \( \Gamma \Rightarrow \varphi \) . By adding a \( \neg \mathrm{L} \) rule, we obtain a derivation of \( \neg \varphi ,\Gamma \Rightarrow \), i.e., \( \Gamma \cup \{ \neg \varphi \} \) is inconsistent.\n\nIf \( \Gamma \cup ... | Yes |
Proposition 9.19. If \( \Gamma \vdash \varphi \) and \( \neg \varphi \in \Gamma \), then \( \Gamma \) is inconsistent. | Proof. Suppose \( \Gamma \vdash \varphi \) and \( \neg \varphi \in \Gamma \) . Then there is a derivation \( \pi \) of a sequent \( {\Gamma }_{0} \Rightarrow \varphi \) . The sequent \( \neg \varphi ,{\Gamma }_{0} \Rightarrow \) is also derivable:\n\n\[ \begin{array}{l} {\Gamma }_{0} \Rightarrow \varphi \;\frac{\varphi... | Yes |
Proposition 9.20. If \( \Gamma \cup \{ \varphi \} \) and \( \Gamma \cup \{ \neg \varphi \} \) are both inconsistent, then \( \Gamma \) is inconsistent. | Proof. There are finite sets \( {\Gamma }_{0} \subseteq \Gamma \) and \( {\Gamma }_{1} \subseteq \Gamma \) and LK-derivations \( {\pi }_{0} \) and \( {\pi }_{1} \) of \( \varphi ,{\Gamma }_{0} \Rightarrow \) and \( \neg \varphi ,{\Gamma }_{1} \Rightarrow \), respectively. We can then derive ![c5962287-92b4-4003-ac67-b5... | Yes |
Proposition 9.21. 1. Both \( \varphi \land \psi \vdash \varphi \) and \( \varphi \land \psi \vdash \psi \). 2. \( \varphi ,\psi \vdash \varphi \land \psi \) . | Proof. 1. Both sequents \( \varphi \land \psi \Rightarrow \varphi \) and \( \varphi \land \psi \Rightarrow \psi \) are derivable:\n\n\[ \frac{\varphi \Rightarrow \varphi }{\varphi \land \psi \Rightarrow \varphi } \land \mathrm{L}\;\frac{\psi \Rightarrow \psi }{\varphi \land \psi \Rightarrow \psi } \land \mathrm{L} \]\n... | Yes |
Proposition 9.22. 1. \( \varphi \vee \psi ,\neg \varphi ,\neg \psi \) is inconsistent. | Proof. 1. We give a derivation of the sequent \( \varphi \vee \psi ,\neg \varphi ,\neg \psi \Rightarrow \) :\n\n\[ \frac{\varphi \Rightarrow \varphi }{\neg \varphi ,\varphi \Rightarrow }\neg \mathrm{L}\;\frac{\psi \Rightarrow \psi }{\neg \psi ,\psi \Rightarrow }\neg \mathrm{L} \]\n\n\[ \frac{\varphi ,\neg \varphi ,\neg... | Yes |
Proposition 9.23. 1. \( \varphi ,\varphi \rightarrow \psi \vdash \psi \) . | 1. The sequent \( \varphi \rightarrow \psi ,\varphi \Rightarrow \psi \) is derivable:\n\n\[ \frac{\varphi \Rightarrow \varphi \;\psi \Rightarrow \psi }{\varphi \rightarrow \psi ,\varphi \Rightarrow \psi } \rightarrow \mathrm{L} \] | No |
Corollary 9.27. If \( \Gamma \vdash \varphi \) then \( \Gamma \vDash \varphi \) . | Proof. If \( \Gamma \vdash \varphi \) then for some finite subset \( {\Gamma }_{0} \subseteq \Gamma \), there is a derivation of \( {\Gamma }_{0} \Rightarrow \varphi \) . By Theorem 9.25, every valuation \( \mathfrak{v} \) either makes some \( \psi \in {\Gamma }_{0} \) false or makes \( \varphi \) true. Hence, if \( \m... | Yes |
Corollary 9.28. If \( \Gamma \) is satisfiable, then it is consistent. | Proof. We prove the contrapositive. Suppose that \( \Gamma \) is not consistent. Then there is a finite \( {\Gamma }_{0} \subseteq \Gamma \) and a derivation of \( {\Gamma }_{0} \Rightarrow \) . By Theorem 9.25, \( {\Gamma }_{0} \Rightarrow \) is valid. In other words, for every valuation \( \mathfrak{v} \), there is \... | Yes |
Every assumption on its own is a derivation. So, e.g., \( \chi \) by itself is a derivation, and so is \( \theta \) by itself. We can obtain a new derivation from these by applying, say, the \( \land \) Intro rule, | \[ \frac{\varphi }{\varphi \land \psi } \land \text{ Intro } \] These rules are meant to be general: we can replace the \( \varphi \) and \( \psi \) in it with any sentences, e.g., by \( \chi \) and \( \theta \) . Then the conclusion would be \( \chi \land \theta \), and so \[ \frac{\chi \;\theta }{\chi \land \theta } ... | Yes |
Let’s give a derivation of the sentence \( \left( {\varphi \land \psi }\right) \rightarrow \varphi \) . | \[ \frac{\frac{{\left\lbrack \varphi \land \psi \right\rbrack }^{1}}{\varphi } \land \text{ Elim }}{1\frac{\left( {\varphi \land \psi }\right) \rightarrow \varphi }{\left( {\varphi \land \psi }\right) \rightarrow \varphi } \rightarrow \text{ Intro }} \] | No |
For instance, suppose we want to derive \( \varphi \vee \neg \varphi \) . Our usual strategy would be to attempt to derive \( \varphi \vee \neg \varphi \) using VIntro. But this would require us to derive either \( \varphi \) or \( \neg \varphi \) from no assumptions, and this can’t be done. \( { \bot }_{C} \) to the r... | Now we’re looking for a derivation of \( \bot \) from \( \neg \left( {\varphi \vee \neg \varphi }\right) \) . Since \( \bot \) is the conclusion of \( \neg \) Elim we might try that:\n\n\n\nOur strategy for finding a... | Yes |
Proposition 10.10 (Reflexivity). If \( \varphi \in \Gamma \), then \( \Gamma \vdash \varphi \) . | Proof. The assumption \( \varphi \) by itself is a derivation of \( \varphi \) where every undischarged assumption (i.e., \( \varphi \) ) is in \( \Gamma \) . | Yes |
Proposition 10.11 (Monotony). If \( \Gamma \subseteq \Delta \) and \( \Gamma \vdash \varphi \), then \( \Delta \vdash \varphi \) . | Proof. Any derivation of \( \varphi \) from \( \Gamma \) is also a derivation of \( \varphi \) from \( \Delta \) . | Yes |
Proposition 10.12 (Transitivity). If \( \Gamma \vdash \varphi \) and \( \{ \varphi \} \cup \Delta \vdash \psi \), then \( \Gamma \cup \Delta \vdash \psi \) . | Proof. If \( \Gamma \vdash \varphi \), there is a derivation \( {\delta }_{0} \) of \( \varphi \) with all undischarged assumptions in \( \Gamma \) . If \( \{ \varphi \} \cup \Delta \vdash \psi \), then there is a derivation \( {\delta }_{1} \) of \( \psi \) with all undischarged assumptions in \( \{ \varphi \} \cup \D... | Yes |
Proposition 10.13. \( \Gamma \) is inconsistent iff \( \Gamma \vdash \varphi \) for every sentence \( \varphi \) . | Proof. Exercise. | No |
Proposition 10.14 (Compactness). 1. If \( \Gamma \vdash \varphi \) then there is a finite subset \( {\Gamma }_{0} \subseteq \Gamma \) such that \( {\Gamma }_{0} \vdash \varphi \). | Proof. 1. If \( \Gamma \vdash \varphi \), then there is a derivation \( \delta \) of \( \varphi \) from \( \Gamma \) . Let \( {\Gamma }_{0} \) be the set of undischarged assumptions of \( \delta \) . Since any derivation is finite, \( {\Gamma }_{0} \) can only contain finitely many sentences. So, \( \delta \) is a deri... | Yes |
Proposition 10.15. If \( \Gamma \vdash \varphi \) and \( \Gamma \cup \{ \varphi \} \) is inconsistent, then \( \Gamma \) is inconsistent. | Proof. Let the derivation of \( \varphi \) from \( \Gamma \) be \( {\delta }_{1} \) and the derivation of \( \bot \) from \( \Gamma \cup \) \( \{ \varphi \} \) be \( {\delta }_{2} \) . We can then derive:\n\n\n\nIn t... | Yes |
Proposition 10.16. \( \Gamma \vdash \varphi \) iff \( \Gamma \cup \{ \neg \varphi \} \) is inconsistent. | Proof. First suppose \( \Gamma \vdash \varphi \), i.e., there is a derivation \( {\delta }_{0} \) of \( \varphi \) from undischarged assumptions \( \Gamma \) . We obtain a derivation of \( \bot \) from \( \Gamma \cup \{ \neg \varphi \} \) as follows:\n\n and \( \neg \varphi \in \Gamma \), then \( \Gamma \) is inconsistent. | Proof. Suppose \( \Gamma \vdash \varphi \) and \( \neg \varphi \in \Gamma \) . Then there is a derivation \( \delta \) of \( \varphi \) from \( \Gamma \) . Consider this simple application of the \( \neg \) Elim rule:\n\n and \( \Gamma \cup \{ \neg \varphi \} \) are both inconsistent, then \( \Gamma \) is inconsistent. | Proof. There are derivations \( {\delta }_{1} \) and \( {\delta }_{2} \) of \( \bot \) from \( \Gamma \cup \{ \varphi \} \) and \( \bot \) from \( \Gamma \cup \{ \neg \varphi \} \) , respectively. We can then derive\n\n and \( \varphi \land \psi \vdash \psi \)\n\n2. \( \varphi ,\psi \vdash \varphi \land \psi \) . | Proof. 1. We can derive both\n\n\[ \frac{\varphi \land \psi }{\varphi } \land \operatorname{Elim}\;\frac{\varphi \land \psi }{\psi } \land \operatorname{Elim} \]\n\n2. We can derive:\n\n\[ \frac{\varphi \;\psi }{\varphi \land \psi } \land \text{Intro} \] | Yes |
Proposition 10.20. 1. \( \varphi \vee \psi ,\neg \varphi ,\neg \psi \) is inconsistent. | Proof. 1. Consider the following derivation:\n\n\[ \frac{\varphi \vee \psi \;\frac{\neg \varphi \;{\left\lbrack \varphi \right\rbrack }^{1}}{ \bot }\neg \operatorname{Elim}\;\frac{\neg \psi \;{\left\lbrack \psi \right\rbrack }^{1}}{ \bot }\neg \operatorname{Elim}}{ \bot }\text{VElim} \]\n\nThis is a derivation of \( \b... | Yes |
Proposition 10.21. 1. \( \varphi ,\varphi \rightarrow \psi \vdash \psi \) . 2. Both \( \neg \varphi \vdash \varphi \rightarrow \psi \) and \( \psi \vdash \varphi \rightarrow \psi \) . | Proof. 1. We can derive: \[ \frac{\varphi \rightarrow \psi \;\varphi }{\psi } \rightarrow \text{Elim} \] 2. This is shown by the following two derivations: \[ \begin{array}{l} \frac{\neg \varphi \;{\left\lbrack \varphi \right\rbrack }^{1}}{\frac{\frac{1}{\psi }{ \bot }_{I}}{\varphi \rightarrow \psi } \rightarrow \text{... | Yes |
Corollary 10.24. If \( \Gamma \) is satisfiable, then it is consistent. | Proof. We prove the contrapositive. Suppose that \( \Gamma \) is not consistent. Then \( \Gamma \vdash \bot \), i.e., there is a derivation of \( \bot \) from undischarged assumptions in \( \Gamma \). By Theorem 10.22, any valuation \( \mathfrak{v} \) that satisfies \( \Gamma \) must satisfy \( \bot \). Since \( \mathf... | Yes |
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