Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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Corollary 24.15. If \( \left\lbrack x\right\rbrack \neq \left\lbrack y\right\rbrack ,\left\lbrack x\right\rbrack \cap \left\lbrack y\right\rbrack = \varnothing \) . | Proof. Suppose \( z \in \left\lbrack x\right\rbrack \) and \( x \otimes y \) . Then \( z \otimes u \) for all \( u \in \left\lbrack y\right\rbrack \) . If \( z \in \left\lbrack y\right\rbrack \), we would have \( z \otimes z \) . Similarly if \( y \otimes x \) . | No |
Proposition 24.16. If \( x \) and \( y \) are non-standard, then \( x \otimes x \oplus y \) and \( x \oplus y \notin \left\lbrack x\right\rbrack \) . | Proof. If \( y \) is nonstandard, then \( y \neq \mathbf{z} \) . \( \mathbf{{PA}} \vdash \forall x\left( {y \neq 0 \rightarrow x < \left( {x + y}\right) }\right) \) . Now suppose \( x \oplus y \in \left\lbrack x\right\rbrack \) . Since \( x \otimes x \oplus y \), we would have \( x \oplus {n}^{ * } = x \oplus y \) . Bu... | Yes |
Proposition 24.17. There is no least non-standard block. | Proof. \( \mathbf{{PA}} \vdash \forall x\exists y\left( {\left( {y + y}\right) = x \vee {\left( y + y\right) }^{\prime } = x}\right) \), i.e., that every \( x \) is divisible by 2 (possibly with remainder 1). If \( x \) is non-standard, so is \( y \) . By the preceding proposition, \( y \otimes y \oplus y \) and \( y \... | No |
Proposition 24.19. The ordering of the blocks is dense. That is, if \( x \otimes y \) and \( \left\lbrack x\right\rbrack \neq \) \( \left\lbrack y\right\rbrack \), then there is a block \( \left\lbrack z\right\rbrack \) distinct from both that is between them. | Proof. Suppose \( x \otimes y \) . As before, \( x \oplus y \) is divisible by two (possibly with remainder): there is a \( z \in \left| \mathfrak{M}\right| \) such that either \( x \oplus y = z \oplus z \) or \( x \oplus y = {\left( z \oplus z\right) }^{ * } \) . The element \( z \) is the \ | No |
Example 24.21. Recall the structure \( \mathfrak{K} \) from Example 24.8. Its domain was \( \left| \mathfrak{K}\right| = \mathbb{N} \cup \{ a\} \) and interpretations\n\n\[{\mathrm{o}}^{\mathfrak{K}} = 0\]\n\n\[{\prime }^{\mathfrak{K}}\left( x\right) = \left\{ \begin{array}{ll} x + 1 & \text{ if }x \in \mathbb{N} \\ a ... | All of these functions are computable functions of natural numbers and \( { < }^{{\mathfrak{K}}^{\prime }} \) is a decidable relation on \( \mathbb{N} \) -but they are not the same functions as successor, addition, and multiplication on \( \mathbb{N} \), and \( { < }^{{\mathfrak{K}}^{\prime }} \) is not the same relati... | Yes |
Prove that \( \mathfrak{K} \) from Example 24.8 satisifies the remaining axioms of \( \mathbf{Q} \) | \[ \forall x\left( {x \times 0}\right) = 0 \] \( \left( {Q}_{6}\right) \) \[ \forall x\forall y\left( {x \times {y}^{\prime }}\right) = \left( {\left( {x \times y}\right) + x}\right) \] \( \left( {Q}_{7}\right) \) \[ \forall x\forall y\left( {x < y \leftrightarrow \exists z\left( {{z}^{\prime } + x}\right) = y}\right) ... | No |
Problem 24.4. Expand \( \mathfrak{L} \) of Example 24.9 to include \( \otimes \) and \( \otimes \) that interpret \( \times \) and \( < \) . Show that your structure satisifies the remaining axioms of \( \mathbf{Q} \), | \[ \forall x\left( {x \times 0}\right) = 0 \] \( \left( {Q}_{6}\right) \)\[ \forall x\forall y\left( {x \times {y}^{\prime }}\right) = \left( {\left( {x \times y}\right) + x}\right) \] \( \left( {Q}_{7}\right) \)\[ \forall x\forall y\left( {x < y \leftrightarrow \exists z\left( {{z}^{\prime } + x}\right) = y}\right) \]... | No |
Lemma 25.2. Suppose \( {\mathcal{L}}_{0} \) is the language containing every constant symbol, function symbol and predicate symbol (other than \( \doteq \) ) that occurs in both \( \Gamma \) and \( \Delta \), and let \( {\mathcal{L}}_{0}^{\prime } \) be obtained by the addition of infinitely many new constant symbols \... | Proof. We proceed indirectly: suppose by way of contradiction that \( \Gamma \) and \( \Delta \) are separated in \( {\mathcal{L}}_{0}^{\prime } \) . Then \( \Gamma \vDash \chi \left\lbrack {c/x}\right\rbrack \) and \( \Delta \vDash \neg \chi \left\lbrack {c/x}\right\rbrack \) for some \( \chi \in {\mathcal{L}}_{0} \) ... | Yes |
Lemma 25.3. Suppose that \( \Gamma \cup \{ \exists {x\sigma }\} \) and \( \Delta \) are inseparable, and \( c \) is a new constant symbol not in \( \Gamma ,\Delta \), or \( \sigma \) . Then \( \Gamma \cup \{ \exists {x\sigma },\sigma \left\lbrack {c/x}\right\rbrack \} \) and \( \Delta \) are also inseparable. | Proof. Suppose for contradiction that \( \chi \) separates \( \Gamma \cup \{ \exists {x\sigma },\sigma \left\lbrack {c/x}\right\rbrack \} \) and \( \Delta \) , while at the same time \( \Gamma \cup \{ \exists {x\sigma }\} \) and \( \Delta \) are inseparable. We distinguish two cases:\n\n1. \( c \) does not occur in \( ... | Yes |
Lemma 26.8. Suppose \( \alpha \in L\left( \mathcal{L}\right) \), with \( \mathcal{L} \) finite, and assume also that there is an \( n \in \mathbb{N} \) such that for any two structures \( \mathfrak{M} \) and \( \mathfrak{N} \), if \( \mathfrak{M}{ \equiv }_{n}\mathfrak{N} \) and \( \mathfrak{M}{ \vDash }_{L}\alpha \) t... | Proof. Let \( n \) be such that any two \( n \) -equivalent structures \( \mathfrak{M} \) and \( \mathfrak{N} \) agree on the value assigned to \( \alpha \) . Recall Proposition 23.19: there are only finitely many first-order sentences in a finite language that have quantifier rank no greater than \( n \), up to logica... | Yes |
Proposition 27.4. The addition function \( \operatorname{add}\left( {x, y}\right) = x + y \) is primitive recursive. | Proof. We already have a primitive recursive definition of add in terms of two functions \( f \) and \( g \) which matches the format of Definition 27.1:\n\n\[ \operatorname{add}\left( {{x}_{0},0}\right) = f\left( {x}_{0}\right) = {x}_{0} \]\n\n\[ \operatorname{add}\left( {{x}_{0}, y + 1}\right) = g\left( {{x}_{0}, y,\... | Yes |
Proposition 27.5. The multiplication function \( \operatorname{mult}\left( {x, y}\right) = x \cdot y \) is primitive recursive. | Proof. Exercise. | No |
Here's our very first example of a primitive recursive definition:\n\n\[ h\left( 0\right) = 1 \]\n\n\[ h\left( {y + 1}\right) = 2 \cdot h\left( y\right) \text{.} \] | This function cannot fit into the form required by Definition 27.1, since \( k = 0 \) . The definition also involves the constants 1 and 2 . To get around the first problem, let’s introduce a dummy argument and define the function \( {h}^{\prime } \) :\n\n\[ {h}^{\prime }\left( {{x}_{0},0}\right) = f\left( {x}_{0}\righ... | Yes |
Proposition 27.7. The exponentiation function \( \exp \left( {x, y}\right) = {x}^{y} \) is primitive recursive. | Proof. We can define exp primitive recursively as\n\n\[ \exp \left( {x,0}\right) = 1 \]\n\n\[ \exp \left( {x, y + 1}\right) = \operatorname{mult}\left( {x,\exp \left( {x, y}\right) }\right) . \]\n\nStrictly speaking, this is not a recursive definition from primitive recursive functions. Officially, though, we have:\n\n... | Yes |
Proposition 27.8. The predecessor function \( \operatorname{pred}\left( y\right) \) defined by\n\n\[ \operatorname{pred}\left( y\right) = \left\{ \begin{array}{ll} 0 & \text{ if }y = 0 \\ y - 1 & \text{ otherwise } \end{array}\right.\n\]\n\nis primitive recursive. | Proof. Note that\n\n\[ \operatorname{pred}\left( 0\right) = 0\text{and} \]\n\n\[ \operatorname{pred}\left( {y + 1}\right) = y. \]\n\nThis is almost a primitive recursive definition. It does not, strictly speaking, fit into the pattern of definition by primitive recursion, since that pattern requires at least one extra ... | Yes |
Proposition 27.9. The factorial function \( \operatorname{fac}\left( x\right) = x! = 1 \cdot 2 \cdot 3\cdots \cdot x \) is primitive recursive. | Proof. The obvious primitive recursive definition is\n\n\[ \operatorname{fac}\left( 0\right) = 1 \]\n\n\[ \operatorname{fac}\left( {y + 1}\right) = \operatorname{fac}\left( y\right) \cdot \left( {y + 1}\right) . \]\n\nOfficially, we have to first define a two-place function \( h \)\n\n\[ h\left( {x,0}\right) = {\operat... | Yes |
Proposition 27.10. Truncated subtraction, \( x - y \), defined by\n\n\[ x - y = \left\{ \begin{array}{ll} 0 & \text{ if }x > y \\ x - y & \text{ otherwise } \end{array}\right. \]\nis primitive recursive. | Proof. We have:\n\n\[ x - 0 = x \]\n\n\[ x - \left( {y + 1}\right) = \operatorname{pred}\left( {x - y}\right) \] | Yes |
Proposition 27.11. The distance between \( x \) and \( y,\left| {x - y}\right| \), is primitive recursive. | Proof. We have \( \left| {x - y}\right| = \left( {x - y}\right) + \left( {y - x}\right) \), so the distance can be defined by composition from + and - , which are primitive recursive. | Yes |
Proposition 27.12. The maximum of \( x \) and \( y,\max \left( {x, y}\right) \), is primitive recursive. | Proof. We can define \( \max \left( {x, y}\right) \) by composition from + and - by\n\n\[ \max \left( {x, y}\right) = x + \left( {y - x}\right) . \]\n\nIf \( x \) is the maximum, i.e., \( x \geq y \), then \( y - x = 0 \), so \( x + \left( {y - x}\right) = x + 0 = x \) . If \( y \) is the maximum, then \( y - x = y - x... | Yes |
Proposition 27.13. The minimum of \( x \) and \( y,\min \left( {x, y}\right) \), is primitive recursive. | Proof. Exercise. | No |
Proposition 27.14. The set of primitive recursive functions is closed under the following two operations:\n\n1. Finite sums: if \( f\left( {\\overrightarrow{x}, z}\\right) \) is primitive recursive, then so is the function\n\n\[ g\\left( {\\overrightarrow{x}, y}\\right) = \\mathop{\\sum }\\limits_{{z = 0}}^{y}f\\left( ... | Proof. For example, finite sums are defined recursively by the equations\n\n\[ g\\left( {\\overrightarrow{x},0}\\right) = f\\left( {\\overrightarrow{x},0}\\right) \]\n\n\[ g\\left( {\\overrightarrow{x}, y + 1}\\right) = g\\left( {\\overrightarrow{x}, y}\\right) + f\\left( {\\overrightarrow{x}, y + 1}\\right) . \] | No |
Proposition 27.16. The set of primitive recursive relations is closed under boolean operations, that is, if \( P\left( \overrightarrow{x}\right) \) and \( Q\left( \overrightarrow{x}\right) \) are primitive recursive, so are\n\n1. \( \neg P\left( \overrightarrow{x}\right) \)\n\n2. \( P\left( \overrightarrow{x}\right) \l... | Proof. Suppose \( P\left( \overrightarrow{x}\right) \) and \( Q\left( \overrightarrow{x}\right) \) are primitive recursive, i.e., their characteristic functions \( {\chi }_{P} \) and \( {\chi }_{Q} \) are. We have to show that the characteristic functions of \( \neg P\left( \overrightarrow{x}\right) \), etc., are also ... | Yes |
Proposition 27.17. The set of primitive recursive relations is closed under bounded quantification, i.e., if \( R\left( {\overrightarrow{x}, z}\right) \) is a primitive recursive relation, then so are the relations \( \left( {\forall z < y}\right) R\left( {\overrightarrow{x}, z}\right) \) and \( \left( {\exists z < y}\... | Proof. By convention, we take \( \left( {\forall z < 0}\right) R\left( {\overrightarrow{x}, z}\right) \) to be true (for the trivial reason that there are no \( z \) less than 0 ) and \( \left( {\exists z < 0}\right) R\left( {\overrightarrow{x}, z}\right) \) to be false. A universal quantifier functions just like a fin... | Yes |
If \( {g}_{0}\left( \overrightarrow{x}\right) ,\ldots ,{g}_{m}\left( \overrightarrow{x}\right) \) are primitive recursive functions, and \( {R}_{0}\left( \overrightarrow{x}\right) \) , \( \ldots ,{R}_{m - 1}\left( \overrightarrow{x}\right) \) are primitive recursive relations, then the function \( f \) defined by\n\n\[... | Proof. When \( m = 1 \), this is just the function defined by\n\n\[ f\left( \overrightarrow{x}\right) = \operatorname{cond}\left( {{\chi }_{\neg {R}_{0}}\left( \overrightarrow{x}\right) ,{g}_{0}\left( \overrightarrow{x}\right) ,{g}_{1}\left( \overrightarrow{x}\right) }\right) .\n\nFor \( m \) greater than 1, one can ju... | Yes |
Proposition 27.19. If \( R\left( {\overrightarrow{x}, z}\right) \) is primitive recursive, so is the function \( {m}_{R}\left( {\overrightarrow{x}, y}\right) \) which returns the least \( z \) less than \( y \) such that \( R\left( {\overrightarrow{x}, z}\right) \) holds, if there is one, and \( y \) otherwise. We will... | Proof. Note than there can be no \( z < 0 \) such that \( R\left( {\overrightarrow{x}, z}\right) \) since there is no \( z < 0 \) at all. So \( {m}_{R}\left( {\overrightarrow{x},0}\right) = 0 \) .\n\nIn case the bound is of the form \( y + 1 \) we have three cases: (a) There is a \( z < y \) such that \( R\left( {\over... | Yes |
Proposition 27.20. The function \( \operatorname{len}\left( s\right) \), which returns the length of the sequence \( s \) , is primitive recursive. | Proof. Let \( R\left( {i, s}\right) \) be the relation defined by\n\n\[ R\left( {i, s}\right) \text{ iff }{p}_{i} \mid s \land {p}_{i + 1} \nmid s \]\n\n\( R \) is clearly primitive recursive. Whenever \( s \) is the code of a non-empty sequence, i.e.,\n\n\[ s = {p}_{0}^{{a}_{0} + 1}\cdots \cdot \cdot {p}_{k}^{{a}_{k} ... | Yes |
Proposition 27.21. The function append \( \left( {s, a}\right) \), which returns the result of appending a to the sequence \( s \), is primitive recursive. | Proof. append can be defined by:\n\n\[ \operatorname{append}\left( {s, a}\right) = \left\{ \begin{array}{ll} {2}^{a + 1} & \text{ if }s = 0\text{ or }s = 1 \\ s \cdot {p}_{\operatorname{len}\left( s\right) }^{a + 1} & \text{ otherwise. } \end{array}\right. \] | Yes |
Proposition 27.22. The function \( \operatorname{element}\left( {s, i}\right) \), which returns the ith element of \( s \) (where the initial element is called the 0th), or 0 if \( i \) is greater than or equal to the length of \( s \), is primitive recursive. | Proof. Note that \( a \) is the \( i \) th element of \( s \) iff \( {p}_{i}^{a + 1} \) is the largest power of \( {p}_{i} \) that divides \( s \), i.e., \( {p}_{i}^{a + 1} \mid s \) but \( {p}_{i}^{a + 2} \nmid s \) . So:\n\n\[ \text{ element }\left( {s, i}\right) = \left\{ \begin{array}{ll} 0 & \text{ if }i \geq \ope... | Yes |
Proposition 27.23. The function \( \operatorname{concat}\left( {s, t}\right) \), which concatenates two sequences, is primitive recursive. | Proof. We want a function concat with the property that\n\n\[ \operatorname{concat}\left( {\left\langle {{a}_{0},\ldots ,{a}_{k}}\right\rangle ,\left\langle {{b}_{0},\ldots ,{b}_{l}}\right\rangle }\right) = \left\langle {{a}_{0},\ldots ,{a}_{k},{b}_{0},\ldots ,{b}_{l}}\right\rangle . \]\n\nWe’ll use a \ | No |
Proposition 27.24. The function subseq \( \left( {s, i, n}\right) \) which returns the subsequence of \( s \) of length \( n \) beginning at the ith element, is primitive recursive. | Proof. Exercise. | No |
Proposition 27.25. The function SubtreeSeq \( \left( t\right) \), which returns the code of a sequence the elements of which are the codes of all subtrees of the tree with code \( t \), is primitive recursive. | Proof. First note that ISubtrees \( \left( t\right) = \operatorname{subseq}\left( {t,1,{\left( t\right) }_{0}}\right) \) is primitive recursive and returns the codes of the immediate subtrees of a tree \( t \) . Now we can define a helper function hSubtreeSeq \( \left( {t, n}\right) \) which computes the sequence of al... | Yes |
Theorem 27.28 (Kleene's Normal Form Theorem). There is a primitive recursive relation \( T\left( {e, x, s}\right) \) and a primitive recursive function \( U\left( s\right) \), with the following property: if \( f \) is any partial recursive function, then for some \( e \) , \[ f\left( x\right) \simeq U\left( {{\mu sT}\... | The proof of the normal form theorem is involved, but the basic idea is simple. Every partial recursive function has an index \( e \), intuitively, a number coding its program or definition. If \( f\left( x\right) \downarrow \), the computation can be recorded systematically and coded by some number \( s \), and that \... | No |
Theorem 27.29. The halting function \( h \) is not partial recursive. | Proof. If \( h \) were partial recursive, we could define\n\n\[ d\left( y\right) = \left\{ \begin{array}{ll} 1 & \text{ if }h\left( {y, y}\right) = 0 \\ {\mu xx} \neq x & \text{ otherwise. } \end{array}\right. \]\n\nFrom this definition it follows that\n\n1. \( d\left( y\right) \downarrow \) iff \( {\varphi }_{y}\left(... | Yes |
Theorem 28.1 (Kleene's Normal Form Theorem). There are a primitive recursive relation \( T\left( {k, x, s}\right) \) and a primitive recursive function \( U\left( s\right) \), with the following property: if \( f \) is any partial computable function, then for some \( k \) ,\n\n\[ f\left( x\right) \simeq U\left( {{\mu ... | Proof Sketch. For any model of computation one can rigorously define a description of the computable function \( f \) and code such description using a natural number \( k \) . One can also rigorously define a notion of \ | No |
Theorem 28.2. Every partial computable function has infinitely many indices. | Again, this is intuitively clear. Given any (description of) a computable function, one can come up with a different description which computes the same function (input-output pair) but does so, e.g., by first doing something that has no effect on the computation (say, test if \( 0 = 0 \), or count to 5, etc.). The ind... | No |
Theorem 28.3. For each pair of natural numbers \( n \) and \( m \), there is a primitive recursive function \( {s}_{n}^{m} \) such that for every sequence \( x,{a}_{0},\ldots ,{a}_{m - 1},{y}_{0},\ldots ,{y}_{n - 1} \), we have\n\n\[{\varphi }_{{s}_{n}^{m}\left( {x,{a}_{0},\ldots ,{a}_{m - 1}}\right) }^{n}\left( {{y}_{... | It is helpful to think of \( {s}_{n}^{m} \) as acting on programs. That is, \( {s}_{n}^{m} \) takes a program, \( x \), for an \( \left( {m + n}\right) \) -ary function, as well as fixed inputs \( {a}_{0},\ldots ,{a}_{m - 1} \) ; and it returns a program, \( {s}_{n}^{m}\left( {x,{a}_{0},\ldots ,{a}_{m - 1}}\right) \), ... | Yes |
Theorem 28.4. There is a universal partial computable function \( \operatorname{Un}\left( {k, x}\right) \) . In other words, there is a function \( \operatorname{Un}\left( {k, x}\right) \) such that:\n\n1. \( \operatorname{Un}\left( {k, x}\right) \) is partial computable.\n\n2. If \( f\left( x\right) \) is any partial ... | Proof. Let \( \operatorname{Un}\left( {k, x}\right) \simeq U\left( {{\mu sT}\left( {k, x, s}\right) }\right) \) in Kleene’s normal form theorem. | Yes |
Theorem 28.5. There is no universal computable function. In other words, the universal function \( {\operatorname{Un}}^{\prime }\left( {k, x}\right) = {\varphi }_{k}\left( x\right) \) is not computable. | Proof. This theorem says that there is no total computable function that is universal for the total computable functions. The proof is a simple diagonalization: if \( {\operatorname{Un}}^{\prime }\left( {k, x}\right) \) were total and computable, then\n\n\[ d\left( x\right) = {\operatorname{Un}}^{\prime }\left( {x, x}\... | Yes |
Theorem 28.6. Let\n\n\\[ \nh\\left( {k, x}\\right) = \\left\\{ \\begin{array}{ll} 1 & \\text{ if }\\operatorname{Un}\\left( {k, x}\\right) \\text{ is defined } \\\\ 0 & \\text{ otherwise. } \\end{array}\\right.\n\\]\n\nThen \\( h \\) is not computable. | Proof. If \\( h \\) were computable, we would have a universal computable function, as follows. Suppose \\( h \\) is computable, and define\n\n\\[ \n{\\operatorname{Un}}^{\\prime }\\left( {k, x}\\right) = \\left\\{ \\begin{array}{ll} \\operatorname{fnUn}\\left( {k, x}\\right) & \\text{ if }h\\left( {k, x}\\right) = 1 \... | Yes |
Theorem 28.9. Let \( S \) be a set of natural numbers. Then the following are equivalent:\n\n1. \( S \) is computably enumerable.\n\n2. \( S \) is the range of a partial computable function.\n\n3. \( S \) is empty or the range of a primitive recursive function.\n\n4. \( S \) is the domain of a partial computable functi... | Proof. Since every primitive recursive function is computable and every computable function is partial computable, (3) implies (1) and (1) implies (2). (Note that if \( S \) is empty, \( S \) is the range of the partial computable function that is nowhere defined.) If we show that (2) implies (3), we will have shown th... | Yes |
Theorem 28.10. A set \( S \) is computably enumerable if and only if there is a computable relation \( R\left( {x, y}\right) \) such that\n\n\[ S = \{ x : \exists {yR}\left( {x, y}\right) \} . \] | Proof. In the forward direction, suppose \( S \) is computably enumerable. Then for some \( e, S = {W}_{e} \) . For this value of \( e \) we can write \( S \) as\n\n\[ S = \{ x : \exists {yT}\left( {e, x, y}\right) \} . \]\n\nIn the reverse direction, suppose \( S = \{ x : \exists {yR}\left( {x, y}\right) \} \) . Defin... | Yes |
Theorem 28.11. Suppose \( A \) and \( B \) are computably enumerable. Then so are \( A \cap B \) and \( A \cup B \) . | Proof. Theorem 28.9 allows us to use various characterizations of the computably enumerable sets. By way of illustration, we will provide a few different proofs.\n\nFor the first proof, suppose \( A \) is enumerated by a computable function \( f \) , and \( B \) is enumerated by a computable function \( g \) . Let\n\n\... | Yes |
Theorem 28.12. Let \( A \) be any set of natural numbers. Then \( A \) is computable if and only if both \( A \) and \( \bar{A} \) are computably enumerable. | Proof. The forwards direction is easy: if \( A \) is computable, then \( \bar{A} \) is computable as well \( \left( {{\chi }_{A} = 1 - {\chi }_{\bar{A}}}\right) \), and so both are computably enumerable.\n\nIn the other direction, suppose \( A \) and \( \bar{A} \) are both computably enumerable. Let \( A \) be the doma... | Yes |
Corollary 28.13. \( \overline{{K}_{0}} \) is not computably enumerable. | Proof. We know that \( {K}_{0} \) is computably enumerable, but not computable. If \( \overline{{K}_{0}} \) were computably enumerable, then \( {K}_{0} \) would be computable by Theorem 28.12. | Yes |
Proposition 28.15. If \( A{ \leq }_{m}B \) and \( B{ \leq }_{m}C \), then \( A{ \leq }_{m}C \) . | Proof. Composing a reduction of \( A \) to \( B \) with a reduction of \( B \) to \( C \) yields a reduction of \( A \) to \( C \) . (You should check the details!) | No |
Proposition 28.16. Let \( A \) and \( B \) be any sets, and suppose \( A \) is many-one reducible to \( B \). 1. If \( B \) is computably enumerable, so is \( A \). 2. If \( B \) is computable, so is \( A \). | Proof. Let \( f \) be a many-one reduction from \( A \) to \( B \) . For the first claim, just check that if \( B \) is the domain of a partial function \( g \), then \( A \) is the domain of \( g \circ f \) : \[ x \in A\text{iff}f\left( x\right) \in B \] \[ \text{iff}g\left( {f\left( x\right) }\right) \downarrow \text... | Yes |
Theorem 28.18. \( K,{K}_{0} \), and \( {K}_{1} \) are all complete computably enumerable sets. | Proof. To see that \( {K}_{0} \) is complete, let \( B \) be any computably enumerable set. Then for some index \( e \) ,\n\n\[ B = {W}_{e} = \left\{ {x : {\varphi }_{e}\left( x\right) \downarrow }\right\} \]\n\nLet \( f \) be the function \( f\left( x\right) = \langle e, x\rangle \) . Then for every natural number \( ... | Yes |
Proposition 28.19. Let\n\n\[ \n{K}_{1} = \left\{ {e : {\varphi }_{e}\left( 0\right) \downarrow }\right\} \n\]\n\nThen \( {K}_{1} \) is computably enumerable but not computable. | Proof. Since \( {K}_{1} = \{ e : \exists {sT}\left( {e,0, s}\right) \} ,{K}_{1} \) is computably enumerable by Theorem 28.10.\n\nTo show that \( {K}_{1} \) is not computable, let us show that \( {K}_{0} \) is reducible to it.\n\nThis is a little bit tricky, since using \( {K}_{1} \) we can only ask questions about comp... | No |
Proposition 28.20. Tot is not computable. | Proof. To see that Tot is not computable, it suffices to show that \( K \) is reducible to it. Let \( h\left( {x, y}\right) \) be defined by\n\n\[ h\left( {x, y}\right) \simeq \left\{ \begin{array}{ll} 0 & \text{ if }x \in K \\ \text{ undefined } & \text{ otherwise } \end{array}\right. \]\n\nNote that \( h\left( {x, y}... | Yes |
Theorem 28.21 (Rice’s Theorem). Let \( C \) be any set of partial computable functions, and let \( A = \left\{ {n : {\varphi }_{n} \in C}\right\} \) . If \( A \) is computable, then either \( C \) is \( \varnothing \) or \( C \) is the set of all the partial computable functions. | Proof of Rice’s theorem. Suppose \( C \) is neither \( \varnothing \) nor the set of all the partial computable functions, and let \( A \) be the set of indices of functions in \( C \) . We will show that if \( A \) were computable, we could solve the halting problem; so \( A \) is not computable.\n\nWithout loss of ge... | Yes |
is there a computable function \( h \), with the following property? For every \( x \) and \( y \) ,\n\n\[ h\left( {\ulcorner {\varphi }_{x}\left( y\right) \urcorner }\right) = \left\{ \begin{array}{ll} 1 & \text{ if }{\varphi }_{x}\left( y\right) \downarrow \\ 0 & \text{ otherwise. } \end{array}\right. \] | No; otherwise, the partial function\n\n\[ g\left( x\right) \simeq \left\{ \begin{array}{ll} 0 & \text{ if }h\left( {\ulcorner {\varphi }_{x}\left( x\right) \urcorner }\right) = 0 \\ \text{ undefined } & \text{ otherwise } \end{array}\right. \]\n\nwould be computable, and so have some index \( e \) . But then we have\n\... | Yes |
Lemma 28.23. The following statements are equivalent:\n\n1. For every partial computable function \( g\left( {x, y}\right) \), there is an index \( e \) such that for every \( y \), \n\n\[ \n{\varphi }_{e}\left( y\right) \simeq g\left( {e, y}\right) \n\] \n\n2. For every computable function \( f\left( x\right) \), ther... | Proof. \( \left( 1\right) \Rightarrow \left( 2\right) \) : Given \( f \), define \( g \) by \( g\left( {x, y}\right) \simeq \operatorname{Un}\left( {f\left( x\right), y}\right) \) . Use (1) to get an index \( e \) such that for every \( y \), \n\n\[ \n{\varphi }_{e}\left( y\right) = \operatorname{Un}\left( {f\left( e\r... | Yes |
The two statements in Lemma 28.23 are true. Specifically, for every partial computable function \( g\left( {x, y}\right) \), there is an index e such that for every \( y \) ,\n\n\[ \n{\varphi }_{e}\left( y\right) \simeq g\left( {e, y}\right) \n\] | Proof. The ingredients are already implicit in the discussion of the halting problem above. Let \( \operatorname{diag}\left( x\right) \) be a computable function which for each \( x \) returns an index for the function \( {f}_{x}\left( y\right) \simeq {\varphi }_{x}\left( {x, y}\right) \), i.e.\n\n\[ \n{\varphi }_{\ope... | Yes |
Theorem 28.25. There is no partial computable function \( f \) with the following property: whenever \( {W}_{e} \) is computable, then \( f\left( e\right) \) is defined and \( {\varphi }_{f\left( e\right) } \) is its characteristic function. | Proof. Let \( f \) be any computable function; we will construct an \( e \) such that \( {W}_{e} \) is computable, but \( {\varphi }_{f\left( e\right) } \) is not its characteristic function. Using the fixed point theorem, we can find an index \( e \) such that\n\n\[ \n{\varphi }_{e}\left( y\right) \simeq \left\{ \begi... | Yes |
Lemma 28.26. Suppose \( f\left( {x, y}\right) \) is primitive recursive. Let \( g \) be defined by\n\n\[ g\left( x\right) \simeq {\mu yf}\left( {x, y}\right) = 0. \]\n\nThen \( g \) is represented by a lambda term. | Proof. The idea is roughly as follows. Given \( x \), we will use the fixed-point lambda term \( Y \) to define a function \( {h}_{x}\left( n\right) \) which searches for a \( y \) starting at \( n \) ; then \( g\left( x\right) \) is just \( {h}_{x}\left( 0\right) \) . The function \( {h}_{x} \) can be expressed as the... | Yes |
Even Machine: The following Turing machine halts if, and only if, there are an even number of 1's on the tape (under the assumption that all 1's come before the first 0 on the tape). | The state diagram corresponds to the following transition function:\n\n\[ \delta \left( {{q}_{0},1}\right) = \left\langle {{q}_{1},1, R}\right\rangle \]\n\n\[ \delta \left( {{q}_{1},1}\right) = \left\langle {{q}_{0},1, R}\right\rangle \]\n\n\[ \delta \left( {{q}_{1},0}\right) = \left\langle {{q}_{1},0, R}\right\rangle ... | Yes |
The machine table for the even machine is: | <table><thead><tr><th></th><th>0</th><th>1</th></tr></thead><tr><td>\\( {q}_{0} \\)</td><td></td><td>\\( 1,{q}_{1}, R \\)</td></tr><tr><td>\\( {q}_{1} \\)</td><td>\\( 0,{q}_{1},0 \\)</td><td>\\( 1,{q}_{0}, R \\)</td></tr></table>\n\nAs we can see, the machine halts when scanning a blank in state \\( {q}_{0} \\) . | No |
Before building a doubler machine, it is important to come up with a strategy for solving the problem. Since the machine (as we have formulated it) cannot remember how many 1 's it has read, we need to come up with a way to keep track of all the 1 's on the tape. One such way is to separate the output from the input wi... | The state diagram of the resulting Turing machine is depicted in Figure 29.2. | No |
Example 29.12. Addition: Build a machine that, when given an input of two non-empty strings of \( 1 \) ’s of length \( n \) and \( m \), computes the function \( f\left( {n, m}\right) = \) \( n + m \) . | We want to come up with a machine that starts with two blocks of strokes on the tape and halts with one block of strokes. We first need a method to carry out. The input strokes are separated by a blank, so one method would be to write a stroke on the square containing the blank, and erase the first (or last) stroke. Th... | No |
Example 29.13. Halting States. To elucidate this concept, let us begin with an alteration of the even machine. Instead of having the machine halt in state \( {q}_{0} \) if the input is even, we can add an instruction to send the machine into a halt state. | Let us further expand the example. When the machine determines that the input is odd, it never halts. We can alter the machine to include a reject state by replacing the looping instruction with an instruction to go to a reject state \( r \) . | No |
Example 29.14. Combining Machines: Design a machine that computes the function \( f\left( {m, n}\right) = 2\left( {m + n}\right) \) . | In order to build this machine, we can combine two machines we are already familiar with: the addition machine, and the doubler. We begin by drawing a state diagram for the addition machine.\n\n\n\nInstead of halting... | Yes |
Theorem 30.1. There are functions from \( \mathbb{N} \) to \( \mathbb{N} \) which are not Turing computable. | Proof. We know that the set of finite strings of symbols from a denumerable alphabet is enumerable. This gives us that the set of descriptions of Turing machines, as a subset of the finite strings from the enumerable vocabulary \( \left\{ {{q}_{0},{q}_{1},\ldots ,\vartriangleright ,{\sigma }_{1},{\sigma }_{2},\ldots }\... | Yes |
Lemma 30.5. The function \( s \) is not Turing computable. | Proof. We suppose, for contradiction, that the function \( s \) is Turing computable. Then there would be a Turing machine \( S \) that computes \( s \) . We may assume, without loss of generality, that when \( S \) halts, it does so while scanning the first square. This machine can be \ | No |
Theorem 30.6 (Unsolvability of the Halting Problem). The halting problem is unsolvable, i.e., the function \( h \) is not Turing computable. | Proof. Suppose \( h \) were Turing computable, say, by a Turing machine \( H \) . We could use \( H \) to build a Turing machine that computes \( s \) : First, make a copy of the input (separated by a blank). Then move back to the beginning, and run \( H \) . We can clearly make a machine that does the former, and if \... | No |
Proposition 30.8. If \( m < k \), then \( \tau \left( {M, w}\right) \vDash \bar{m} < \bar{k} \) | Proof. Exercise. | No |
Lemma 30.10. If \( M \) run on input \( w \) is in a halting configuration after \( n \) steps, then \( \chi \left( {M, w, n}\right) \vDash \alpha \left( {M, w}\right) \) . | Proof. Suppose that \( M \) halts for input \( w \) after \( n \) steps. There is some state \( q \) , square \( m \), and symbol \( \sigma \) such that:\n\n1. After \( n \) steps, \( M \) is in state \( q \) scanning square \( m \) on which \( \sigma \) appears.\n\n2. The transition function \( \delta \left( {q,\sigma... | Yes |
Lemma 30.12. If \( M \) halts on input \( w \), then \( \tau \left( {M, w}\right) \rightarrow \alpha \left( {M, w}\right) \) is valid. | Proof. By Lemma 30.11, we know that, for any time \( n \), the description \( \chi \left( {M, w, n}\right) \) of the configuration of \( M \) at time \( n \) is entailed by \( \tau \left( {M, w}\right) \) . Suppose \( M \) halts after \( k \) steps. It will be scanning square \( m \), say. Then \( \chi \left( {M, w, k}... | Yes |
Lemma 30.13. If \( \vDash \tau \left( {M, w}\right) \rightarrow \alpha \left( {M, w}\right) \), then \( M \) halts on input \( w \) . | Proof. Consider the \( {\mathcal{L}}_{M} \) -structure \( \mathfrak{M} \) with domain \( \mathbb{N} \) which interprets 0 as 0, \( {}^{\prime } \) as the successor function, and \( < \) as the less-than relation, and the predicates \( {Q}_{q} \) and \( {S}_{\sigma } \) as follows:\n\n\[ \begin{array}{l} {Q}_{q}^{\mathf... | Yes |
Any theory axiomatized by a finite set of sentences is axiomatizable, since any finite set is decidable. Thus, \( \mathbf{Q} \), for instance, is axiomatizable. | Schematically axiomatized theories like PA are also axiomatizable. For to test if \( \psi \) is among the axioms of PA, i.e., to compute the function \( {\chi }_{X} \) where \( {\chi }_{X}\left( \psi \right) = 1 \) if \( \psi \) is an axiom of \( \mathbf{{PA}} \) and \( = 0 \) otherwise, we can do the following: First,... | No |
Theorem 31.14. If \( \Gamma \) is a consistent and axiomatizable theory in \( {\mathcal{L}}_{A} \) which represents all computable functions and decidable relations, then \( \Gamma \) is not complete. | To say that \( \Gamma \) is not complete is to say that for at least one sentence \( \varphi \) , \( \Gamma \nvdash \varphi \) and \( \Gamma \nvdash \neg \varphi \) . Such a sentence is called independent (of \( \Gamma \) ). We can in fact relatively quickly prove that there must be independent sentences. But the power... | Yes |
Theorem 31.15. If \( \Gamma \) is a consistent theory that represents every decidable relation, then \( \Gamma \) is not decidable. | Proof. Suppose \( \Gamma \) were decidable. We show that if \( \Gamma \) represents every decidable relation, it must be inconsistent.\n\nDecidable properties (one-place relations) are represented by formulas with one free variable. Let \( {\varphi }_{0}\left( x\right) ,{\varphi }_{1}\left( x\right) ,\ldots \), be a co... | Yes |
Theorem 31.16. If \( \Gamma \) is axiomatizable and complete it is decidable. | Proof. Any inconsistent theory is decidable, since inconsistent theories contain all sentences, so the answer to the question \ | No |
If \( \Gamma \) is consistent, axiomatizable, and represents every decidable property, it is not complete. | Proof. If \( \Gamma \) were complete, it would be decidable by the previous theorem (since it is axiomatizable and consistent). But since \( \Gamma \) represents every decidable property, it is not decidable, by the first theorem. | Yes |
Recall that if \( {k}_{0},\ldots ,{k}_{n - 1} \) is a sequence of numbers, then the code of the sequence \( \left\langle {{k}_{0},\ldots ,{k}_{n - 1}}\right\rangle \) in the power-of-primes coding is\n\n\[ \n{2}^{{k}_{0} + 1} \cdot {3}^{{k}_{1} + 1}\cdots \cdot {p}_{n - 1}^{{k}_{n - 1}} \n\]\n\nwhere \( {p}_{i} \) is t... | So for instance, the formula \( {v}_{0} = \mathrm{o} \), or, more explicitly, \( = \left( {{v}_{0},{c}_{0}}\right) \), has the Gödel number\n\n\[ \n\left\langle {{\mathrm{c}}_{ = },{\mathrm{c}}_{\left( \prime \right. },{\mathrm{c}}_{{v}_{0}},{\mathrm{c}}_{,},{\mathrm{c}}_{{c}_{0}},{\mathrm{c}}_{)}}\right\rangle .\n\]\n... | Yes |
Proposition 32.5. The relations \( \operatorname{Term}\left( x\right) \) and \( \operatorname{ClTerm}\left( x\right) \) which hold iff \( x \) is the Gödel number of a term or a closed term, respectively, are primitive recursive. | Proof. A sequence of symbols \( s \) is a term iff there is a sequence \( {s}_{0},\ldots ,{s}_{k - 1} = s \) of terms which records how the term \( s \) was formed from constant symbols and variables according to the formation rules for terms. To express that such a putative formation sequence follows the formation rul... | Yes |
Proposition 32.6. The function \( \operatorname{num}\left( n\right) = {}^{\# }{n}^{\# } \) is primitive recursive. | Proof. We define num \( \left( n\right) \) by primitive recursion:\n\n\[ \operatorname{num}\left( 0\right) = {}^{ * }{\mathrm{o}}^{\# } \]\n\n\[ \operatorname{num}\left( {n + 1}\right) = {}^{\# }/{\left( {}^{\# } \frown \operatorname{num}\left( n\right) \frown {}^{\# }\right) }^{\# }.\] | Yes |
Proposition 32.7. The relation \( \operatorname{Atom}\left( x\right) \) which holds iff \( x \) is the Gödel number of an atomic formula, is primitive recursive. | Proof. The number \( x \) is the Gödel number of an atomic formula iff one of the following holds:\n\n1. There are \( n, j < x \), and \( z < x \) such that for each \( i < n,\operatorname{Term}\left( {\left( z\right) }_{i}\right) \) and \( x = \)\n\n\[ \n{}^{\# }{P}_{j}^{n}{\left( {}^{\# } \frown \text{ flatten }\left... | No |
Proposition 32.8. The relation \( \operatorname{Frm}\left( x\right) \) which holds iff \( x \) is the Gödel number of a formula is primitive recursive. | Proof. A sequence of symbols \( s \) is a formula iff there is formation sequence \( {s}_{0} \) , \( \ldots ,{s}_{k - 1} = s \) of formula which records how \( s \) was formed from atomic formulas according to the formation rules. The code for each \( {s}_{i} \) (and indeed of the code of the sequence \( \left\langle {... | No |
Proposition 32.9. The relation \( \operatorname{FreeOcc}\left( {x, z, i}\right) \), which holds iff the \( i \) -th symbol of the formula with Gödel number \( x \) is a free occurrence of the variable with Gödel number \( z \), is primitive recursive. | Proof. Exercise. | No |
Proposition 32.10. The property \( \operatorname{Sent}\left( x\right) \) which holds iff \( x \) is the Gödel number of a sentence is primitive recursive. | Proof. A sentence is a formula without free occurrences of variables. So \( \operatorname{Sent}\left( x\right) \) holds iff\n\n\[ \left( {\forall i < \operatorname{len}\left( x\right) }\right) \left( {\forall z < x}\right) \]\n\n\[ \left( {\left( {\exists j < z}\right) z = {}^{ * }{v}_{j}{}^{\# } \rightarrow \neg \oper... | No |
Proposition 32.11. There is a primitive recursive function \( \operatorname{Subst}\left( {x, y, z}\right) \) with the property that\n\n\[ \operatorname{Subst}\left( {{}^{\# }{\varphi }^{\# },{}^{\# }{t}^{\# },{}^{\# }{u}^{\# }}\right) = {}^{\# }\varphi {\left\lbrack t/u\right\rbrack }^{\# } \] | Proof. We can then define a function hSubst by primitive recursion as follows:\n\n\[ \operatorname{hSubst}\left( {x, y, z,0}\right) = \Lambda \]\n\n\( \operatorname{hSubst}\left( {x, y, z, i + 1}\right) = \)\n\n\[ \left\{ \begin{array}{ll} \operatorname{hSubst}\left( {x, y, z, i}\right) \frown y & \text{ if }\operatorn... | Yes |
Proposition 32.12. The relation \( \operatorname{FreeFor}\left( {x, y, z}\right) \), which holds iff the term with Gödel number \( y \) is free for the variable with Gödel number \( z \) in the formula with Gödel number \( x \), is primitive recursive. | Proof. Exercise. | No |
Consider the very simple derivation\n\n\[ \n\\begin{aligned} \\frac{\\varphi \\Rightarrow \\varphi }{\\varphi \\land \\psi } & \\Rightarrow \\varphi \\land \\mathrm{L} \\\\ & \\Rightarrow \\left( {\\varphi \\land \\psi }\\right) \\rightarrow \\varphi \\end{aligned} \\rightarrow \\mathrm{R} \n\] | The Gödel number of the initial sequent would be \( {p}_{0} = \\left\\langle {0,{}^{\\# }\\varphi \\Rightarrow {\\varphi }^{\\# }}\\right\\rangle \) . The Gödel number of the derivation ending in the conclusion of \( \\land \\mathrm{L} \) would be \( {p}_{1} = \) \( \\left\\langle {1,{p}_{0},{}^{ * }\\varphi \\land \\p... | Yes |
The property \( \operatorname{Correct}\left( p\right) \) which holds iff the last inference in the derivation \( \pi \) with Gödel number \( p \) is correct, is primitive recursive. | Proof. \( \Gamma \Rightarrow \Delta \) is an initial sequent if either there is a sentence \( \varphi \) such that \( \Gamma \Rightarrow \Delta \) is \( \varphi \Rightarrow \varphi \), or there is a term \( t \) such that \( \Gamma \Rightarrow \Delta \) is \( \varnothing \Rightarrow t = t \) . In terms of Gödel numbers... | Yes |
Proposition 32.16. The relation \( \operatorname{Deriv}\left( p\right) \) which holds if \( p \) is the Gödel number of a correct derivation \( \pi \), is primitive recursive. | Proof. A derivation \( \pi \) is correct if every one of its inferences is a correct application of a rule, i.e., if every one of its sub-derivations ends in a correct inference. So, \( \operatorname{Deriv}\left( d\right) \) iff\n\n\[ \left( {\forall i < \operatorname{len}\left( {\operatorname{SubtreeSeq}\left( p\right... | No |
Consider the very simple derivation\n\n\[ \frac{\frac{{\left\lbrack \varphi \land \psi \right\rbrack }^{1}}{\varphi } \land \text{ Elim }}{1\frac{\left( {\varphi \land \psi }\right) \rightarrow \varphi }{\left( {\varphi \land \psi }\right) \rightarrow \varphi } \rightarrow \text{ Intro }} \] | The Gödel number of the assumption would be \( {d}_{0} = \left\langle {0,{}^{\# }\varphi \land {\psi }^{\# },1}\right\rangle \) . The Gödel number of the derivation ending in the conclusion of \( \land \) Elim would be \( {d}_{1} = \left\langle {1,{d}_{0},{}^{ * }{\varphi }^{\# },0,2}\right\rangle \) (1 since \( \land ... | Yes |
Proposition 32.20. The following relations are primitive recursive:\n\n1. \( \varphi \) occurs as an assumption in \( \delta \) with label \( n \) .\n\n2. All assumptions in \( \delta \) with label \( n \) are of the form \( \varphi \) (i.e., we can discharge the assumption \( \varphi \) using label \( n \) in \( \delt... | Proof. We have to show that the corresponding relations between Gödel numbers of formulas and Gödel numbers of derivations are primitive recursive.\n\n1. We want to show that \( \operatorname{Assum}\left( {x, d, n}\right) \), which holds if \( x \) is the Gödel number of an assumption of the derivation with Gödel numbe... | Yes |
Proposition 32.21. The property \( \operatorname{Correct}\left( d\right) \) which holds iff the last inference in the derivation \( \delta \) with Gödel number \( d \) is correct, is primitive recursive. | Proof. Here we have to show that for each rule of inference \( R \) the relation \( {\text{FollowsBy}}_{R}\left( d\right) \) is primitive recursive, where FollowsBy \( {}_{R}\left( d\right) \) holds iff \( d \) is the Gödel number of derivation \( \delta \), and the end-formula of \( \delta \) follows by a correct appl... | Yes |
Proposition 32.22. The relation \( \operatorname{Deriv}\left( d\right) \) which holds if \( d \) is the Gödel number of a correct derivation \( \delta \), is primitive recursive. | Proof. A derivation \( \delta \) is correct if every one of its inferences is a correct application of a rule, i.e., if every one of its sub-derivations ends in a correct inference. So, \( \operatorname{Deriv}\left( d\right) \) iff\n\n\[ \left( {\forall i < \operatorname{len}\left( {\operatorname{SubtreeSeq}\left( d\ri... | Yes |
Proposition 32.23. The relation OpenAssum \( \left( {z, d}\right) \) that holds if \( z \) is the Gödel number of an undischarged assumption \( \varphi \) of the derivation \( \delta \) with Gödel number \( d \), is primitive recursive. | Proof. An occurrence of an assumption is discharged if it occurs with label \( n \) in a sub-derivation of \( \delta \) that ends in a rule with discharge label \( n \) . So \( \varphi \) is an undischarged assumption of \( \delta \) if at least one of its occurrences is not discharged in \( \delta \) . We must be care... | Yes |
Consider the very simple derivation\n\n1. \( \psi \rightarrow \left( {\psi \vee \varphi }\right) \)\n\n2. \( \left( {\psi \rightarrow \left( {\psi \vee \varphi }\right) }\right) \rightarrow \left( {\varphi \rightarrow \left( {\psi \rightarrow \left( {\psi \vee \varphi }\right) }\right) }\right) \)\n\n3. \( \varphi \rig... | The Gödel number of this derivation would simply be\n\n\[ \n{\langle }^{\# }\psi \rightarrow {\left( \psi \vee \varphi \right) }^{\# },{}^{\# }\left( {\psi \rightarrow \left( {\psi \vee \varphi }\right) }\right) \rightarrow {\left( \varphi \rightarrow \left( \psi \rightarrow \left( \psi \vee \varphi \right) \right) \ri... | Yes |
Theorem 33.2. A function is representable in \( \mathbf{Q} \) if and only if it is computable. | There are two directions to proving the theorem. The left-to-right direction is fairly straightforward once arithmetization of syntax is in place. The other direction requires more work. Here is the basic idea: we pick \ | No |
Lemma 33.3. Every function that is representable in \( \mathbf{Q} \) is computable. | Proof. Let’s first give the intuitive idea for why this is true. If \( f\left( {{x}_{0},\ldots ,{x}_{k}}\right) \) is representable in \( \mathbf{Q} \), there is a formula \( \varphi \left( {{x}_{0},\ldots ,{x}_{k}, y}\right) \) such that\n\n\[ \mathbf{Q} \vdash {\varphi }_{f}\left( {\overline{{n}_{0}},\ldots ,\overlin... | Yes |
Theorem 33.7. Suppose \( {x}_{0},\ldots ,{x}_{n} \) are (pairwise) relatively prime. Let \( {y}_{0},\ldots ,{y}_{n} \) be any numbers. Then there is a number \( z \) such that\n\n\[ z \equiv {y}_{0}{\;\operatorname{mod}\;{x}_{0}} \]\n\n\[ z \equiv {y}_{1}{\;\operatorname{mod}\;{x}_{1}} \]\n\n\[ \vdots \]\n\n\[ z \equiv... | Here is how we will use the Chinese Remainder theorem: if \( {x}_{0},\ldots ,{x}_{n} \) are bigger than \( {y}_{0},\ldots ,{y}_{n} \) respectively, then we can take \( z \) to code the sequence \( \left\langle {{y}_{0},\ldots ,{y}_{n}}\right\rangle \) . To recover \( {y}_{i} \), we need only divide \( z \) by \( {x}_{i... | Yes |
Lemma 33.8. If \( h \) can be defined from \( f \) and \( g \) using primitive recursion, it can be defined from \( f, g \), the functions zero, succ, \( {P}_{i}^{n} \), add, mult, \( {\chi }_{ = } \), using composition and regular minimization. | Proof. First, define an auxiliary function \( \widehat{h}\left( {\overrightarrow{x}, y}\right) \) which returns the least number \( d \) such that \( d \) codes a sequence which satisfies\n\n1. \( {\left( d\right) }_{0} = f\left( \overrightarrow{x}\right) \), and\n\n2. for each \( i < y,{\left( d\right) }_{i + 1} = g\l... | Yes |
Lemma 33.13. Given natural numbers \( n \) and \( m \), if \( n \neq m \), then \( \mathbf{Q} \vdash \bar{n} \neq \bar{m} \) . | Proof. Use induction on \( n \) to show that for every \( m \), if \( n \neq m \), then \( Q \vdash \bar{n} \neq \bar{m} \) .\n\nIn the base case, \( n = 0 \) . If \( m \) is not equal to 0, then \( m = k + 1 \) for some natural number \( k \) . We have an axiom that says \( \forall {x0} \neq {x}^{\prime } \) . By a qu... | Yes |
Proposition 33.14. The addition function \( \operatorname{add}\left( {{x}_{0},{x}_{1}}\right) = {x}_{0} + {x}_{1} \) is represented in \( \mathbf{Q} \) by | \[ y = \left( {{x}_{0} + {x}_{1}}\right) \] | No |
Lemma 33.15. \( \mathbf{Q} \vdash \left( {\bar{n} + \bar{m}}\right) = \overline{n + m} \) | Proof. We prove this by induction on \( m \) . If \( m = 0 \), the claim is that \( \mathbf{Q} \vdash (\bar{n} + \) o) \( = \bar{n} \) . This follows by axiom \( {Q}_{4} \) . Now suppose the claim for \( m \) ; let’s prove the claim for \( m + 1 \), i.e., prove that \( \mathbf{Q} \vdash \left( {\bar{n} + \overline{m + ... | Yes |
Proposition 33.16. The multiplication function \( \operatorname{mult}\left( {{x}_{0},{x}_{1}}\right) = {x}_{0} \cdot {x}_{1} \) is represented in \( \mathbf{Q} \) by | \[ y = \left( {{x}_{0} \times {x}_{1}}\right) \] | No |
Lemma 33.17. \( \mathbf{Q} \vdash \left( {\bar{n} \times \bar{m}}\right) = \overline{n \cdot m} \) | Proof. Exercise. | No |
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