Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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Proposition 33.18. If \( h\left( n\right) = m \), then \( \mathbf{Q} \vdash {\varphi }_{h}\left( {\bar{n},\bar{m}}\right) \) . | Proof. Suppose \( h\left( n\right) = m \), i.e., \( f\left( {g\left( n\right) }\right) = m \) . Let \( k = g\left( n\right) \) . Then\n\n\[ \mathbf{Q} \vdash {\varphi }_{g}\left( {\bar{n},\bar{k}}\right) \]\n\nsince \( {\varphi }_{g} \) represents \( g \), and\n\n\[ \mathbf{Q} \vdash {\varphi }_{f}\left( {\bar{k},\bar{... | Yes |
Proposition 33.19. If \( h\left( n\right) = m \), then \( \mathbf{Q} \vdash \forall z\left( {{\varphi }_{h}\left( {\bar{n}, z}\right) \rightarrow z = \bar{m}}\right) \) . | Proof. Suppose \( h\left( n\right) = m \), i.e., \( f\left( {g\left( n\right) }\right) = m \) . Let \( k = g\left( n\right) \) . Then\n\n\[ \mathbf{Q} \vdash \forall y\left( {{\varphi }_{g}\left( {\bar{n}, y}\right) \rightarrow y = \bar{k}}\right) \]\n\nsince \( {\varphi }_{g} \) represents \( g \), and\n\n\[ \mathbf{Q... | Yes |
Proposition 33.20. If \( {\varphi }_{f}\left( {{y}_{0},\ldots ,{y}_{k - 1}, z}\right) \) represents \( f\left( {{y}_{0},\ldots ,{y}_{k - 1}}\right) \) in \( \mathbf{Q} \), and \( {\varphi }_{{g}_{i}}\left( {{x}_{0},\ldots ,{x}_{l - 1}, y}\right) \) represents \( {g}_{i}\left( {{x}_{0},\ldots ,{x}_{l - 1}}\right) \) in ... | Proof. Exercise. | No |
For every constant symbol a and every natural number \( n \) , \n\n\[ \n\mathbf{Q} \vdash \left( {{a}^{\prime } + \bar{n}}\right) = {\left( a + \bar{n}\right) }^{\prime }. \n\] | Proof. The proof is, as usual, by induction on \( n \) . In the base case, \( n = 0 \), we need to show that \( \mathbf{Q} \) proves \( \left( {{a}^{\prime } + \mathrm{o}}\right) = {\left( a + \mathrm{o}\right) }^{\prime } \) . But we have: \n\n\[ \n\mathbf{Q} \vdash \left( {{a}^{\prime } + \mathrm{o}}\right) = {a}^{\p... | Yes |
Lemma 33.22. \( \mathbf{Q} \vdash \forall x\neg x < 0 \) . | Proof. We give the proof informally (i.e., only giving hints as to how to construct the formal derivation).\n\nWe have to prove \( \neg a < \) 0 for an arbitrary \( a \) . By the definition of \( < \), we need to prove \( \neg \exists y\left( {{y}^{\prime } + a}\right) = 0 \) in \( \mathbf{Q} \) . We’ll assume \( \exis... | Yes |
For every natural number \( n \) , \[ \mathbf{Q} \vdash \forall x\left( {x < \overline{n + 1} \rightarrow \left( {x = 0 \vee \cdots \vee x = \bar{n}}\right) }\right) . | Proof. We use induction on \( n \) . Let us consider the base case, when \( n = 0 \) . In that case, we need to show \( a < \overline{1} \rightarrow a = 0 \), for arbitrary \( a \) . Suppose \( a < \overline{1} \) . Then by the defining axiom for \( < \), we have \( \exists y\left( {{y}^{\prime } + a}\right) = {\mathrm... | Yes |
Proposition 33.25. If \( {\varphi }_{g}\left( {x, z, y}\right) \) represents \( g\left( {x, z}\right) \) in \( \mathbf{Q} \), then\n\n\[ \n{\varphi }_{f}\left( {z, y}\right) \equiv {\varphi }_{g}\left( {y, z,\mathrm{o}}\right) \land \forall w\left( {w < y \rightarrow \neg {\varphi }_{g}\left( {w, z,\mathrm{o}}\right) }... | Proof. First we show that if \( f\left( n\right) = m \), then \( \mathbf{Q} \vdash {\varphi }_{f}\left( {\bar{n},\bar{m}}\right) \), i.e.,\n\n\[ \n\mathbf{Q} \vdash {\varphi }_{g}\left( {\bar{m},\bar{n},\mathrm{o}}\right) \land \forall w\left( {w < \bar{m} \rightarrow \neg {\varphi }_{g}\left( {w,\bar{n},\mathrm{o}}\ri... | No |
Theorem 33.28. A relation is representable in \( \mathbf{Q} \) if and only if it is computable. | Proof. For the forwards direction, suppose \( R\left( {{x}_{0},\ldots ,{x}_{k}}\right) \) is represented by the formula \( {\varphi }_{R}\left( {{x}_{0},\ldots ,{x}_{k}}\right) \) . Here is an algorithm for computing \( R \) : on input \( {n}_{0} \) , \( \ldots ,{n}_{k} \), simultaneously search for a proof of \( {\var... | No |
Theorem 33.29. Q is undecidable, i.e., the relation\n\n\[ \n{\operatorname{Prov}}_{\mathbf{Q}}\left( y\right) \Leftrightarrow \operatorname{Sent}\left( y\right) \land \exists x{\operatorname{Prf}}_{\mathbf{Q}}\left( {x, y}\right) \n\]\n\nis not recursive. | Proof. Suppose it were. Then we could solve the halting problem as follows: Given \( e \) and \( n \), we know that \( {\varphi }_{e}\left( n\right) \downarrow \) iff there is an \( s \) such that \( T\left( {e, n, s}\right) \), where \( T \) is Kleene’s predicate from Theorem 27.28. Since \( T \) is primitive recursiv... | Yes |
Corollary 33.30. First-order logic is undecidable. | Proof. If first-order logic were decidable, provability in \( \mathbf{Q} \) would be as well, since \( \mathbf{Q} \vdash \varphi \) iff \( \vdash \omega \rightarrow \varphi \), where \( \omega \) is the conjunction of the axioms of \( \mathbf{Q} \) . | Yes |
Theorem 34.1. Q is c.e. but not decidable. In fact, it is a complete c.e. set. | Proof. It is not hard to see that \( \mathbf{Q} \) is c.e., since it is the set of (codes for) sentences \( y \) such that there is a proof \( x \) of \( y \) in \( \mathbf{Q} \) :\n\n\[ Q = \left\{ {y : \exists x{\operatorname{Prf}}_{\mathbf{Q}}\left( {x, y}\right) }\right\} \]\n\nBut we know that \( {\operatorname{Pr... | Yes |
Theorem 34.3. Let \( \mathbf{T} \) be any \( \omega \) -consistent theory that includes \( \mathbf{Q} \) . Then \( \mathbf{T} \) is not decidable. | Proof. If \( \mathbf{T} \) includes \( \mathbf{Q} \), then \( \mathbf{T} \) represents the computable functions and relations. We need only modify the previous proof. As above, if \( x \in K \), then \( \mathbf{T} \) proves \( \exists s{\varphi }_{T}\left( {\bar{x},\bar{x}, s}\right) \) . Conversely, suppose \( \mathbf... | Yes |
Theorem 34.5. Let \( \mathbf{T} \) be any consistent theory that includes \( \mathbf{Q} \) . Then \( \mathbf{T} \) is not decidable. | Proof. Suppose \( \mathbf{T} \) is a consistent, decidable extension of \( \mathbf{Q} \) . We will obtain a contradiction by using \( \mathbf{T} \) to define a universal computable relation.\n\nLet \( R\left( {x, y}\right) \) hold if and only if\n\n\( x \) codes a formula \( \theta \left( u\right) \), and \( \mathbf{T}... | Yes |
Lemma 34.7. Suppose \( \mathbf{T} \) is axiomatizable. Then \( \mathbf{T} \) is computably enumerable. | Proof. Suppose \( A \) is a computable set of axioms for \( \mathbf{T} \) . To determine if \( \varphi \in T \) , just search for a proof of \( \varphi \) from the axioms.\n\nPut slightly differently, \( \varphi \) is in \( \mathbf{T} \) if and only if there is a finite list of axioms \( {\psi }_{1},\ldots ,{\psi }_{k}... | No |
Lemma 34.8. Suppose a theory \( \mathbf{T} \) is complete and axiomatizable. Then \( \mathbf{T} \) is decidable. | Proof. Suppose \( \mathbf{T} \) is complete and \( A \) is a computable set of axioms. If \( \mathbf{T} \) is inconsistent, it is clearly computable. (Algorithm: \ | No |
Theorem 34.9. There is no complete, consistent, axiomatizable extension of \( \mathbf{Q} \) . | Proof. We already know that there is no consistent, decidable extension of \( \mathbf{Q} \) . But if \( \mathbf{T} \) is complete and axiomatized, then it is decidable. | Yes |
Lemma 34.10. Q and \( \overline{\mathbf{Q}} \) are computably inseparable. | Proof. Suppose \( C \) is a computable set such that \( \mathbf{Q} \subseteq C \) and \( \overline{\mathbf{Q}} \subseteq \bar{C} \) . Let \( R\left( {x, y}\right) \) be the relation\n\n\( x \) codes a formula \( \theta \left( u\right) \) and \( \theta \left( \bar{y}\right) \) is in \( C \) .\n\nWe will show that \( R\l... | Yes |
Theorem 34.11. Let \( \mathbf{T} \) be any theory in the language of arithmetic that is consistent with \( \mathbf{Q} \) (i.e., \( \mathbf{T} \cup \mathbf{Q} \) is consistent). Then \( \mathbf{T} \) is undecidable. | Proof. Remember that \( \mathbf{Q} \) has a finite set of axioms, \( {Q}_{1},\ldots ,{Q}_{8} \) . We can even replace these by a single axiom, \( \alpha = {Q}_{1} \land \cdots \land {Q}_{8} \) . Suppose \( \mathbf{T} \) is a decidable theory consistent with \( \mathbf{Q} \) . Let \[ C = \{ \varphi : \mathbf{T} \vdash \... | Yes |
Corollary 34.12. First-order logic for the language of arithmetic (that is, the set \( \{ \varphi \) : \( \varphi \) is provable in first-order logic \( \} \) ) is undecidable. | Proof. First-order logic is the set of consequences of \( \varnothing \), which is consistent with \( \mathbf{Q} \) . | No |
Theorem 34.13. Suppose \( \mathbf{T} \) is a theory in a language in which one can interpret the language of arithmetic, in such a way that \( \mathbf{T} \) is consistent with the interpretation of \( \mathbf{Q} \). Then \( \mathbf{T} \) is undecidable. If \( \mathbf{T} \) proves the interpretation of the axioms of \( ... | The proof is just a small modification of the proof of the last theorem; one could use a counterexample to get a separation of \( \mathbf{Q} \) and \( \overline{\mathbf{Q}} \). One can take \( \mathbf{{ZFC}} \), Zermelo-Fraenkel set theory with the axiom of choice, to be an axiomatic foundation that is powerful enough ... | No |
First-order logic for any language with a binary relation symbol is undecidable. | This result extends to any language with two unary function symbols, since one can use these to simulate a binary relation symbol. The results just cited are tight: it turns out that first-order logic for a language with only unary relation symbols and at most one unary function symbol is decidable. | No |
Lemma 35.1. Let \( \mathbf{T} \) be any theory extending \( \mathbf{Q} \), and let \( \psi \left( x\right) \) be any formula with only the variable \( x \) free. Then there is a sentence \( \varphi \) such that \( \mathbf{T} \vdash \varphi \leftrightarrow \psi \left( {\ulcorner \varphi \urcorner }\right) \) . | The lemma asserts that given any property \( \psi \left( x\right) \), there is a sentence \( \varphi \) that asserts \ | No |
Lemma 35.2. Let \( \psi \left( x\right) \) be any formula with one free variable \( x \) . Then there is a sentence \( \varphi \) such that \( \mathbf{Q} \vdash \varphi \leftrightarrow \psi \left( {\ulcorner \varphi \urcorner }\right) \) . | Proof. Given \( \psi \left( x\right) \), let \( \alpha \left( x\right) \) be the formula \( \exists y\left( {{\theta }_{\text{diag }}\left( {x, y}\right) \land \psi \left( y\right) }\right) \) and let \( \varphi \) be its diagonalization, i.e., the formula \( \alpha \left( {\ulcorner \alpha \left( x\right) \urcorner }\... | Yes |
Lemma 35.3. If \( \mathbf{T} \) is a consistent, axiomatizable theory extending \( \mathbf{Q} \), then \( \mathbf{T} \nvdash {\gamma }_{\mathbf{T}} \) . | Proof. Suppose \( \mathbf{T} \) derives \( {\gamma }_{\mathbf{T}} \) . Then there is a derivation, and so, for some number \( m \), the relation \( {\operatorname{Prf}}_{T}\left( {m,{}^{\# }{\gamma }_{\mathrm{T}}{}^{\# }}\right) \) holds. But then \( \mathbf{Q} \) derives the sentence \( \mathop{\operatorname{Prf}}\lim... | Yes |
Lemma 35.5. If \( \mathbf{T} \) is an \( \omega \) -consistent, axiomatizable theory extending \( \mathbf{Q} \), then \( \mathbf{T} \nvdash \) \( \neg {\gamma }_{\mathrm{T}} \) . | Proof. We show that if \( \mathbf{T} \) derives \( \neg {\gamma }_{\mathbf{T}} \), then it is \( \omega \) -inconsistent. Suppose \( \mathbf{T} \) derives \( \neg {\gamma }_{\mathrm{T}} \) . If \( \mathbf{T} \) is inconsistent, it is \( \omega \) -inconsistent, and we are done. Otherwise, \( \mathbf{T} \) is consistent... | Yes |
Theorem 35.6. Let \( \mathbf{T} \) be any \( \omega \) -consistent, axiomatizable theory extending \( \mathbf{Q} \) . Then \( \mathbf{T} \) is not complete. | Proof. If \( \mathbf{T} \) is \( \omega \) -consistent, it is consistent, so \( \mathbf{T} \nvdash {\gamma }_{\mathbf{T}} \) by Lemma 35.3. By Lemma 35.5, \( \mathbf{T} \nvdash \neg {\gamma }_{\mathbf{T}} \) . This means that \( \mathbf{T} \) is incomplete, since it derives neither \( {\gamma }_{\mathrm{T}} \) nor \( \... | Yes |
Theorem 35.7. Let \( \mathbf{T} \) be any consistent, axiomatizable theory extending \( \mathbf{Q} \) . Then \( \mathbf{T} \) is not complete. | Proof. Recall that \( {\operatorname{Prov}}_{T}\left( y\right) \) is defined as \( \exists x{\operatorname{Prf}}_{T}\left( {x, y}\right) \), where \( {\operatorname{Prf}}_{T}\left( {x, y}\right) \) represents the decidable relation which holds iff \( x \) is the Gödel number of a derivation of the sentence with Gödel n... | Yes |
Theorem 35.8. Assuming PA is consistent, then PA does not derive \( {\operatorname{Con}}_{\mathrm{{PA}}} \) . | It is informative to read Gödel's sketch of an argument, since the theorem follows like a good punch line. It goes like this. Let \( {\gamma }_{\mathrm{{PA}}} \) be the Gödel sentence that we constructed in the proof of Theorem 35.6. We have shown \ | No |
Theorem 35.10. Let \( \mathbf{T} \) be an axiomatizable theory extending \( \mathbf{Q} \), and suppose \( {\operatorname{Prov}}_{T}\left( y\right) \) is a formula satisfying conditions P1-P3 from section 35.7. If \( \mathbf{T} \) derives \( {\operatorname{Prov}}_{T}\left( {\ulcorner \varphi \urcorner }\right) \rightarr... | Proof of Theorem 35.10. Suppose \( \varphi \) is a sentence such that \( \mathbf{T} \) derives \( {\operatorname{Prov}}_{T}\left( {\ulcorner \varphi \urcorner }\right) \rightarrow \) \( \varphi \) . Let \( \psi \left( y\right) \) be the formula \( {\operatorname{Prov}}_{T}\left( y\right) \rightarrow \varphi \), and use... | Yes |
Lemma 35.12. Every computable relation is definable in \( \mathfrak{N} \) . | Proof. It is easy to check that the formula representing a relation in \( \mathbf{Q} \) defines the same relation in \( \mathfrak{N} \) . | No |
Lemma 35.13. The halting relation is definable in \( \mathfrak{N} \) . | Proof. Let \( H \) be the halting relation, i.e.,\n\n\[ H = \{ \langle e, x\rangle : \exists {sT}\left( {e, x, s}\right) \} . \]\n\nLet \( {\theta }_{T} \) define \( T \) in \( \mathfrak{N} \) . Then\n\n\[ H = \left\{ {\langle e, x\rangle : \mathfrak{N} \vDash \exists s{\theta }_{T}\left( {\bar{e},\bar{x}, s}\right) }\... | Yes |
The set of true sentences of arithmetic is not definable in arithmetic. | Proof. Suppose \( \theta \left( x\right) \) defined it, i.e., \( \mathfrak{N} \vDash \varphi \) iff \( \mathfrak{N} \vDash \theta \left( {\ulcorner \varphi \urcorner }\right) \) . By the fixed-point lemma, there is a formula \( \varphi \) such that \( \mathbf{Q} \vdash \varphi \leftrightarrow \neg \theta \left( {\ulcor... | Yes |
Consider the formula \( \forall z\left( {X\left( z\right) \leftrightarrow \neg Y\left( z\right) }\right) \). It contains no second-order quantifiers, but does contain the second-order variables \( X \) and \( Y \) (here understood to be one-place). | \( \mathfrak{M}, s \vDash \forall z\left( {X\left( z\right) \leftrightarrow \neg Y\left( z\right) }\right) \) whenever the elements of \( s\left( X\right) \) are not elements of \( s\left( Y\right) \), and vice versa, i.e., iff \( s\left( Y\right) = \left| \mathfrak{M}\right| \smallsetminus s\left( X\right) \). So for ... | Yes |
Example 36.8. \( \mathfrak{M}, s \vDash \exists Y\left( {\exists {yY}\left( y\right) \land \forall z\left( {X\left( z\right) \leftrightarrow \neg Y\left( z\right) }\right) }\right) \) if there is an \( {s}^{\prime }{ \sim }_{Y} \) \( s \) such that \( \mathfrak{M},{s}^{\prime } \vDash \left( {\exists {yY}\left( y\right... | And that is the case iff \( {s}^{\prime }\left( Y\right) \neq \varnothing \) (so that \( \mathfrak{M},{s}^{\prime } \vDash \exists {yY}\left( y\right) \) ) and, as in the previous example, \( {s}^{\prime }\left( Y\right) = \) \( \left| \mathfrak{M}\right| \backslash {s}^{\prime }\left( X\right) .\; \) In other words \(... | Yes |
In first-order logic we can define the identity relation \( {\operatorname{Id}}_{\left| \mathfrak{M}\right| } \) (i.e., \( \{ \langle a, a\rangle : a \in \left| \mathfrak{M}\right| \} \) ) by the formula \( x = y \) . In second-order logic, we can define this relation without \( = \) . | For if \( a \) and \( b \) are the same element of \( \left| \mathfrak{M}\right| \), then they are elements of the same subsets of \( \left| \mathfrak{M}\right| \) (since sets are determined by their elements). Conversely, if \( a \) and \( b \) are different, then they are not elements of the same subsets: e.g., \( a ... | Yes |
If \( R \) is a two-place predicate symbol, \( {R}^{\mathfrak{M}} \) is a two-place relation on \( \left| \mathfrak{M}\right| \) . Perhaps somewhat confusingly, we’ll use \( R \) as the predicate symbol for \( R \) and for the relation \( {R}^{\mathfrak{M}} \) itself. The transitive closure \( {R}^{ * } \) of \( R \) i... | \[ {\psi }_{R}\left( X\right) \equiv \forall x\forall y\left( {R\left( {x, y}\right) \rightarrow X\left( {x, y}\right) }\right) \land \forall x\forall y\forall z\left( {\left( {X\left( {x, y}\right) \land X\left( {y, z}\right) }\right) \rightarrow X\left( {x, z}\right) }\right) . \] The first conjunct says that \( R \s... | Yes |
Proposition 36.15. \( \mathfrak{M} \vDash \operatorname{Inf} \) iff \( \left| \mathfrak{M}\right| \) is infinite. | Proof. \( \mathfrak{M} \vDash \) Inf iff \( \mathfrak{M}, s \vDash \forall x\forall y\left( {u\left( x\right) = u\left( y\right) \rightarrow x = y}\right) \land \exists y\forall {xy} \neq u\left( x\right) \) for some \( s \) . If it does, \( s\left( u\right) \) is an injective function, and some \( y \in \left| \mathfr... | Yes |
Proposition 36.16. \( \mathfrak{M} \vDash \) Count iff \( \left| \mathfrak{M}\right| \) is enumerable. | Proof. Suppose \( \left| \mathfrak{M}\right| \) is enumerable, and let \( {m}_{0},{m}_{1},\ldots \), be an enumeration. By removing repetions we can guarantee that no \( {m}_{k} \) appears twice. Define \( f\left( {m}_{k}\right) = {m}_{k + 1} \) and let \( s\left( z\right) = {m}_{0} \) and \( s\left( u\right) = f \) . ... | Yes |
Theorem 37.1. If \( \mathfrak{M} \vDash {\mathbf{{PA}}}^{2} \) then \( \left| \mathfrak{M}\right| = \left\{ {{\operatorname{Val}}^{\mathfrak{M}}\left( \bar{n}\right) : n \in \mathbb{N}}\right\} \) . | Proof. Let \( N = \left\{ {{\operatorname{Val}}^{\mathfrak{M}}\left( \bar{n}\right) : n \in \mathbb{N}}\right\} \), and suppose \( \mathfrak{M} \vDash {\mathbf{{PA}}}^{2} \) . Of course, for any \( n \in \mathbb{N},{\operatorname{Val}}^{\mathfrak{M}}\left( \bar{n}\right) \in \left| \mathfrak{M}\right| \), so \( N \subs... | Yes |
Corollary 37.2. Any two models of \( {\mathbf{{PA}}}^{2} \) are isomorphic. | Proof. By Theorem 37.1, the domain of any model of \( {\mathbf{{PA}}}^{2} \) is exhausted by \( {\operatorname{Val}}^{\mathfrak{M}}\left( \bar{n}\right) \). Any such model is also a model of \( \mathbf{Q} \). By Proposition 24.3, any such model is standard, i.e., isomorphic to \( \mathfrak{N} \). | Yes |
Proposition 37.3. Let \( {\mathbf{{PA}}}^{2 + } \) be the second-order theory containing the first two arithmetical axioms (the successor axioms) and the second-order induction axiom. Then \( \leq , + \), and \( \times \) are definable in \( {\mathbf{{PA}}}^{2 + } \) . | Proof. To show that \( \leq \) is definable, we have to find a formula \( {\varphi }_{ \leq }\left( {x, y}\right) \) such that \( \mathfrak{N} \vDash {\varphi }_{ \leq }\left( {\bar{n},\bar{m}}\right) \) iff \( n \leq m \) . Consider the formula\n\n\[ \psi \left( {x, Y}\right) \equiv Y\left( x\right) \land \forall y\le... | Yes |
Corollary 37.4. \( \mathfrak{M} \vDash {\mathbf{{PA}}}^{2} \) iff \( \mathfrak{M} \vDash {\mathbf{{PA}}}^{2 + } \) . | Proof. Immediate from Proposition 37.3. | No |
Theorem 37.5. Second-order logic is undecidable. | Proof. A first-order sentence is valid in first-order logic iff it is valid in second-order logic, and first-order logic is undecidable. | Yes |
Theorem 37.6. There is no sound and complete proof system for second-order logic. | Proof. Let \( \varphi \) be a sentence in the language of arithmetic. \( \mathfrak{N} \vDash \varphi \) iff \( {\mathbf{{PA}}}^{2} \vDash \varphi \) . Let \( P \) be the conjunction of the nine axioms of \( {\mathbf{{PA}}}^{2}.{\mathbf{{PA}}}^{2} \vDash \varphi \) iff \( \vDash P \rightarrow \varphi \), i.e., \( \mathf... | Yes |
Theorem 37.7. Second-order logic is not compact. | Proof. Recall that\n\n\[ \text{Inf} \equiv \exists u\left( {\forall x\forall y\left( {u\left( x\right) = u\left( y\right) \rightarrow x = y}\right) \land \exists y\forall {xy} \neq u\left( x\right) }\right) \]\n\nis satisfied in a structure iff its domain is infinite. Let \( {\varphi }^{ \geq n} \) be a sentence that a... | Yes |
Theorem 37.8. The Löwenheim-Skolem Theorem fails for second-order logic: There are sentences with infinite models but no enumerable models. | Proof. Recall that\n\n\[ \text{Count} \equiv \exists z\exists u\forall X\left( {\left( {X\left( z\right) \land \forall x\left( {X\left( x\right) \rightarrow X\left( {u\left( x\right) }\right) }\right) }\right) \rightarrow \forall {xX}\left( x\right) }\right) \]\n\nis true in a structure \( \mathfrak{M} \) iff \( \left|... | Yes |
Theorem 37.9. There are sentences with denumerable but no non-enumerable models. | Proof. Count \( \land \) Inf is true in \( \mathbb{N} \) but not in any structure \( \mathfrak{M} \) with \( \left| \mathfrak{M}\right| \) non-enumerable. | No |
Proposition 38.6. The sentence \( \forall X\forall Y\left( {\left( {X \preccurlyeq Y \land Y \preccurlyeq X}\right) \rightarrow X \approx Y}\right) \) is valid. | Proof. The sentence is satisfied in a structure \( \mathfrak{M} \) if, for any subsets \( X \subseteq \left| \mathfrak{M}\right| \) and \( Y \subseteq \left| \mathfrak{M}\right| \), if \( X \preccurlyeq Y \) and \( Y \preccurlyeq X \) then \( X \approx Y \) . But this holds for any sets \( X \) and \( Y \) -it is the S... | Yes |
Proposition 38.11. The sentence\n\n\[ \forall X\forall Y\forall R(\operatorname{Pow}\left( {Y, R, X}\right) \rightarrow \neg \exists u(\forall x\forall y\left( {u\left( x\right) = u\left( y\right) \rightarrow x = y}\right) \land \left. {\forall x\left( {Y\left( x\right) \rightarrow X\left( {u\left( x\right) }\right) }\... | The power set of a denumerable set is non-enumerable, and so its cardinality is larger than that of any denumerable set (which is \( {\aleph }_{0} \) ). The size of \( \wp \left( \mathbb{N}\right) \) is called the \ | No |
Proposition 38.12. If \( \mathbb{R} \preccurlyeq \left| \mathfrak{M}\right| \), then the formula\n\n\[ \n\operatorname{Cont}\left( Y\right) \equiv \exists X\exists R\left( {\left( {{\operatorname{Aleph}}_{0}\left( X\right) \land \operatorname{Pow}\left( {Y, R, X}\right) }\right) \land }\right.\n\]\n\n\[ \n\forall x\for... | Proof. Pow \( \left( {Y, R, X}\right) \) expresses that \( s\left( Y\right) s\left( R\right) \) -codes the power set of \( s\left( X\right) \), which \( {\operatorname{Aleph}}_{0}\left( X\right) \) says is countable. So \( s\left( Y\right) \) is at least as large as the power of the continuum, although it may be larger... | Yes |
Corollary 39.2. Suppose \( M \) can be reduced to normal form. Then this normal form is unique. | Proof. If \( M \rightarrow {N}_{1} \) and \( M \rightarrow {N}_{2} \), by the previous theorem there is a term \( P \) such that \( {N}_{1} \) and \( {N}_{2} \) both reduce to \( P \) . If \( {N}_{1} \) and \( {N}_{2} \) are both in normal form, this can only happen if \( {N}_{1} \equiv P \equiv {N}_{2} \) . | Yes |
Theorem 39.6. If a partial function \( f \) is \( \lambda \) -defined by a lambda term, it is computable. | Proof. Suppose a function \( f \) is \( \lambda \) -defined by a lambda term \( X \) . Let us describe an informal procedure to compute \( f \) . On input \( {m}_{0},\ldots ,{m}_{n - 1} \), write down the term \( X{\bar{m}}_{0}\ldots {\bar{m}}_{n - 1} \) . Build a tree, first writing down all the one-step reductions of... | No |
Theorem 39.7. Every computable partial function is \( \lambda \) -definable. | Proof. Wwe need to show that every partial computable function \( f \) is \( \lambda \) -defined by a lambda term \( F \) . By Kleene’s normal form theorem, it suffices to show that every primitive recursive function is \( \lambda \) -defined by a lambda term, and then that the functions \( \lambda \) -definable are cl... | No |
Lemma 39.8. The functions zero, succ, and \( {P}_{i}^{n} \) are \( \lambda \) -definable. | Proof. zero is just \( {\lambda x}.{\lambda y}.y \) .\n\nThe successor function succ, is defined by \( \operatorname{Succ}\left( u\right) = {\lambda x}.{\lambda y}.x\left( {uxy}\right) \) . You should think about why this works; for each numeral \( \bar{n} \), thought of as an iterator, and each function \( f,\operator... | No |
Lemma 39.9. The \( \lambda \) -definable functions are closed under composition. | Proof. Suppose \( f \) is defined by composition from \( h,{g}_{0},\ldots ,{g}_{k - 1} \) . Assuming \( h \) , \( {g}_{0},\ldots ,{g}_{k - 1} \) are \( \lambda \) -defined by \( H,{G}_{0},\ldots ,{G}_{k - 1} \), respectively, we need to find a term \( F \) that \( \lambda \) -defines \( f \) . But we can simply define ... | Yes |
Lemma 39.10. There is a lambda term \( D \) such that for each pair of lambda terms \( M \) and \( N, D\left( {M, N}\right) \left( \overline{0}\right) \rightarrow M \) and \( D\left( {M, N}\right) \left( \overline{1}\right) \rightarrow N \) . | Proof. First, define the lambda term \( K \) by\n\n\[ K\left( y\right) = {\lambda x}.y. \]\n\nIn other words, \( K \) is the term \( {\lambda y}.{\lambda x}.y \) . Looking at it differently, for every \( M \) , \( K\left( M\right) \) is a constant function that returns \( M \) on any input.\n\nNow define \( D\left( {x,... | Yes |
Lemma 39.12. Suppose \( f\left( {x, y}\right) \) is primitive recursive. Let \( g \) be defined by\n\n\[ g\left( x\right) \simeq {\mu yf}\left( {x, y}\right) . \]\n\nThen \( g \) is \( \lambda \) -definable. | 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 |
Lemma 40.2. A term starts with either a variable or a parenthesis. | Proof. Something counts as a term only if it is constructed according to Definition 40.1. If it is the result of (1), it must be a variable. If it is the result of (2) or (3), it starts with a parenthesis. | Yes |
Lemma 40.3. The result of an application starts with either two parentheses or a parenthesis and a variable. | Proof. If \( M \) is the result of an application, it is of the form \( \left( {PQ}\right) \), so it begins with a parenthesis. Since \( P \) is a term, by Lemma 40.2, it begins either with a parenthesis or a variable. | Yes |
Proposition 40.5 (Unique Readability). There is a unique formation for each term. In other words, if a term \( M \) is formed by a formation, then it is the only formation that can form this term. | Proof. We prove this by induction on the formation of terms.\n\n1. \( M \) is of the form \( x \), where \( x \) is some variable. Since the results of abstractions and applications always start with parentheses, they cannot have been used to construct \( M \) ; Thus, the formation of \( M \) must be a single step of D... | Yes |
Lemma 40.12. 1. If \( y \neq x \), then \( y \in \mathrm{{FV}}\left( {{\lambda x}.N}\right) \) iff \( y \in \mathrm{{FV}}\left( N\right) \) . | Proof. Exercise. | No |
Theorem 40.14. If \( x \notin \mathrm{{FV}}\left( M\right) \), then \( \mathrm{{FV}}\left( {M\left\lbrack {N/x}\right\rbrack }\right) = \mathrm{{FV}}\left( M\right) \), if the left-hand side is defined. | Proof. By induction on the formation of \( M \) .\n\n1. \( M \) is a variable: exercise.\n\n2. \( M \) is of the form \( \left( {PQ}\right) \) : exercise.\n\n3. \( M \) is of the form \( {\lambda y}.P \), and since \( {\lambda y}.P\left\lbrack {N/x}\right\rbrack \) is defined, it has to be \( {\lambda y}.P\left\lbrack ... | No |
Theorem 40.15. If \( x \in \mathrm{{FV}}\left( M\right) \), then \( \mathrm{{FV}}\left( {M\left\lbrack {N/x}\right\rbrack }\right) = \left( {\mathrm{{FV}}\left( M\right) \smallsetminus \{ x\} }\right) \cup \mathrm{{FV}}\left( N\right) \), provided the left hand is defined. | Proof. By induction on the formation of \( M \).\n\n1. \( M \) is a variable: exercise.\n\n2. \( M \) is of the form \( {PQ} \): Since \( \left( {PQ}\right) \left\lbrack {N/y}\right\rbrack \) is defined, it has to be \( \left( {P\left\lbrack {N/x}\right\rbrack }\right) \left( {Q\left\lbrack {N/x}\right\rbrack }\right) ... | No |
Theorem 40.16. \( x \notin \operatorname{FV}\left( {M\left\lbrack {N/x}\right\rbrack }\right) \), if the right-hand side is defined and \( x \notin \) \( \mathrm{{FV}}\left( N\right) \) . | Proof. Exercise. | No |
Theorem 40.17. If \( M\left\lbrack {y/x}\right\rbrack \) is defined and \( y \notin \mathrm{{FV}}\left( M\right) \), then \( M\left\lbrack {y/x}\right\rbrack \left\lbrack {x/y}\right\rbrack = M \) . | Proof. By induction on the formation of \( M \) .\n\n1. \( M \) is a variable \( z \) : Exercise.\n\n2. \( M \) is of the form \( \left( {PQ}\right) \) . Then:\n\n\[ \left( {PQ}\right) \left\lbrack {y/x}\right\rbrack \left\lbrack {x/y}\right\rbrack = \left( {\left( {P\left\lbrack {y/x}\right\rbrack }\right) \left( {Q\l... | No |
Lemma 40.25. If \( P\overset{\alpha }{ \rightarrow }Q \) then \( \mathrm{{FV}}\left( P\right) = \mathrm{{FV}}\left( Q\right) \) . | Proof. By induction on the derivation of \( P\overset{\alpha }{ \rightarrow }Q \) .\n\n1. If the last rule is (4), then \( P \) is of the form \( {\lambda x}.N \) and \( Q \) of the form \( {\lambda y}.N\left\lbrack {y/x}\right\rbrack \), with \( x \neq y, y \notin \mathrm{{FV}}\left( N\right) \) and \( N\left\lbrack {... | No |
Lemma 40.26. If \( P\overset{\alpha }{ \rightarrow }Q \) then \( Q\overset{\alpha }{ \rightarrow }P \) . | Proof. Induction on the derivation of \( P\overset{\alpha }{ \rightarrow }Q \) .\n\n1. If the last rule is (4), then \( P \) is of the form \( {\lambda x}.N \) and \( Q \) of the form\n\n\( {\lambda y}.N\left\lbrack {y/x}\right\rbrack \), where \( x \neq y, y \notin \mathrm{{FV}}\left( N\right) \) and \( N\left\lbrack ... | Yes |
Theorem 40.27. \( \\alpha \) -Conversion is an equivalence relation on terms, i.e., it is reflexive, symmetric, and transitive. | Proof. 1. For each term \( M, M \) can be changed to \( M \) by zero changes of bound variables.\n\n2. If \( P \) is \( \\alpha \) -converts to \( Q \) by a series of changes of bound variables, then from \( Q \) we can just inverse these changes (by Lemma 40.26) in opposite order to obtain \( P \).\n\n3. If \( {P\\alp... | Yes |
Theorem 40.28. If \( M\overset{\alpha }{ = }N \), then \( \mathrm{{FV}}\left( M\right) = \mathrm{{FV}}\left( N\right) \) . | Proof. Immediate from Lemma 40.25. | No |
Lemma 40.29. If \( R\overset{\alpha }{ = }{R}^{\prime } \) and \( M\left\lbrack {R/y}\right\rbrack \) is defined, then \( M\left\lbrack {{R}^{\prime }/y}\right\rbrack \) is defined and \( \alpha \) -equivalent to \( M\left\lbrack {R/y}\right\rbrack \) . | Proof. Exercise. | No |
Theorem 40.30. For any \( M, R \), and \( y \), there exists \( {M}^{\prime } \) such that \( M\overset{\alpha }{ = }{M}^{\prime } \) and \( {M}^{\prime }\left\lbrack {R/y}\right\rbrack \) is defined. Moreover, if there is another pair \( {M}^{\prime \prime }\overset{\alpha }{ = }M \) and \( {R}^{\prime \prime } \) whe... | Proof. By induction on the formation of \( M \) :\n\n1. \( M \) is a variable \( z \) : Exercise.\n\n2. Suppose \( M \) is of the form \( {\lambda x}.N \) . Select a variable \( z \) other than \( x \) and \( y \) and such that \( z \notin \mathrm{{FV}}\left( N\right) \) and \( z \notin \mathrm{{FV}}\left( R\right) \) ... | No |
For any \( M, R \), and \( y \), there exists a pair of \( {M}^{\prime } \) and \( {R}^{\prime } \) such that \( {M}^{\prime } \triangleq M, R \triangleq {R}^{\prime } \) and \( {M}^{\prime }\left\lbrack {{R}^{\prime }/y}\right\rbrack \) is defined. Moreover, if there is another pair \( {M}^{\prime \prime } \triangleq ... | Immediate from Theorem 40.30. | No |
Theorem 40.47. \( M\overset{\text{ ext }}{ = }N \) if and only if \( M\overset{\eta }{ = }N \) . | Proof. First we prove that \( \frac{\eta }{ - } \) is closed under the extensionality rule. That is, ext rule doesn’t add anything to \( \frac{\eta }{ - } \) . We then have \( \frac{\eta }{ - } \) contains \( \frac{ext}{ - } \), and if \( M\overset{ext}{ = }N \) , then \( M\overset{\eta }{ = }N \) .\n\nTo prove \( \fra... | Yes |
Problem 40.6. What is the result of the following substitutions? | 1. \( {\lambda y}.x\left( {{\lambda w}.{vwx}}\right) \left\lbrack {\left( {uv}\right) /x}\right\rbrack \)\n2. \( {\lambda y}.x\left( {{\lambda x}.x}\right) \left\lbrack {\left( {{\lambda y}.{xy}}\right) /x}\right\rbrack \)\n3. \( y\left( {{\lambda v}.{xv}}\right) \left\lbrack {\left( {{\lambda y}.{vy}}\right) /x}\right... | No |
Problem 40.11. Are the following pairs of terms \( \alpha \) -convertible? | 1. \( {\lambda x}.{\lambda y}.x \) and \( {\lambda y}.{\lambda x}.y \)\n2. \( {\lambda x}.{\lambda y}.x \) and \( {\lambda c}.{\lambda b}.a \)\n3. \( {\lambda x}.{\lambda y}.x \) and \( {\lambda c}.{\lambda b}.a \) | No |
Theorem 41.2. If a relation \( \overset{X}{ \rightarrow } \) satisfies the Church-Rosser property, and \( \overset{X}{ \rightarrow } \) is the smallest transitive relation containing \( \overset{X}{ \rightarrow } \), then \( \overset{X}{ \rightarrow } \) satisfies the Church-Rosser property too. | Proof. Suppose\n\n\[ \nM\overset{X}{ \rightarrow }{P}_{1}\overset{X}{ \rightarrow }\ldots \overset{X}{ \rightarrow }{P}_{m}\text{and} \]\n\n\[ \nM\overset{X}{ \rightarrow }{Q}_{1}\overset{X}{ \rightarrow }\ldots \overset{X}{ \rightarrow }{Q}_{n} \]\n\nWe will prove the theorem by constructing a grid \( N \) of terms of... | Yes |
Theorem 41.4. \( M\overset{\beta }{ \Rightarrow }M \) . | Proof. Exercise. | No |
Lemma 41.6. If \( M\overset{\beta }{ \Rightarrow }{M}^{\prime } \) and \( R\overset{\beta }{ \Rightarrow }{R}^{\prime } \), then \( M\left\lbrack {R/y}\right\rbrack \overset{\beta }{ \Rightarrow }{M}^{\prime }\left\lbrack {{R}^{\prime }/y}\right\rbrack \) . | Proof. By induction on the derivation of \( M\overset{\beta }{ \Rightarrow }{M}^{\prime } \) .\n\n1. The last step is (1): Exercise.\n\n2. The last step is (2): Then \( M \) is \( {\lambda x}.N \) and \( {M}^{\prime } \) is \( {\lambda x}.{N}^{\prime } \), where \( N\overset{\beta }{ \Rightarrow }{N}^{\prime } \) . We ... | No |
Lemma 41.7. If \( M\overset{\beta }{ \Rightarrow }{M}^{\prime } \) then \( {M}^{\prime }\overset{\beta }{ \Rightarrow }{M}^{*\beta } \) . | Proof. By induction on the derivation of \( M\overset{\beta }{ \Rightarrow }{M}^{\prime } \) .\n\n1. The last rule is (1): Exercise.\n\n2. The last rule is (2): \( M \) is \( {\lambda x}.N \) and \( {M}^{\prime } \) is \( {\lambda x}.{N}^{\prime } \) with \( N\overset{\beta }{ \Rightarrow }{N}^{\prime } \) . We want to... | No |
Theorem 41.8. \( \overset{\beta }{ \Rightarrow } \) has the Church-Rosser property. | Proof. Immediate from Lemma 41.7. | No |
Lemma 41.9. If \( M\overset{\beta }{ \rightarrow }{M}^{\prime } \), then \( M\overset{\beta }{ \Rightarrow }{M}^{\prime } \) . | Proof. If \( M\overset{\beta }{ \rightarrow }{M}^{\prime } \), then \( M \) is \( \left( {{\lambda x}.N}\right) Q,{M}^{\prime } \) is \( N\left\lbrack {Q/x}\right\rbrack \), for some \( x, N \), and \( Q \) . Since \( N\overset{\beta }{ \Rightarrow }N \) and \( Q\overset{\beta }{ \Rightarrow }Q \) by Theorem 41.4, we i... | Yes |
Lemma 41.10. If \( M\overset{\beta }{ \Rightarrow }{M}^{\prime } \), then \( M\xrightarrow[]{\beta }{M}^{\prime } \) . | Proof. By induction on the derivation of \( M\overset{\beta }{ \Rightarrow }{M}^{\prime } \) .\n\n1. The last rule is (1): Then \( M \) and \( {M}^{\prime } \) are just \( x \), and \( x\overset{\beta }{ \rightarrow }x \) .\n\n2. The last rule is (2): \( M \) is \( {\lambda x}.N \) and \( {M}^{\prime } \) is \( {\lambd... | Yes |
Lemma 41.11. \( \overset{\beta }{ \rightarrow } \) is the smallest transitive relation containing \( \overset{\beta }{ \rightarrow } \) . | Proof. Let \( \overset{X}{ \rightarrow } \) be the smallest transitive relation containing \( \overset{\beta }{ \Rightarrow } \) . \n\n\( \overset{\beta }{ \rightarrow } \subseteq \overset{X}{ \rightarrow } \) : Suppose \( M\overset{\beta }{ \rightarrow }{M}^{\prime } \), i.e., \( M \equiv {M}_{1}\overset{\beta }{ \rig... | Yes |
Theorem 41.12. \( \overset{\beta }{ \rightarrow } \) satisfies the Church-Rosser property. | Proof. Immediate from Theorem 41.2, Theorem 41.8, and Lemma 41.11. 599 | No |
Theorem 41.14. \( M\overset{\beta \eta }{ \Rightarrow }M \) . | Proof. Exercise. | No |
Lemma 41.16. If \( M\overset{\beta \eta }{ \rightarrow }{M}^{\prime } \) and \( R\overset{\beta \eta }{ \rightarrow }{R}^{\prime } \), then \( M\left\lbrack {R/y}\right\rbrack \overset{\beta \eta }{ \rightarrow }{M}^{\prime }\left\lbrack {{R}^{\prime }/y}\right\rbrack \) . | Proof. By induction on the derivation of \( M\overset{\beta \eta }{ \Rightarrow }{M}^{\prime } \) .\n\nThe first four cases are exactly like those in Lemma 41.6. If the last rule is (5), then \( M \) is \( {\lambda x}.{Nx},{M}^{\prime } \) is \( {N}^{\prime } \) for some \( x \) and \( {N}^{\prime } \) where \( x \noti... | Yes |
Lemma 41.17. If \( M\overset{\beta \eta }{ \Rightarrow }{M}^{\prime } \) then \( {M}^{\prime }\overset{\beta \eta }{ \Rightarrow }{M}^{*{\beta \eta }} \) . | Proof. By induction on the derivation of \( M\overset{\beta \eta }{ \Rightarrow }{M}^{\prime } \). The first four cases are like those in Lemma 41.7. If the last rule is (5), then \( M \) is \( {\lambda x}.{Nx} \) and \( {M}^{\prime } \) is \( {N}^{\prime } \) for some \( x, N,{N}^{\prime } \) where \( x \notin {FV}\le... | Yes |
Theorem 41.18. \( \overset{\beta \eta }{ \rightarrow } \) has the Church-Rosser property. | Proof. Immediate from Lemma 41.17. | No |
Lemma 41.19. If \( M\xrightarrow[]{\beta \eta }{M}^{\prime } \), then \( M\overset{\beta \eta }{ \Rightarrow }{M}^{\prime } \) . | Proof. By induction on the derivation of \( M\overset{\beta \eta }{ \rightarrow }{M}^{\prime } \) . If \( M\overset{\beta }{ \rightarrow }{M}^{\prime } \) by \( \eta \) -conversion (i.e., Definition 40.43), we use Theorem 41.14. The other cases are as in Lemma 41.9. \( ▱ \) | No |
Lemma 41.20. If \( M\overset{\beta \eta }{ \Rightarrow }{M}^{\prime } \), then \( M\xrightarrow[]{\beta \eta }{M}^{\prime } \) . | Proof. Induction on the derivation of \( M\overset{\beta \eta }{ \rightarrow }{M}^{\prime } \) .\n\nIf the last rule is (5), then \( M \) is \( {\lambda x}.{Nx} \) and \( {M}^{\prime } \) is \( {N}^{\prime } \) for some \( x, N,{N}^{\prime } \) where \( x \notin {FV}\left( N\right) \) and \( N\overset{\beta \eta }{ \ri... | Yes |
Lemma 41.21. \( \overset{\beta \eta }{ \rightarrow } \) is the smallest transitive relation containing \( \overset{\beta \eta }{ \rightarrow } \) . | Proof. As in Lemma 41.11 | No |
Theorem 41.22. \( \xrightarrow[]{\beta \eta } \) satisfies Church-Rosser property. | Proof. By Theorem 41.2, Theorem 41.18 and Lemma 41.21. | No |
Proposition 42.3. The successor function succ is \( \lambda \) -definable. | Proof. A term that \( \lambda \) -defines the successor function is\n\n\[ \text{ Succ } \equiv {\lambda a}.{\lambda fx}.f\left( {afx}\right) . \]\n\nSucc is a function that accepts as argument a number \( a \), and evaluates to another function, \( {\lambda fx}.f\left( {afx}\right) \) . That function is not itself a Ch... | Yes |
Proposition 42.4. The addition function add is \( \lambda \) -definable. | Proof. Addition is \( \lambda \) -defined by the terms\n\n\[ \text{Add} \equiv {\lambda ab}.{\lambda fx}.{af}\left( {bfx}\right) \]\n\nor, alternatively,\n\n\[ {\operatorname{Add}}^{\prime } \equiv {\lambda ab} \cdot a\operatorname{Succ}b. \]\n\nThe first addition works as follows: Add first accept two numbers \( a \) ... | Yes |
Proposition 42.5. Multiplication is \( \lambda \) -definable by the term\n\n\[ \n\text{Mult} \equiv {\lambda ab}.{\lambda fx}.a\left( {bf}\right) x \n\] | Proof. To see how this works, suppose we apply Mult to Church numerals \( \bar{n} \) and \( \bar{m} \) : Mult \( \bar{n}\bar{m} \) reduces to \( {\lambda fx}.\bar{n}\left( {\bar{m}f}\right) x \) . The term \( \bar{m}f \) defines a function which applies \( f \) to its argument \( m \) times. Consequently, \( \bar{n}\le... | No |
Lemma 42.8. The basic primitive recursive functions zero, succ, and projections \( {P}_{i}^{n} \) are \( \lambda \) -definable. | Proof. They are \( \lambda \) -defined by the following terms:\n\n\[ \text{Zero} \equiv {\lambda a}.{\lambda fx}.x \]\n\n\[ \text{Succ} \equiv {\lambda a}.{\lambda fx}.f\left( {afx}\right) \]\n\n\[ {\operatorname{Proj}}_{i}^{n} \equiv \lambda {x}_{0}\ldots {x}_{n - 1}.{x}_{i} \] | Yes |
Lemma 42.9. Suppose the \( k \) -ary function \( f \), and \( n \) -ary functions \( {g}_{0},\ldots ,{g}_{k - 1} \), are \( \lambda \) -definable by terms \( F,{G}_{0},\ldots ,{G}_{k} \), and \( h \) is defined from them by composition. Then \( H \) is \( \lambda \) -definable. | Proof. \( h \) can be \( \lambda \) -defined by the term\n\n\[ H \equiv \lambda {x}_{0}\ldots {x}_{n - 1}.F\left( {{G}_{0}{x}_{0}\ldots {x}_{n - 1}}\right) \ldots \left( {{G}_{k - 1}{x}_{0}\ldots {x}_{n - 1}}\right) \]\n\nWe leave verification of this fact as an exercise. | No |
Proposition 42.11. Every primitive recursive function is \( \lambda \) -definable. | Proof. By Lemma 42.8, all basic functions are \( \lambda \) -definable, and by Lemma 42.9 and Lemma 42.10, the \( \lambda \) -definable functions are closed under composition and primitive recursion. | Yes |
Theorem 42.13. \( Y \) has the property that \( {Yg} \rightarrow g\left( {Yg}\right) \) for any term \( g \) . Thus, \( {Yg} \) is always a fixpoint of \( g \) . | Proof. Let’s abbreviate \( \left( {{\lambda ux}.x\left( {uux}\right) }\right) \) by \( U \), so that \( Y \equiv {UU} \) . Then\n\n\[ \n{Yg} \equiv \left( {{\lambda ux} \cdot x\left( {uux}\right) }\right) {Ug} \n\]\n\n\[ \n\rightarrow \left( {{\lambda x}.x\left( {UUx}\right) }\right) g \n\]\n\n\[ \n\rightarrow g\left( ... | Yes |
Lemma 42.14. If \( f\left( {{x}_{1},\ldots ,{x}_{k}, y}\right) \) is regular and \( \lambda \) -definable, then \( g \) defined by\n\n\[ g\left( {{x}_{1},\ldots ,{x}_{k}}\right) = {\mu yf}\left( {{x}_{1},\ldots ,{x}_{k}, y}\right) = 0 \]\n\nis also \( \lambda \) -definable. | Proof. Suppose the lambda term \( {F\lambda } \) -defines the regular function \( f\left( {\overrightarrow{x}, y}\right) \) . To \( \lambda \) -define \( h \) we use a search function and a fixpoint combinator:\n\n\[ \text{ Search } \equiv {\lambda g}.{\lambda f}\overrightarrow{x}y\text{. IsZero }\left( {f\overrightarr... | Yes |
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