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Lemma 4.8. \( {R}_{0} \) is elementary. | Proof \( \left( {m, n}\right) \in {R}_{0} \) iff \( m \) is the Gödel number of a word, \( n \) is the Gödel number of a word, and \( \exists x \leq n\exists y \leq n\left\lbrack {\operatorname{Cat}\left( {\operatorname{Cat}\left( {x, m}\right), y}\right) = n}\right\rbrack \) . (Recall from 3.30 the definition of Cat.) | Yes |
Lemma 4.10. \( {R}_{1} \) is elementary. | Proof. \( \left( {m, n, p, q}\right) \in {R}_{1} \) if \( m, n, p, q \) are Gödel numbers of words and Cat \( \left( {\text{Cat}\left( {m, n}\right), p}\right) = q \) and \( \forall x \leq q\forall y \leq q\lbrack {1x} < {1m}\& x \) and \( y \) are Gödel numbers of words \( \Rightarrow \operatorname{Cat}\left( {\operat... | No |
Lemma 4.12. \( {R}_{2} \) is elementary. | Proof. \( \left( {p, m, n}\right) \in {R}_{2} \) iff \( p \) is the Gödel number of a Markov algorithm, \( m \) and \( n \) are Gödel numbers of words, \( \exists i \leq \lg \) such that \( \left( {{\left( {\left( p\right) }_{i}\right) }_{0}, m}\right) \in {R}_{0} \), and \( \forall i \leq \lg \forall x \leq m\;\forall... | Yes |
Lemma 4.16. \( {R}_{4} \) is elementary. | Proof. \( \left( {m, n}\right) \in {R}_{4} \) iff \( m \) is a Gödel number of a Markov algorithm, \( \ln \geq 1 \) , and \( \forall i < \ln - 1\left\lbrack {\left( {m,{\left( n\right) }_{i},{\left( n\right) }_{i + 1}}\right) \in {R}_{2}}\right\rbrack \) and \( \left( {m,{\left( n\right) }_{\ln + 1},{\left( n\right) }_... | Yes |
Lemma 4.26. Every algorithmic function is recursive. | Proof. Say \( f \) is \( m \) -ary and is computed by a Markov algorithm \( A \) . Let \( e \) be the Gödel number of \( A \) . Then for any \( {x}_{0},\ldots ,{x}_{m - 1} \in \omega \) we have\n\n\[ f\left( {{x}_{0},\ldots ,{x}_{m - 1}}\right) = {V}^{\prime }{\mu z}\left( {\left\langle {e,{x}_{0},\ldots ,{x}_{m - 1}, ... | Yes |
Theorem 5.5. Partial Turing computable \( = \) partial recursive. | Proof. Partial recursive \( \Rightarrow \) Partial Turing computable. Here it is only necessary to read again the proofs of Lemmas 3.10-3.16 and check that they adapt to the situation of partial functions and the new Definitions 5.1 and 5.3.\n\nPartial Turing computable \( \Rightarrow \) Partial recursive. Again one ne... | No |
Theorem 5.7. There is a partial recursive function \( f \) such that \( f \) cannot be extended to a recursive function. | Proof. The rule for computing \( f \) is as follows. For a given \( x \in \omega \), determine whether or not \( x \) is the Gödel number of a Turing machine. If it is not, set \( {fx} = 0 \) . If it is, test in succession whether or not \( \left( {x, x,0}\right) \in {T}_{1},\left( {x, x,1}\right) \in {T}_{1} \) , \( \... | Yes |
Theorem 5.9 (Normal form theorem). For any partial recursive function \( f \) (say \( m \) -ary) there is an \( e \in \omega \) such that \( f = {\mathbf{\varphi }}_{e}^{m} \) . | Proof. By the proof of 5.5, second part. | No |
Corollary 5.10. For each \( m \in \omega \sim 1 \) there exist a 1-place elementary function \( f \) and an \( \left( {m + 2}\right) \) -place elementary function \( g \) such that for any \( m \) -ary partial recursive function \( h \) there is an \( e \in \omega \) such that for all \( {x}_{0},\ldots ,{x}_{m - 1} \in... | Proof. Let \( f = \mathrm{V} \) and \( g = \overline{\mathrm{{sg}}} \circ {\chi }_{\mathrm{T}m} \) . | No |
Corollary 5.11 (Universal Turing machines). There is a Turing machine \( M \) with the following property. If \( f \) is any unary partial Turing computable function and a Turing machine \( N \) computes it, and if \( e \) is the Gödel number of \( N \), then if \( {\left\lbrack \begin{array}{llllll} 0 & {1}^{\left( e ... | Proof. Let \( g \) be the partial recursive function defined by\n\n\[ g\left( {e, x}\right) \simeq \mathrm{V}{\mu u}\left\lbrack {\left( {e, x, u}\right) \in {\mathrm{T}}_{1}}\right\rbrack \]\n\nfor all \( e, x \in \omega \) . Let \( M \) compute \( g \) . Clearly \( M \) is as desired.\n\nIn more intuitive terms we ca... | Yes |
Corollary 5.12 (Universal partial recursive function). There is a partial recursive function \( g \) of two variables such that for any partial recursive function f of one variable there is an \( e \in \omega \) such that for all \( x \in \omega, g\left( {e, x}\right) \simeq {fx} \) . | Proof. Let \( g \) be as in the proof of 5.11.\n\nIn view of the proof of 3.4, the reader might view 5.12 with some suspicion. Let us see what happens if we try the diagonal method on the \( g \) of 5.12 . For any \( x \in \omega \), let \( {fx} \simeq g\left( {x, x}\right) + 1 \) . Then \( f \) is partial recursive, s... | Yes |
Theorem 5.13 (Iteration theorem). For any \( m, n \in \omega \sim 1 \) there is an \( \left( {m + 1}\right) \) - ary recursive function \( {\mathrm{s}}_{n}^{m} \) such that for all \( e,{y}_{1},\ldots ,{y}_{m},{x}_{1},\ldots ,{x}_{n} \in \omega \) , \[ {\varphi }_{e}^{m + n}\left( {{x}_{1},\ldots ,{x}_{n},{y}_{1},\ldot... | Proof. If \( M \) is any Turing machine and \( {y}_{1},\ldots ,{y}_{m} \in \omega \), let \( {M}_{{y1},\ldots ,{ym}}^{ * } \) be the following Turing machine: \[ \text{ Start } \rightarrow {\left( {T}_{\text{left }} \rightarrow {T}_{1}\right) }^{y1} \rightarrow {\left( {T}_{\text{left }} \rightarrow {T}_{1}\right) }^{y... | Yes |
Corollary 5.14. There is no binary function \( f \) such that for all \( x, y \in \omega \) ,\n\n\[ f\left( {x, y}\right) = 1\;\text{ if }y \in \operatorname{Dmn}{\varphi }_{x}^{1}, \]\n\n\[ f\left( {x, y}\right) = 0\;\text{if}\;y \notin \mathrm{D}\mathrm{m}\mathrm{n}\;{\varphi }_{x}^{1}. \] | Proof. Suppose there is such an \( f \) ; say \( f = {\mathbf{\varphi }}_{e}^{2} \) . Now for any \( x, y \in \omega \) let\n\n\[ g\left( {x, y}\right) \simeq {\mu z}\left\lbrack {\mathrm{V}{\mu u}\left( {\left( {y, x, x, u}\right) \in {\mathrm{T}}_{2}}\right) = 0}\right\rbrack . \]\n\nHence for any \( x, y \in \omega ... | Yes |
Theorem 5.15 (Recursion theorem). If \( m > 1 \) and \( f \) is an \( m \) -ary partial recursive function, then there is an \( e \in \omega \) such that for all \( {x}_{0},\ldots ,{x}_{m - 2} \in \omega \) , \[ f\left( {{x}_{0},\ldots ,{x}_{m - 2}, e}\right) \simeq {\mathbf{\varphi }}_{e}^{m - 1}\left( {{x}_{0},\ldots... | Proof. For any \( {x}_{0},\ldots ,{x}_{m - 1} \in \omega \) let \[ g\left( {{x}_{0},\ldots ,{x}_{m - 1}}\right) \simeq f\left( {{x}_{0},\ldots ,{x}_{m - 2},{\mathrm{\;s}}_{n - 1}^{1}\left( {{x}_{m - 1},{x}_{m - 1}}\right) }\right) . \] Thus \( g \) is partial recursive; say \( g = {\varphi }_{r}^{m} \) . Let \( e = {\m... | Yes |
Theorem 5.16 (Fixed point theorem). If \( f \) is a unary recursive function then there is an \( e \in \omega \) such that \( {\mathbf{\varphi }}_{e}^{1} = {\mathbf{\varphi }}_{fe}^{1} \) . | Proof. For any \( x, y \in \omega \) let\n\n\[ g\left( {x, y}\right) \simeq \mathrm{V}{\mu u}\left( {\left( {{fy}, x, u}\right) \in {\mathrm{T}}_{1}}\right) . \]\n\nThus \( g \) is partial recursive, and \( g\left( {x, y}\right) \simeq {\mathbf{\varphi }}_{fy}^{1}x \) for all \( x, y \in \omega \) . Now we apply the re... | Yes |
Theorem 5.17 (Rice). Let \( F \) be a set of one-place partial recursive functions such that \( 0 \neq F \) and \( F \) does not consist of all one-place partial recursive functions. Then \( A = \left\{ {e : {\mathbf{\varphi }}_{e}^{1} \in F}\right\} \) is not recursive. | Proof. Suppose it is. Let \( a \in A \) and \( b \notin A \) . Now define\n\n\[ \n{gx} = a\;x \notin A, \n\]\n\n\[ \n{gx} = b\;x \in A. \n\]\n\nThen \( g \) is recursive. By 5.16 choose \( e \) such that \( {\mathbf{\varphi }}_{e}^{1} = {\mathbf{\varphi }}_{ge}^{1} \) . Then if \( e \in A \) we see that \( {\mathbf{\va... | Yes |
Corollary 5.20. \( \left\{ {\left( {x, y}\right) : y}\right. \) is in the range of \( \left. {\mathbf{\varphi }}_{x}^{1}\right\} \) is not recursive. | Proof. If the given set is recursive, then clearly so is\n\n\[ \left\{ {x : 0}\right. \text{is in the range of}\left. {\mathbf{\varphi }}_{x}^{1}\right\} \text{,}\]\n\ncontradicting 5.17. | Yes |
Corollary 5.21. \( \left\{ {\left( {x, y}\right) : {\mathbf{\varphi }}_{x}^{1} = {\mathbf{\varphi }}_{y}^{1}}\right\} \) is not recursive. | Proof. If the given set is recursive, and \( e \in \omega \), then\n\n\[ \left\{ {x : {\varphi }_{x}^{1} = {\varphi }_{e}^{1}}\right\} \]\n\n is recursive, contradicting 5.17. | Yes |
Theorem 5.22. There is no binary recursive function \( f \) such that for all \( e \) , \( x \in \omega ,\exists u\left( {\left( {e, x, u}\right) \in {\mathrm{T}}_{1}}\right) \) iff \( \exists u \leq f\left( {e, x}\right) \left( {\left( {e, x, u}\right) \in {\mathrm{T}}_{1}}\right) \) . | Proof. Suppose there is such an \( f \) . Let\n\n\[ g\left( {e, x}\right) = 1\;\text{ if }\exists u \leq f\left( {e, x}\right) \left( {\left( {e, x, u}\right) \in {\mathrm{T}}_{1}}\right) ,\]\n\n\[ g\left( {e, x}\right) = 0\;\text{ otherwise. }\]\n\nThus \( g \) is recursive and\n\n\[ g\left( {e, x}\right) = 1\;\text{ ... | Yes |
Theorem 5.23. Let \( m > 1 \) . If \( R \) is an \( m \) -ary recursive relation, then there exist \( e,{e}^{\prime } \in \omega \) such that for all \( {x}_{0},\ldots ,{x}_{m - 2} \in \omega \) , (i) \( \exists y\left( {\left( {{x}_{0},\ldots ,{x}_{m - 2}, y}\right) \in R}\right) \; \) iff \( \exists y\left( {\left( {... | Proof. For any \( {x}_{0},\ldots ,{x}_{m - 2} \in \omega \) let \[ f\left( {{x}_{0},\ldots ,{x}_{m - 2}}\right) \simeq {\mu y}\left( {\left( {{x}_{0},\ldots ,{x}_{m - 2}, y}\right) \in R}\right) . \] Thus \( f \) is partial recursive, so by 5.9 there is an \( e \in \omega \) such that for all \( {x}_{0},\ldots \) , \( ... | Yes |
Proposition 5.29. If \( R \) and \( S \) are n-ary \( {\sum }_{m} \) -relations, then so are \( R \cup S \) and \( R \cap S \) . Similarly for \( {\Pi }_{m} \) and \( {\Delta }_{m} \) . | Proof. The assertions for \( {\Delta }_{m} \) follow from those for \( {\sum }_{m} \) and \( {\Pi }_{m} \) . The assertions for \( {\sum }_{m} \) and \( {\Pi }_{m} \) are proved simultaneously by induction on \( m \) . The case \( m = 0 \) is trivial. Now assume the assertions for \( m \) . We take just one typical ass... | Yes |
Proposition 5.31. If \( R \) is an \( n \) -ary \( {\sum }_{m} \) -relation, then so are the two relations\n\n\[ S = \\left\\{ {\\left( {{x}_{0},\\ldots ,{x}_{n - 1}}\\right) : \\exists y < {x}_{n - 1}\\left\\[ {\\left( {{x}_{0},\\ldots ,{x}_{n - 2}, y}\\right) \\in R}\\right\\] ,}\\right.\n\]\n\n\[ T = \\left\\{ {\\le... | Proof. Again we prove all cases simultaneously by induction on \( m \) . The case \( m = 0 \) is trivial. Assume that all of the statements are true for \( m \) . We take one typical case for \( m + 1 \) :\n\nLet \( R \) be an \( n \) -ary \( {\\sum }_{m + 1} \) -relation, and let \( T \) be as above. By 5.25, let \( {... | Yes |
Proposition 5.33. \( {\sum }_{m} \cup {\Pi }_{m} \subseteq {\Delta }_{m + 1} \). | Proof. Let \( R \in {\sum }_{m} \), say \( R \) is \( n \) -ary. Let \( S = \left\{ {\left( {{x}_{0},\ldots ,{x}_{n}}\right) : \left( {{x}_{0},\ldots ,{x}_{n - 1}}\right) \in R}\right\} \). Then \( S \in {\sum }_{m} \) by 5.26 and 5.27. Clearly \( R = \left\{ {\left( {{x}_{0},\ldots ,{x}_{n - 1}}\right) : \forall y \in... | Yes |
Theorem 5.34. \( {\Delta }_{1} = {\Delta }_{0} \) . | Proof. We know that \( {\Delta }_{0} \subseteq {\Delta }_{1} \) . Suppose \( R \in {\Delta }_{1} \), say \( R \) is \( n \) -ary. Then there are recursive \( S, T\left( {\left( {n + 1}\right) \text{-ary}}\right) \) such that for all \( {x}_{0},\ldots ,{x}_{n - 1} \in \omega \) ,\n\n\[ \left( {{x}_{0},\ldots ,{x}_{n - 1... | Yes |
Theorem 5.35. For \( m, n > 0 \) there is an \( \left( {n + 1}\right) \) -ary \( {\sum }_{m} \) -relation \( {R}_{m}^{n} \) with the following properties:\n\n(i) for every n-ary \( {\sum }_{m} \) -relation \( S \) there is an \( e \in \omega \) such that \( S = \) \( \left\{ {\left( {{x}_{0},\ldots ,{x}_{n - 1}}\right)... | Proof. We construct \( {R}_{m}^{n} \) by recursion on \( m \) . Let\n\n\[ \n{R}_{1}^{n} = \left\{ {\left( {{x}_{0},\ldots ,{x}_{n}}\right) : \exists y \in \omega \left\lbrack {\left( {{x}_{0},\ldots ,{x}_{n}, y}\right) \in {\mathrm{T}}_{n}}\right\rbrack }\right\} .\n\]\n\nIf \( {R}_{m}^{n} \) has been defined for all \... | Yes |
Theorem 5.36 (Hierarchy theorem). For any \( m, n > 0 \) there exists an \( n \) -ary relation \( T \in {\sum }_{m} \sim {\Pi }_{m} \) . Hence \( {}^{n}\omega \sim T \in {\Pi }_{m} \sim {\sum }_{m} \) . Furthermore, there is an \( n \) -ary relation \( W \in {\Delta }_{m + 1} \sim \left( {{\sum }_{m} \cup {\Pi }_{m}}\r... | Proof. Let \( {R}_{m}^{n} \) be as in 5.35. Let\n\n\[ T = \left\{ {\left( {{x}_{0},\ldots ,{x}_{n - 1}}\right) : \left( {{x}_{0},{x}_{0},{x}_{1},{x}_{2},\ldots ,{x}_{n - 1}}\right) \in {R}_{m}^{n}}\right\} . \]\n\nThus \( T \in {\sum }_{m} \) . If \( T \in {\Pi }_{m} \), by 5.35 choose \( e \in \omega \) so that \( T =... | Yes |
Theorem 6.2. For \( A \subseteq \omega \) the following are equivalent;\n\n(i) \( A = 0 \) or \( A \) is the range of an elementary function;\n\n(ii) \( A = 0 \) or \( A \) is the range of a primitive recursive function;\n\n(iii) \( A \) is recursively enumerable;\n\n(iv) \( A \) is the range of a partial recursive fun... | Proof. Obviously \( \left( i\right) \Rightarrow \left( {ii}\right) \Rightarrow \left( {iii}\right) \) . To show that \( \left( {iii}\right) \Rightarrow \left( {iv}\right) \) we just need to show that 0 (the empty set) is the range of some partial recursive function; and obviously the only possibility for such a functio... | Yes |
Theorem 6.5. \( K \) is recursively enumerable but not recursive. | Proof. Obviously \( \mathrm{K} \in {\sum }_{1} \) so \( \mathrm{K} \) is recursively enumerable. Suppose \( \mathrm{K} \) is recursive. Then so is \( \omega \sim \mathrm{K} \), so by \( {6.2}\left( v\right) \) there is an \( e \in \omega \) such that \( \omega \sim \mathrm{K} = \) Dmn \( {\varphi }_{e}^{1} \) . Then\n\... | Yes |
Theorem 6.6. Let \( A \subseteq \omega \) . The following conditions are equivalent:\n\n(i) \( A \) is recursive;\n\n(ii) \( A \) and \( \omega \sim A \) are recursively enumerable. | This theorem can be seen in the following fashion, working directly from Definition 6.1: Of course \( \left( {ii}\right) \Rightarrow \left( i\right) \) is the main part of 6.6. Assume \( \left( {ii}\right) \) . We may suppose \( 0 \neq A \neq \omega \) . Then let \( f \) and \( g \) be recursive functions with \( \oper... | No |
Theorem 6.7. Let \( A \subseteq \omega \) . The following are equivalent:\n\n(i) \( A \) is infinite and recursive:\n\n(ii) there is a recursive function \( f \) with \( \operatorname{Rng}f = A \) and \( \forall x \in \omega \left( {{fx} < f\left( {x + 1}\right) }\right) \) . | Proof. \( \left( i\right) \Rightarrow \left( {ii}\right) \) . Let \( a \) be the least member of \( A \) . Define\n\n\[ \n{f0} = a \n\]\n\n\[ \nf\left( {x + 1}\right) = {\mu y}\left( {y \in A\text{ and }y > {fx}}\right) . \n\]\n\nClearly \( f \) is as desired.\n\n(ii) \( \Rightarrow \) (i). Assume \( f \) as in (ii). T... | Yes |
Theorem 6.8. Any infinite recursively enumerable set has an infinite recursive subset. | Proof. Let \( A \) be infinite r.e., say \( A = \operatorname{Rng}f, f \) recursive. We define \( g \) by induction:\n\n\[ \n{g0} = {f0} \n\]\n\n\[ \ng\left( {x + 1}\right) = {f\mu y}\left( {{fy} > {gx}}\right) . \n\]\n\nThus \( {gx} < g\left( {x + 1}\right) \) for all \( x \in \omega \), and hence, by 6.7, Rng \( g \)... | Yes |
Theorem 6.9. If \( A \) is r.e. and \( f \) is partial recursive, then \( {f}^{ * }A \) is r.e. | Proof. We may assume that \( A \neq 0 \) . Say \( A = \operatorname{Rng}g, g \) recursive. Clearly \( {f}^{ * }A = \operatorname{Rng}\left( {f \circ g}\right) \) and \( f \circ g \) is partial recursive. | Yes |
Theorem 6.10. If \( A \) is r.e. and \( f \) is partial recursive, then \( {f}^{-1 * }A \) is r.e. | Proof. Say \( A = \operatorname{Dmn}{\mathbf{\varphi }}_{e}^{1} \) . Then \( {f}^{-1 * }A = \operatorname{Dmn}\left( {{\mathbf{\varphi }}_{e}^{1} \circ f}\right) \) as desired. | No |
Theorem 6.11. If \( A \) is r.e., then \( \mathop{\bigcup }\limits_{{x \in A}}\operatorname{Rng}{\varphi }_{e}^{1} \) is r.e. | Proof. For any \( y \in \omega \) ,\n\n\[ y \in \mathop{\bigcup }\limits_{{\mid x \in A}}\operatorname{Rng}{\varphi }_{x}^{1}\;\text{ iff }\exists x \in A\left( {y \in \operatorname{Rng}{\varphi }_{x}^{1}}\right) .\n\]\nSince both \( A \) and each Rng \( {\varphi }_{x}^{1} \) are in \( {\sum }_{1} \), it follows easily... | Yes |
Theorem 6.14. Every recursive relation is recursively enumerable. For each positive \( m \) there is a recursively enumerable m-ary relation which is not recursive. | Proof. The first part is true by \( {6.13}\left( x\right) \) and \( {5.33} \) ; for the second part, use 5.36. | No |
Theorem 6.16. Let \( f \) be a unary partial function. Then the following conditions are equivalent:\n\n(i) \( f \) is partial recursive;\n\n(ii) \( \{ \left( {x,{fx}}\right) : x \in \operatorname{Dmn}f\} \) is r.e. | Proof. \( \;\left( i\right) \Rightarrow \left( {ii}\right) \) . Assume \( \left( i\right) \) . For any \( x, y \in \omega \) let\n\n\[ g\left( {x, y}\right) \simeq {\mu z}\left( {\left| {y - {fx}}\right| = 0}\right) . \]\n\nClearly \( g \) is partial recursive and \( \operatorname{Dmn}g = \{ \left( {x,{fx}}\right) : x ... | Yes |
Theorem 6.18. \( \mathrm{K} \) is creative. | Proof. By 6.5, K is r.e. Now \( {\mathrm{U}}_{0}^{1} \) is a productive function for \( \omega \sim \mathrm{K} \) . For if \( e \in \omega \) and Dmn \( {\mathbf{\varphi }}_{e}^{1} \subseteq \omega \sim \mathrm{K} \), then \( e \in \left( {\omega \sim \mathrm{K}}\right) \sim \operatorname{Dmn}{\mathbf{\varphi }}_{e}^{1... | Yes |
Theorem 6.19. If \( A \) is r.e. and \( C \) is creative, then there is a recursive function \( f \) such that \( A = {f}^{-1 * }C \) . | Proof. Say \( A = \operatorname{Dmn}{\mathbf{\varphi }}_{d}^{1} \), and let \( g \) be a productive function for \( \omega \sim C \) . For any \( x, y, z \in \omega \) let\n\n\[ l\left( {z, y, x}\right) \simeq {\mu u}\left\lbrack {z = g{\mathrm{\;s}}_{1}^{1}\left( {x, y}\right) }\right\rbrack + {\mathbf{\varphi }}_{d}^... | Yes |
Theorem 6.20. If \( A \) is productive, then \( A \) has an infinite recursive subset. | Proof. By 6.8 it suffices to show that \( A \) has an infinite r.e. subset. Let \( f \) be a productive function for \( A \). For any \( x, y \), let\n\n\[ k\left( {y, x}\right) \simeq {\mu i}\left( {i \leq {1x}\text{ and }y = {\left( x\right) }_{i} \div 1}\right) .\n\]\n\nClearly \( k \) is partial recursive; say \( k... | Yes |
Theorem 6.23. If \( A \) and \( B \) are recursively enumerable and effectively inseparable, then both \( A \) and \( B \) are creative. | Proof. By symmetry it suffices to show that \( A \) is creative, i.e. that \( \omega \sim A \) is productive. Let \( f \) be as in \( {6.21} \) (iii). Say \( A = \operatorname{Dmn}{\mathbf{\varphi }}_{u}^{1} \) and \( B = \operatorname{Dmn}{\mathbf{\varphi }}_{s}^{1} \) . For any \( e, x \in \omega \) let\n\n\[ g\left(... | Yes |
Theorem 6.24. There exist two recursively enumerable effectively inseparable sets. | Proof. Let\n\n\[ \n{K}_{1} = \left\{ {x : \exists y\left\lbrack {\left( {{\left( x\right) }_{0}, x, y}\right) {\mathrm{T}}_{1}\text{ and }\forall z \leq y\left( {\left( {{\left( x\right) }_{1}, x, z}\right) \notin {\mathrm{T}}_{1}}\right) }\right\rbrack }\right\} ,\n\]\n\n\[ \n{K}_{2} = \left\{ {x : \exists y\left\lbra... | Yes |
Theorem 6.25. Suppose that \( A \) and \( B \) are effectively inseparable, \( f \) is a unary recursive function, \( C, D \subseteq \omega, C \cap D = 0, A \subseteq {f}^{-1 * }C \), and \( B \subseteq {f}^{-1 * }D \) . Then \( C \) and \( D \) are effectively inseparable. | Proof. Let \( h \) be a function given by \( {6.21} \) (iii) because \( A \) and \( B \) are effectively inseparable. For any \( e, x \in \omega \), let \( g\left( {x, e}\right) \simeq {\mu y}\left( {\left( {e,{fx}, y}\right) \in {\mathrm{T}}_{1}}\right) \) . Thus \( g \) is partial recursive; say \( g = {\mathbf{\varp... | Yes |
Theorem 6.27. A simple set is neither recursive nor creative. | Proof. If \( A \) is simple and recursive, then \( \omega \sim A \) is an infinite r.e. set and \( A \cap \left( {\omega \sim A}\right) = 0 \), contradiction. If \( A \) is simple and creative, by 6.20 choose \( B \) infinite recursive such that \( B \subseteq \omega \sim A \) . Contradiction. | Yes |
Theorem 6.28. Simple sets exist. | Proof. Let \( g \) be a recursive function universal for unary primitive recursive functions (see Lemma 3.5). For any \( e \in \omega \) let\n\n\[ \n{fe} \simeq {\left( \mu y\left\lbrack g\left( e,{\left( y\right) }_{0}\right. = {\left( y\right) }_{1}\text{ and }{\left( y\right) }_{1} > 2e\right\rbrack \right) }_{1}.\n... | Yes |
Proposition 7.17. Addition and multiplication of RET's are commutative and associative. Multiplication is distributive over addition. | The structure (RET, +, .) is not, however, a ring, and it cannot be embedded in a ring. This can be seen for example, from the fact that \( \alpha + \beta = \alpha \) where \( \alpha \) and \( \beta \) are respectively the recursive equivalence types of \( \omega \) and of 1 . Since \( \beta + \beta \neq \beta ,\beta \... | No |
Theorem 7.20. If \( r \in {}^{ \circ }\mathbb{Q} \) is recursive, strictly monotone, converges to a recursive real number \( \alpha \), then \( r \) recursively converges to \( \alpha \) . | A sequence \( r \in {}^{\omega }F \) is recursive provided there are binary recursive functions \( f, g, h, k \) such that for all \( m, n \in \omega \) and all \( p \geq k\left( {m, n}\right) \) we have\n\n\[ \left| {{r}_{n}-\{ \left\lbrack {f\left( {p, n}\right) - g\left( {p, n}\right) }\right\rbrack /\left\lbrack {1... | No |
Theorem 8.5 (Unique readability)\n\n(i) Every sentence is of positive length.\n\n(ii) If \( \varphi \) is a sentence, then either \( \varphi = \langle s\rangle \) for some \( s \in P,\varphi = \neg \psi \) for some sentence \( \psi \), or \( \varphi = \psi \rightarrow \chi \) for some sentences \( \psi ,\chi \) .\n\n(i... | Proof. Conditions \( \left( i\right) \) and \( \left( {ii}\right) \) are easily established using 8.3; for example, to prove \( \left( i\right) \) we let \( \Gamma \) be the set of all expressions of positive length. We establish (iii) by induction on \( m \) . If \( m = 1 \), then by (ii) and (i), \( \varphi = \langle... | Yes |
Theorem 8.8. \( \Gamma \vdash \varphi \) iff there is a finite sequence \( \left\langle {{\psi }_{0},\ldots ,{\psi }_{m - 1}}\right\rangle, m > 0 \), of sentences of \( \mathcal{P} \) such that \( {\psi }_{m - 1} = \varphi \) and for each \( i < m \) one of the following holds:\n\n(i) \( {\psi }_{i} \) is a logical axi... | A sequence \( \left\langle {{\psi }_{0},\ldots ,{\psi }_{m - 1}}\right\rangle \) as in 8.8 is called a formal proof of \( \varphi \) from the hypotheses \( \Gamma \) . | No |
Lemma 8.10. \( \vdash \varphi \rightarrow \varphi \) . | Proof. Just this once we give a formal proof, with justifications listed in the column to the right.\n\n(1) \( \{ \varphi \rightarrow \left\lbrack {\left( {\varphi \rightarrow \varphi }\right) \rightarrow \varphi }\right\rbrack \} \rightarrow \{ \left\lbrack {\varphi \rightarrow \left( {\varphi \rightarrow \varphi }\ri... | Yes |
Theorem 8.11 (Deduction theorem). If \( \Gamma \cup \{ \varphi \} \vdash \psi \), then \( \Gamma \vdash \varphi \rightarrow \psi \) . | Proof. By induction on \( m \) we show that for every nonzero \( m \in \omega \), if \( \left\langle {{\chi }_{0},\ldots ,{\chi }_{m - 1}}\right\rangle \) is a formal proof of \( \psi \) from \( \Gamma \cup \{ \varphi \} \), then \( \Gamma \vdash \varphi \rightarrow \psi \) . Suppose this is true for all \( n < m \), a... | Yes |
Lemma 8.12. \( \; \vdash \left( {\varphi \rightarrow \psi }\right) \rightarrow \left\lbrack {\left( {\psi \rightarrow \chi }\right) \rightarrow \left( {\varphi \rightarrow \chi }\right) }\right\rbrack \) . | Proof\n\n\[ \{ \varphi \rightarrow \psi ,\psi \rightarrow \chi ,\varphi \} \vdash \varphi \]\n\n\[ \{ \varphi \rightarrow \psi ,\psi \rightarrow \chi ,\varphi \} \vdash \varphi \rightarrow \psi \]\n\n\[ \{ \varphi \rightarrow \psi ,\psi \rightarrow \chi ,\varphi \} \vdash \psi \]\n\n\[ \{ \varphi \rightarrow \psi ,\psi... | Yes |
Lemma 8.14. \( \; \vdash \varphi \rightarrow \left( {\neg \varphi \rightarrow \psi }\right) \) . | Proof\n\n\[ \{ \varphi ,\neg \varphi \} \vdash \neg \varphi \rightarrow \left( {\neg \psi \rightarrow \neg \varphi }\right) \]\n\nA1\n\n\[ \{ \varphi ,\neg \varphi \} \vdash \neg \varphi \]\n\n\[ \{ \varphi ,\neg \varphi \} \vdash \neg \psi \rightarrow \neg \varphi \]\n\n\[ \{ \varphi ,\;\neg \varphi \} \vdash \left( {... | Yes |
Lemma 8.16. \( \vdash \neg \neg \varphi \rightarrow \varphi \) . | Proof\n\n\[ \{ \neg \neg \varphi \} \vdash \neg \neg \neg \neg \neg \varphi \rightarrow \neg \neg \varphi \]\n\nusing \( \mathrm{A}1 \)\n\n\[ \{ \neg \neg \varphi \} \vdash \neg \varphi \rightarrow \neg \neg \neg \varphi \]\n\nA3\n\n\[ \{ \neg \neg \varphi \} \vdash \neg \neg \varphi \rightarrow \varphi \]\n\nA3\n\n\[ ... | Yes |
Lemma 8.19. \( F\left( {\varphi \rightarrow \neg \varphi }\right) \rightarrow \neg \varphi \) . | \[ \vdash \left( {\varphi \rightarrow \neg \varphi }\right) \rightarrow \neg \varphi \] | No |
Lemma 8.23. If \( \Gamma \vdash \varphi \), then \( \Gamma \vDash \varphi \) . | Proof. Let \( \Delta = \{ \varphi \) : every model of \( \Gamma \) is a model of \( \varphi \} \) . It is easy to check, using truth tables for the logical axioms, that \( \Gamma \subseteq \Delta \), every logical axiom is in \( \Delta \), and \( \Delta \) is closed under detachment. Hence all \( \Gamma \) -theorems ar... | No |
Theorem 8.25. The following conditions are equivalent:\n\n(i) \( \Gamma \) is inconsistent.\n\n(ii) \( \Gamma \vdash \neg \left( {\varphi \rightarrow \varphi }\right) \) for every sentence \( \varphi \) .\n\n(iii) \( \Gamma \vdash \neg \left( {\varphi \rightarrow \varphi }\right) \) for some sentence \( \varphi \) . | Proof. Obviously \( \left( i\right) \Rightarrow \left( {ii}\right) \Rightarrow \left( {iii}\right) \) . Now suppose \( \Gamma \vdash \neg \left( {\varphi \rightarrow \varphi }\right) \) for a certain sentence \( \varphi \) . Let \( \psi \) be any sentence. By \( \mathrm{{Al}},\Gamma \vdash \left( {\varphi \rightarrow \... | Yes |
Theorem 8.26. \( \Gamma \cup \{ \varphi \} \) is inconsistent iff \( \Gamma \vdash \neg \varphi \) . | Proof. \( \Rightarrow \) : Since \( \Gamma \cup \{ \varphi \} \vdash \psi \) for any sentence \( \psi \), we have \( \Gamma \cup \{ \varphi \} \vdash \neg \varphi \), so by the deduction theorem \( \Gamma \vdash \varphi \rightarrow \neg \varphi \) . By 8.19, \( \Gamma \vdash \neg \varphi \) . \( \Leftarrow : \Gamma \cu... | Yes |
Theorem 8.27. 0 is consistent. | Proof. Since \( \neg \left( {\varphi \rightarrow \varphi }\right) \) always receives the value 0 under any model, for any sentence \( \varphi \), by 8.23 we have not \( \left( { \vdash \neg \left( {\varphi \rightarrow \varphi }\right) }\right) \) . | Yes |
Theorem 8.29 (Completeness theorem). \( \Gamma \vdash \varphi \) iff \( \Gamma \vDash \varphi \) . | Proof. \( \Rightarrow \) : by 8.23. \( \Leftarrow \) : Suppose not \( \left( {\Gamma \vdash \varphi }\right) \) . Then not \( \left( {\Gamma \vdash \neg \neg \varphi }\right) \) by 8.16, so \( \Gamma \cup \{ \neg \varphi \} \) is consistent, by 8.26. By 8.28, let \( f \) be a model of \( \Gamma \cup \{ \neg \varphi \} ... | Yes |
Theorem 8.36. If \( \Gamma \cup \Delta \vdash \psi \) and \( \Delta \neq 0 \), then there is an \( m \in \omega \) and a \( \varphi \in {}^{m + 1}\Delta \) such that \( \Gamma \vdash {\varphi }_{0} \land \cdots \land {\varphi }_{m} \rightarrow \psi \) . | Proof. By \( {8.9}\left( {ii}\right) \) we may assume that \( \Delta \) is finite. Hence it is enough to prove by induction on \( m \) that for all \( m \in \omega \) and all \( \varphi \in {}^{m + 1}{\operatorname{Sent}}_{\mathcal{P}} \), if \( \Gamma \cup \) \( \left\{ {{\varphi }_{i} : i \leq m}\right\} \vdash \psi ... | Yes |
Theorem 8.37. For any set \( \Gamma \) of sentences the following conditions are equivalent:\n\n(i) \( \Gamma \) is inconsistent;\n\n(ii) there is an \( m \in \omega \) and a \( \varphi \in {}^{m + 1}\Gamma \) such that \( \vdash \mathop{\bigvee }\limits_{{i \leq m}}\neg {\varphi }_{i} \) . | Proof. \( \left( i\right) \Rightarrow \left( {ii}\right) \) . Assuming \( \left( i\right) \), we have \( \Gamma \vdash \psi \land \neg \psi \) for any sentence \( \psi \) ; we fix \( \psi \) . From 8.27 we know that \( \Gamma \neq 0 \) . Hence by Theorem 8.36 there is an \( m \in \omega \) and a \( \varphi \in {}^{m + ... | Yes |
Theorem 8.38 (Distinguished disjunctive normal form). Let \( \mathcal{P} = \langle n, c \) , \( \left\{ {{s}_{0},\ldots ,{s}_{m}}\right\} \rangle \) be a sentential language, with \( m \in \omega \) . Let \( \varphi \) be a sentence which has at least one model. Then there is a function \( \psi \) such that:\n\n(i) the... | Proof. Let \( M \) be the set of all models of \( \varphi ;M \neq 0 \) by assumption. Let \( g \) be a one-one function mapping some integer \( p + 1 \) onto \( M \) . For each \( i \leq p \) and \( j \leq m \) let\n\n\[{\psi }_{ij} = \left\{ \begin{matrix} \left\langle {s}_{j}\right\rangle & \text{ if }{g}_{i}{s}_{j} ... | Yes |
Theorem 8.39 (Distinguished conjunctive normal form). Let \( \mathcal{P} = \langle n, c \) , \( \left. \left\{ {{s}_{0},\ldots ,{s}_{m - 1}}\right\} \right\rangle \) be a sentential logic, with \( m \in \omega \) . Let \( \varphi \) be a sentence which is not a tautology. Then there is a function \( \psi \) such that:\... | Proof. Assume the hypothesis. Thus not \( \left( { \vdash \varphi }\right) \), so by \( {8.33}\left( {vi}\right) ,{\varphi }^{\mathrm{d}} \) has a model. Hence by 8.38 we can choose \( \psi \) so that (i) and (ii) hold and\n\n\[ \vdash {\varphi }^{\mathrm{d}} \leftrightarrow \mathop{\bigvee }\limits_{{i \leq p}}\mathop... | Yes |
Theorem 8.41. Let \( \mathcal{P} = \langle \{ n, c\}, P, f, g\rangle \) be a sentential logic in the original sense, with \( P \) finite. Then \( \mathcal{P} \) is functionally complete. | Proof. Note that there are exp exp \( \left| P\right| P \) -truth functions. Hence it suffices to show that there are at least that many truth functions of the form \( {\mathcal{T}}_{\varphi } \) . Let \( \left| P\right| = m + 1 \), say \( P = \left\{ {{s}_{i} : i \leq m}\right\} \) . For each \( h \in {}^{P}2 \) we de... | Yes |
Theorem 8.43. Negation and conjunction give rise to a functionally complete sentential logic. That is, if \( \mathcal{P} = \langle \{ n, k\}, P, f, g\rangle \) is a general sentential logic with \( {fn} = 1,{fk} = 2 \), and \( g \) given by:\n\n\[ \n{g}_{n}\langle 0\rangle = 1,\;{g}_{n}\langle 1\rangle = 0 \]\n\n\[ \n{... | Proof. We may assume that \( P \cap 2 = 0 \) . Let \( \neg \varphi = \langle n\rangle \varphi \) and \( \varphi \land \psi = \) \( \langle k\rangle {\varphi \psi } \) for all sentences \( \varphi ,\psi \) . Clearly \( {neg} = {\mathcal{T}}_{\langle 0\rangle } \) in \( \langle \{ n, k\} ,1, f, g\rangle \), and \( {imp} ... | Yes |
Theorem 8.45. Let \( \mathcal{P} = \langle \{ s\}, P, f, g\rangle \) be a general sentential logic such that \( {fs} = 1 \) and \( g \) is given by\n\n\[ \n{g}_{s}\langle 1,1\rangle = 0 \n\]\n\n\[ \n{g}_{s}\langle 0,1\rangle = {g}_{s}\langle 1,0\rangle = {g}_{s}\langle 0,0\rangle = 1, \n\]\n\nand with \( P \) finite. T... | Proof. \( \varphi \mid \psi = \langle s\rangle {\varphi \psi } \) for any sentences \( \varphi ,\psi \) . (The connective here is called the Sheffer stroke.) Then neg \( = {\mathcal{T}}_{\varphi } \) and imp \( = {\mathcal{T}}_{\psi } \) where \( \varphi \) is \( \langle 0\rangle \mid \langle 0\rangle \) and \( \psi \)... | No |
Proposition 9.8. Let \( \mathfrak{A} \) be a BA, \( X \subseteq A \) . Then the following conditions are equivalent:\n\n(i) \( X \) is a subuniverse of \( \mathfrak{A} \) ;\n\n(ii) \( X \neq 0 \), and \( X \) is closed under + and -;\n\n(iii) \( X \neq 0 \), and \( X \) is closed under \( \cdot \) and - . | Proof. Obviously \( \left( i\right) \Rightarrow \left( {ii}\right) \) . Now assume \( \left( {ii}\right) \) . Then \( X \) is closed under \( \cdot \) , since \( x \cdot y = - \left( {-x + - y}\right) \) ; hence (iii) holds. Assume (iii). Then \( X \) is closed under + by the dual of the argument just given. Choose \( ... | Yes |
Proposition 9.9. If \( \mathcal{A} \) is a nonempty collection of subuniverses of a BA \( \mathfrak{A} \) , then \( \cap \mathcal{A} \) is a subuniverse of \( \mathfrak{A} \) . | The proof is trivial. | No |
Theorem 9.11. If \( X \) is a subset of \( A \), then the subuniverse of \( \mathfrak{A} \) generated by \( X \) consists of 0,1, and all elements of \( A \) of the form\n\n(1)\n\n\[ \mathop{\sum }\limits_{{i < m}}\mathop{\prod }\limits_{{j < {ni}}}{y}_{ij} \]\n\nwhere for each \( i, j \), either \( {y}_{ij} \in X \) o... | Proof. Clearly each such element is in the subuniverse of \( \mathfrak{A} \) generated by \( X \) . Thus it suffices to show that the set \( S \) of all such elements is a subuniverse of \( \mathfrak{A} \) containing \( X \) . Obviously \( X \subseteq S \) and \( S \) is closed under +. By \( {9.8}\left( i\right) \) it... | Yes |
Proposition 9.13. Let \( \mathfrak{A} \) and \( {\mathfrak{A}}^{\prime } \) be BA’s, as in 9.12, and let \( f \) be a map of \( A \) into \( {A}^{\prime } \) . Then the following conditions are equivalent;\n\n(i) \( f \) is a homomorphism;\n\n(ii) \( f\left( {-x}\right) = - {}^{\prime }{fx} \) and \( f\left( {x + y}\ri... | Proof. Obviously \( \left( i\right) \Rightarrow \left( {ii}\right) \) . Now assume \( \left( {ii}\right) \), and let \( x, y \in A \) . Then\n\n\[ f\left( {x \cdot y}\right) = f\left( {-\left( {-x + - y}\right) }\right) = - {}^{\prime }\left( {-{}^{\prime }{fx} + {}^{\prime } - {}^{\prime }{fy}}\right) = {fx} \cdot {}^... | Yes |
Proposition 9.14. Let \( \mathfrak{A} \) and \( {\mathfrak{A}}^{\prime } \) be BA’s, as in 9.12, and let \( f \) be a map of \( A \) into \( {A}^{\prime } \) . Then the following conditions are equivalent:\n\n(i) \( f \) is an isomorphism of \( \mathfrak{A} \) onto \( {\mathfrak{A}}^{\prime } \) ;\n\n(ii) f maps \( A \... | Proof. Obviously \( \left( i\right) \Rightarrow \left( {ii}\right) \) . Now assume \( \left( {ii}\right) \) . Now if \( {fx} = {fy} \), then \( {fx}{ \leq }^{\prime }{fy} \) and \( {fy}{ \leq }^{\prime }{fx} \), so \( x \leq y \) and \( y \leq x \), hence \( x = y \) . Thus \( f \) is one-one. Now we shall apply \( {9.... | Yes |
Proposition 9.16. Let \( I \) be an ideal of \( \mathfrak{A} \) . Let \( R = \{ \left( {x, y}\right) : x \cdot - y + y \cdot - x \in I\} \) . Then \( R \) is an equivalence relation on \( A \) . Moreover, (i) if \( {xRy} \), then \( - {xR} - y \) ; (ii) if \( {xRy} \) and \( {x}^{\prime }R{y}^{\prime } \), then \( \lef... | Proof. \( \;R \) is reflexive on \( A \) : if \( x \in A \), then \( x \cdot - x + - x \cdot x = 0 \in I \) . Obviously \( R \) is symmetric. \( R \) is transitive: assume that \( {xRyRz} \) . Thus \( x \cdot - y + y \cdot - x \in I \) and \( y \cdot - z + z \cdot - y \in I \) . Now \[ x \cdot - z = x \cdot \left( {y +... | Yes |
Theorem 9.19 (The homomorphism theorem). Let \( f \) be a homomorphism from a BA \( \mathfrak{A} \) onto a BA \( \mathfrak{B} \). Let \( I = \{ x : x \in A \) and \( {fx} = 0\} \). Then \( I \) is an ideal of \( \mathfrak{A} \), and \( \mathfrak{A}/I \) is isomorphic to \( \mathfrak{B} \). \( I \) is called the kernel ... | Proof. Since \( {f0} = 0 \), we have \( 0 \in I \) and hence \( I \neq 0 \). Suppose \( x, y \in I \). Then \( f\left( {x + y}\right) = {fx} + {fy} = 0 + 0 = 0 \). Thus \( x + y \in I \). Suppose \( x \leq y \in I \). Then \( {fx} \leq {fy} = 0 \), so \( {fx} = 0 \) and hence \( x \in I \). Thus \( I \) is an ideal. No... | Yes |
Proposition 9.20. There is a one-one correspondence between ideals and filters on a given BA \( \mathfrak{A} \) . In more detail, if \( I \) is an ideal of \( \mathfrak{A} \) and \( F \) is a filter of \( \mathfrak{A} \) , then:\n\n(i) \( {I}^{f} = \{ x : - x \in I\} \) is a filter of \( \mathfrak{A} \) ;\n\n(ii) \( {F... | The following proposition is also clear: | No |
Proposition 9.25. For any ideal \( I \) in a BA A the following conditions are equivalent:\n\n(i) I is maximal;\n\n(ii) \( I \neq A \), and for any \( x, y \in A \), if \( x \cdot y \in I \) then \( x \in I \) or \( y \in I \) ;\n\n(iii) \( I \neq A \), and for any \( x \in A \), either \( x \in I \) or \( - x \in I \)... | Proof\n\n\( \left( i\right) \Rightarrow \left( {ii}\right) \) . Assume that \( I \) is maximal, but that there are elements \( x, y \in A \) with \( x \cdot y \in I \) while \( x \notin I \) and \( y \notin I \) . Then the ideal generated by \( I \cup \{ x\} \) is \( A \) , so by 9.23 there is a \( u \in I \) such that... | Yes |
Theorem 9.26 (Boolean prime ideal theorem). If \( I \) is a proper ideal of \( \mathfrak{A} \) (i.e., \( I \) is an ideal of \( \mathfrak{A} \), and \( I \neq A \) ), then there is a maximal ideal \( J \) of \( \mathfrak{A} \) such that \( I \subseteq J \) . | Proof. Let \( \mathcal{A} = \{ J : I \subseteq J, J \) is a proper ideal of \( \mathfrak{A}\} \) . Then \( \mathcal{A} \neq 0 \), since obviously \( I \in \mathcal{A} \) . Let \( \mathcal{B} \) be a nonempty subset of \( \mathcal{A} \) simply ordered by inclusion. We now show that \( \bigcup \mathcal{B} \in \mathcal{A}... | Yes |
Theorem 9.27 (Boolean representation theorem). Every BA is isomorphic to a Boolean set algebra. | Proof. Let \( \mathfrak{A} \) be a BA. Let \( X = \{ F : F \) is an ultrafilter of \( \mathfrak{A}\} \) . It suffices to find an isomorphism \( f \) of \( \mathfrak{A} \) into the BA \( \langle \mathbf{S}X, \cup , \cap , \sim ,0, X\rangle \) . For each \( a \in A \), let \( {fa} = \{ F : a \in F \in X\} \) . Then for a... | Yes |
Proposition 9.30. If \( \mathfrak{A} \) is a finite BA with \( m \) atoms, then \( \mathfrak{A} \) is isomorphic to the Boolean set algebra of all subsets of \( m \) . | Proof. Let \( a \in {}^{m}A \) enumerate all atoms of \( \mathfrak{A} \) . For each \( x \in A \), let\n\n\[ \n{fx} = \left\{ {i : {a}_{i} \leq x}\right\} \n\]\n\nThus \( f \) maps \( A \) into the Boolean set algebra \( \mathfrak{B} \) of all subsets of \( m \) . Suppose \( i \in f\left( {x + y}\right) \), while \( i ... | Yes |
Proposition 9.34. The following conditions are equivalent:\n\n(i) \( \mathfrak{A} \) is atomic;\n\n(ii) for every \( a \in A,\sum \{ x : x \leq a, x \) is an atom of \( \mathfrak{A}\} \) exists and equals \( a \) . | Proof\n\n\( \left( i\right) \Rightarrow \left( {ii}\right) \) . Assume \( \left( i\right) \), and let \( a \in A \) . Obviously \( a \) is an upper bound for \( X = \{ x : x \leq a, x \) is an atom of \( \mathfrak{A}\} \) . Suppose that \( b \) is another upper bound for \( X \) . If \( a \leq b \), then \( a \cdot - b... | No |
Proposition 9.36. If \( {\sum X} \) exists, then \( \Pi \{ - x : x \in X\} \) exists and equals \( - {\sum X} \) . | Proof. If \( x \in X \), then \( x \leq {\sum X} \) and hence \( - {\sum X} \leq - x \) . Thus \( - {\sum X} \) is a lower bound for \( Y = \{ - x : x \in X\} \) . Let \( a \) be any lower bound for \( Y \) . Thus \( \forall x \in X\left( {a \leq - x}\right) \), so \( \forall x \in X\left( {x \leq - a}\right) \) . Henc... | Yes |
Proposition 9.37. If \( {\sum X} \) exists and \( a \in A \), then \( \sum \{ x \cdot a : x \in X\} \) exists and equals \( a \cdot {\sum X} \) . | Proof. For any \( x \in X, x \leq {\sum X} \) and hence \( a \cdot x \leq a \cdot {\sum X} \) . Now suppose that \( b \) is any upper bound of \( \{ x \cdot a : x \in X\} \) . Then for any \( x \in X, x \cdot a \leq b \) and so \( x \leq - a + b \) . Thus \( {\sum X} \leq - a + b \), so \( a \cdot {\sum X} \leq b \) . | Yes |
Proposition 9.40. The BA \( \langle \mathbf{S}X, \cup , \cap , \sim ,0, X\rangle \) is isomorphic to \( {}^{X}\mathbf{2} \) . | Proof. For each \( Y \subseteq X \), let \( {\chi }_{Y} \) be the characteristic function \( Y \), i.e., for all \( x \in X \) let\n\n\[ \n{\chi }_{Y}x = 1\;\text{ if }x \in Y \n\]\n\n\[ \n{\chi }_{Y}x = 0\;\text{ if }x \in Y. \n\]\n\nIt is easily verified that \( \chi \) is the desired isomorphism. | No |
Proposition 9.44. For any \( a \in A,\mathfrak{A} \) is isomorphic to \( \left( {\mathfrak{A} \upharpoonright a}\right) \times \left( {\mathfrak{A} \upharpoonright - a}\right) \) . | Proof. For any \( x \in A \), let \( {fx} = \langle x \cdot a, x \cdot - a\rangle \) . It is easily checked that \( f \) is the desired isomorphism. | No |
Theorem 9.48. Any two denumerable atomless BA's are isomorphic. | Proof. Let \( R = \{ \left( {\mathfrak{A},\mathfrak{B}}\right) : \left| A\right| = \left| B\right| = 1 \) or \( \mathfrak{A} \) and \( \mathfrak{B} \) are denumerable and atomless\}. The hypothesis of 9.47 is easily verified. | No |
Proposition 9.50. If \( \mathfrak{A} \) and \( \mathfrak{B} \) are each freely generated by \( X \), then they are isomorphic. | Proof. Let \( f \) be a homomorphism from \( \mathfrak{A} \) into \( \mathfrak{B} \) extending Id \( \upharpoonright X \) (identity on \( X \) ), and let \( g \) be a homomorphism from \( \mathfrak{V} \) into \( \mathfrak{A} \) extending \( \operatorname{Id} \upharpoonright X \) . Then \( \{ a \in A : {gfa} = a\} \) is... | Yes |
Lemma 9.51. If \( X \) generates \( \mathfrak{A} \), then \( \left| A\right| \leq \left| X\right| + {\aleph }_{0} \) . | Proof. Let \( {Y}_{0} = X \) and, for \( m \in \omega \) ,\n\n\[ \n{Y}_{m + 1} = {Y}_{m} \cup \left\{ {-x : x \in {Y}_{m}}\right\} \cup \left\{ {x + y : x, y \in {Y}_{m}}\right\} \cup \{ 0,1\} .\n\]\n\nIt is easily checked that \( \mathop{\bigcup }\limits_{{m \in \omega }}{Y}_{m} = A \) . Furthermore, by induction on \... | Yes |
Lemma 9.52. If \( \mathfrak{A} \) is a BA and \( f \) is a one-one function from \( A \) onto \( B \), then there is a BA \( \mathfrak{B} \) with universe B such that \( f \) is an isomorphism from \( \mathfrak{A} \) onto \( \mathfrak{B} \) . | Proof. Let \( \mathfrak{A} = \langle A, + , \cdot , - ,0,1\rangle \) . We define the operations of \( \mathfrak{B} \) as follows. For any \( b, c \in B \) ,\n\n\[ b + {}^{\prime }c = f\left( {{f}^{-1}b + {f}^{-1}c}\right) \]\n\n\[ b \cdot {}^{\prime }c = f\left( {{f}^{-1}b \cdot {f}^{-1}c}\right) ,\]\n\n\[ - {}^{\prime... | Yes |
Theorem 9.53. For any nonempty set \( X \) there is a BA freely generated by \( X \) . | Proof. Let \( A \) be any set such that \( \left| A\right| = \left| X\right| + {\aleph }_{0} \) . Let \( I = \{ \left( {\mathfrak{B}, f}\right) : \mathfrak{B} \) is a BA, \( B \subseteq A \), and \( f \) is a function mapping \( X \) into \( B\} \) . Let \( \mathfrak{C} = {\mathrm{P}}_{\left( {\mathfrak{B}, f}\right) \... | Yes |
Theorem 9.58. If \( \mathcal{P} = \left( {n, c, P}\right) \) is a sentential language, then \( {\mathfrak{M}}_{0} \) is freely generated by \( \{ \left\lbrack {\langle s\rangle }\right\rbrack : s \in P\} \), and \( \langle \left\lbrack {\langle s\rangle }\right\rbrack : s \in P\rangle \) is a one-one function. | Proof. We check the second statement first. If \( s, t \in P \) and \( s \neq t \), clearly \( \sharp \langle s\rangle \leftrightarrow \langle t\rangle \) and hence \( \sharp \langle s\rangle \leftrightarrow \langle t\rangle \), so \( \left\lbrack {\langle s\rangle }\right\rbrack \neq \left\lbrack {\langle t\rangle }\r... | Yes |
Proposition 9.59. Let \( F \) be a filter in a BA \( \mathfrak{M} \), and set \( \Delta = \bigcup F \) . Then \( \Gamma \subseteq \Delta \), and \( {\mathfrak{M}}_{\Gamma }^{\mathcal{P}}/F \) is isomorphic to \( {\mathfrak{M}}_{\Delta }^{\mathcal{P}} \) . | Proof. Note first that \( \Gamma \) is a subset of the unit element 1 of \( \mathfrak{M}F \) ; since \( 1 \in F \) , clearly \( \Gamma \subseteq \Delta \) . We write below \( {\left\lbrack \varphi \right\rbrack }_{\Gamma },{\left\lbrack x\right\rbrack }_{F},{\left\lbrack \varphi \right\rbrack }_{\Delta } \) for the equ... | Yes |
Theorem 9.60. Any BA is isomorphic to \( \mathfrak{M} \) ? for some sentential language \( \mathcal{P} \) and some \( \Gamma \subseteq {\operatorname{Sent}}_{\mathcal{P}} \) . | Proof. Let \( \mathfrak{A} \) be any BA. Let \( \mathcal{P} = \left( {n, c, A}\right) \) be a sentential language, and let \( f \) be the function such that \( f\left\lbrack {\langle a\rangle }\right\rbrack = a \) for \( a \in A \) . By 9.58, we can extend \( f \) to a homomorphism \( {f}^{ + } \) of \( \mathfrak{M}\ma... | Yes |
Proposition 10.5. \( {g}^{+ * }{\operatorname{Expr}}_{\mathcal{L}} \) is recursive. | Proof. For any \( x \in \omega \), we have \( x \in {\mathcal{g}}^{+ * }{\operatorname{Expr}}_{\mathcal{L}} \) iff \( x = 1 \) or else \( x > 1 \) and \( \forall i \leq \operatorname{lx}\left( {{\left( x\right) }_{i} - 1 \in {\mathcal{J}}^{ * }{\operatorname{Sym}}_{\mathcal{L}}}\right) . \n\nThus there is an effective ... | Yes |
Proposition 10.10. The functions \( {\mathrm{{Con}}}_{m},{\mathrm{{Con}}}^{\prime } \), and \( {\mathrm{{Con}}}^{\prime \prime } \) are recursive. The set \( {\mathcal{J}}^{+ * } \) Trm is recursive. | Proof. The first statement is obvious. To prove the second, the reader should check the following statement, using 10.9. For any \( x \in \omega, x \in {y}^{+ * } \) Trm iff \( x > 1 \) and there is a \( y \leq {\mathrm{p}}_{1x}^{x \cdot {1x}} \) such that \( {\left( y\right) }_{1y} = x \) and for each \( i \leq {1y} \... | No |
Proposition 10.12 (Unique readability)\n\n(i) Every term is nonempty.\n\n(ii) If \( \sigma \) is a term, then either \( \sigma = \left\langle {v}_{m}\right\rangle \) for some \( m \in \omega \), or else there exist \( \mathbf{O} \in \operatorname{Dmn}\mathcal{O} \), say with \( \mathcal{O}\mathbf{O} = m \), and \( {\ta... | Proof. In each of the cases \( \left( i\right) \) and \( \left( {ii}\right) \), let \( \Gamma \) be the collection of terms \( \sigma \) for which the desired condition holds, and apply 10.11. We prove (iii) by induction on \( \operatorname{Dmn}\sigma \) . The case \( \operatorname{Dmn}\sigma = 1 \) is clear by \( \lef... | Yes |
Proposition 10.15. \( {g}^{+ * } \) Fmla is recursive. | Proof. The proposition becomes obvious after checking the following statement whose proof is based on a proposition for formulas which is entirely similar to 10.9. For any \( x \in \omega, x \in {\mathcal{g}}^{+ * } \) Fmla iff there is a \( y \leq {\mathrm{p}}_{1x}^{x \cdot {1x}} \) such that \( {\left( y\right) }_{1y... | Yes |
Lemma 10.20. For any formula \( \varphi \) the following conditions are equivalent:\n\n(i) \( \varphi \) is a tautology;\n\n(ii) for any \( f \), iff maps the set \( S \) of subformulas of \( \varphi \) into 2 and satisfies the following conditions;\n\n(a) for any \( \psi \), if \( \psi \in S \) and \( \neg \psi \in S ... | Proof\n\n\( \left( i\right) \Rightarrow \left( {ii}\right) \) . Assume that \( \varphi \) is a tautology, and let \( f \) satisfy the hypothesis of (ii). By recursion on formulas we define a function \( h : \mathrm{{Fmla}} \rightarrow 2 \) as follows:\n\n\[ \n{h\psi } = {f\psi }\;\text{ if }\psi \text{ is an atomic for... | Yes |
Proposition 10.21. \( \left\{ {\left( {{g}^{ + }\psi ,{g}^{ + }\varphi }\right) : \psi }\right. \) is a subformula of \( \left. \varphi \right\} \) is recursive. | Proof. Let \( R \) be the set in question. Then for any \( m, n \in \omega ,\left( {m, n}\right) \in R \) iff \( m, n \in {\mathcal{g}}^{+ * } \) Fmla and \( \exists x, y \leq n\left( {n = \operatorname{Cat}\left( {\operatorname{Cat}\left( {x, m}\right), y}\right) }\right) \). | Yes |
Proposition 10.22. The set of Gödel numbers of tautologies is recursive. | Proof. Let \( R \) be the relation of 10.21. Using 10.20, it is easy to check that for any \( m \in \omega, m \) is the Gödel number of a tautology iff:\n\n\[ \exists n \leq {\mathrm{p}}_{1m}^{m \cdot {1m}}\left\{ {n > 1\;\& \;\forall p \leq m\lbrack \left( {p, m}\right) \in R \Rightarrow \exists i \leq \ln (p = \left(... | Yes |
Proposition 10.24. For \( \Gamma \cup \{ \varphi \} \subseteq \mathrm{{Fmla}},\Gamma \vdash \varphi \) iff there is a finite sequence \( \left\langle {{\psi }_{0},\ldots ,{\psi }_{m - 1}}\right\rangle \) of formulas such that \( {\psi }_{m - 1} = \varphi \) and for each \( i < m \) one of the following conditions holds... | A sequence of the sort described in 10.24 is called a formal proof of \( \varphi \) from \( \Gamma \), or a \( \Gamma \) -formal proof of \( \varphi \) . We denote by \( \Gamma \) -Prf the set of all \( \Gamma \) -proofs. This is our rigorous formulation of the intuitive notion of a proof. In fact, as stated in the int... | No |
Proposition 10.25. \( {g}^{+ * } \) Axm is recursive. | Proof. For any \( x \in \omega, x \in {\mathcal{A}}^{+ * } \) Axm iff one of the following conditions holds:\n\n(1) \( x \) is the Gödel number of a tautology;\n\n(2) \( \exists y, z, w \leq x\left\lbrack {y, z \in {\mathcal{g}}^{+ * }}\right. \) Fmla and \( w \in \operatorname{Rng}\left( {\mathcal{g} \circ v}\right) \... | Yes |
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