1 \documentclass[a4paper]{article}
4 \usepackage{amssymb,amsmath,mathrsfs}
8 \newcommand{\sem}[1]{[\![ #1 ]\!]}
9 \newcommand{\R}{\;\mathscr{R}\;}
10 \newcommand{\N}{\,\mathbb{N}\,}
11 \newcommand{\NT}{\,\mathbb{N}\,}
12 \newcommand{\NH}{\,\mathbb{N}\,}
24 \section{Heyting's arithmetics}
30 \item $nat\_ind: P(0) \to (\forall x.P(x) \to P(S(x))) \to \forall x.P(x)$
31 \item $ex\_ind: (\forall x.P(x) \to Q) \to \exists x.P(x) \to Q$
32 \item $ex\_intro: \forall x.(P \to \exists x.P)$
33 \item $fst: P \land Q \to P$
34 \item $snd: P \land Q \to Q$
35 \item $conj: P \to Q \to P \land Q$
36 \item $false\_ind: \bot \to Q$
37 \item $discriminate:\forall x.0 = S(x) \to \bot$
38 \item $injS: \forall x,y.S(x) = S(y) \to x=y$
39 \item $plus\_O:\forall x.x+0=x$
40 \item $plus\_S:\forall x,y.x+S(y)=S(x+y)$
41 \item $times\_O:\forall x.x*0=0$
42 \item $times\_S:\forall x,y.x*S(y)=x+(x*y)$
49 (\to_i)\frac{\Gamma,x:A \vdash M:Q}{\Gamma \vdash \lambda x:A.M: A \to Q} \hspace{2cm}
50 (\to_e)\frac{\Gamma \vdash M: A \to Q \hspace{1cm}\Gamma \vdash N: A}
51 {\Gamma \vdash M N: Q}
55 % (\land_i)\frac{\Gamma \vdash M:A \hspace{1cm}\Gamma \vdash N:B}
56 % {\Gamma \vdash <M,N> : A \land B}
58 % (\land_{el})\frac{\Gamma \vdash A \land B}{\Gamma \vdash A}
60 % (\land_{er})\frac{\Gamma \vdash A \land B}{\Gamma \vdash B}
64 (\forall_i)\frac{\Gamma \vdash M:P}{\Gamma \vdash
65 \lambda x:\N.M: \forall x.P}(*) \hspace{2cm}
66 (\forall_e)\frac{\Gamma \vdash M :\forall x.P}{\Gamma \vdash M t: P[t/x]}
71 % (\exists_i)\frac{\Gamma \vdash P[t/x]}{\Gamma \vdash \exists x.P}\hspace{2cm}
72 % (\exists_e)\frac{\Gamma \vdash \exists x.P\hspace{1cm}\Gamma \vdash \forall x.P \to Q}
79 \item $\sem{A} = 1$ if A is atomic
80 \item $\sem{A \land B} = \sem{A}\times \sem{B}$
81 \item $\sem{A \to B} = \sem{A}\to \sem{B}$
82 \item $\sem{\forall x.P} = \N \to \sem{P}$
83 \item $\sem{\exists x.P} = \N \times \sem{P}$
87 For any type T of system T $\bot_T: 1 \to T$ is inuctively defined as follows:
89 \item $\bot_1 = \lambda x:1.x$
90 \item $\bot_N = \lambda x:1.0$
91 \item $\bot_{U\times V} = \lambda x:1.<\bot_{U} x,\bot_{V} x>$
92 \item $\bot_{U\to V} = \lambda x:1.\lambda \_:U. \bot_{V} x$
96 \item $\sem{nat\_ind} = R$
97 \item $\sem{ex\_ind} = (\lambda f:(\N \to \sem{P} \to \sem{Q}).
98 \lambda p:\N\times \sem{P}.f (fst \,p) (snd \,p)$.
99 \item $\sem{ex\_intro} = \lambda x:\N.\lambda f:\sem{P}.<x,f>$
100 \item $\sem{fst} = fst$
101 \item $\sem{snd} = snd$
102 \item $\sem{conj} = \lambda x:\sem{P}.\lambda y:\sem{Q}.<x,y>$
103 \item $\sem{false\_ind} = \bot_{\sem{Q}}$
104 \item $\sem{discriminate} = \lambda \_:\N.\lambda \_:1.*$
105 \item $\sem{injS}= \lambda \_:\N. \lambda \_:\N.\lambda \_:1.*$
106 \item $\sem{plus\_O} = \sem{times\_O} = \lambda \_:\N.*$
107 \item $\sem{plus\_S} = \sem{times_S} = \lambda \_:\N. \lambda \_:\N.*$
110 In the case of structured proofs:
112 \item $\sem{M N} = \sem{M} \sem{N}$
113 \item $\sem{\lambda x:A.M} = \lambda x:\sem{A}.\sem{M}$
114 \item $\sem{\lambda x:\N.M} = \lambda x:\N.\sem{M}$
115 \item $\sem{M t} = \sem{M} \sem{t}$
118 \section{Realizability}
119 The realizability relation is a relation $f \R P$ where $f: \sem{P}$.
122 \item $\neg (* \R \bot)$
123 \item $* \R (t_1=t_2)$ iff $t_1=t_2$ is true ...
124 \item $<f,g> \R (P\land Q)$ iff $f \R P$ and $g \R Q$
125 \item $f \R (P\to Q)$ iff for any $m$ such that $m \R P$, $(f \,m) \R Q$
126 \item $f \R (\forall x.P)$ iff for any natural number $n$ $(f n) \R P[\underline{n}/x]$
127 \item $<n,g> \R (\exists x.P)$ iff $g \R P[\underline{n}/x]$
131 We proceed to prove that all axioms $ax:Ax$ are realized by $\sem{ax}$.
135 We must prove that the recursion schema $R$ realizes the induction principle.
136 To this aim we must prove that for any $a$ and $f$ such that $a \R P(0)$ and
137 $f \R \forall x.(P(x) \to P(S(x)))$, and any natural number $n$, $(R \,a \,f
138 \,n) \R P(\underline{n})$.\\
139 We proceed by induction on n.\\
140 If $n=O$, $(R \,a \,f \,O) = a$ and by hypothesis $a \R P(0)$.\\
141 Suppose by induction that
142 $(R \,a \,f \,n) \R P(\underline{n})$, and let us prove that the relation
143 still holds for $n+1$. By definition
144 $(R \,a \,f \,(n+1)) = f \,n \,(R \,a \,f \,n)$,
145 and since $f \R \forall x.(P(x) \to P(S(x)))$,
146 $(f n (R a f n)) \R P(S(\underline{n}))=P(\underline{n+1})$.
149 We must prove that $$\underline{ex\_ind} \R (\forall x:(P x)
150 \to Q) \to (\exists x:(P x)) \to Q$$ Following the definition of $\R$ we have
151 to prove that given\\ $f~\R~\forall~x:((P~x)~\to~Q)$ and
152 $p~\R~\exists~x:(P~x)$, then $\underline{ex\_ind}~f~p \R Q$.\\
153 $p$ is a couple $<n_p,g_p>$ such that $g_p \R P[\underline{n_p}/x]$, while
154 $f$ is a function such that forall $n$ and for all $m \R P[\underline{n}/x]$
155 then $f~n~m \R Q$. Exapanding the definition of $\underline{ex\_ind}$, $fst$
156 and $snd$ we obtain $f~n_p~g_p$ that we know is in relation $\R$ with $Q$
157 since $g_p \R P[\underline{n_p}/x]$.
160 We have to prove that $\bot_{\sem{Q}} \R \bot \to Q$.
161 Trivial, since there is no $m \R \bot$.
163 \item $discriminate$.
164 Since there is no $n$ such that $0 = S n$ is true... \\
165 $\underline{discriminate}~n \R 0 = S~\underline{n} \to \bot$ for each n.
169 \begin{thebibliography}{}
172 \end{thebibliography}