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Commit 1a7af4a2 authored by SimonGuilloud's avatar SimonGuilloud
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Merge conflicts in the manual

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6 merge requests!62Easy tactics,!58Easy tactics,!55Front integration and various changes,!54Front integration,!53Front integration,!52Front integration
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\begin{document}
\maketitle
\chapter*{Introduction}
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This document aims to give a complete documentation on LISA. Tentatively, every chapter and section will explain a part or concept of LISA, and explains both its implementation and its theoretical foundations.
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This document aims to give a complete documentation on LISA. Tentatively, every chapter and section will explain a part or concept of LISA, and explains both its implementation and its theoretical foundations \cite{DBLP:conf/tacas/GuilloudK22}.
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\input{part1.tex}
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An extension by definition is the formal way of introducing new symbols in a mathematical theory.
Theories can be extended into new ones by adding new symbols and new axioms to it. We're interested in a special kind of extension, called \textit{conservative extension}.
\begin{defin}[Conservative Extension]
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A theory $\mathcal{T}_2$ is a conservative extension over a theory $\mathcal{T}_1$ if:
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A theory $\mathcal{T}_2$ is a conservative extension of a theory $\mathcal{T}_1$ if:
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\begin{itemize}
\item $\mathcal{T}_1 \subset \mathcal{T}_2$
\item For any formula $\phi$ in the language of $\mathcal{T}_1$, if $\mathcal{T}_2 \vdash \phi$ then $\mathcal{T}_1 \vdash \phi$
......@@ -33,11 +30,8 @@ Moreover, in that case we require that
$$
\exists ! y. \phi_{y, x_1,...,x_k}
$$
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is a theorem of $\mathcal{T}_1$
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is a theorem of $\mathcal{T}_1$.
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\end{itemize}
\end{itemize}
\end{defin}
......@@ -46,11 +40,9 @@ We also say that a theory $\mathcal{T}_k$ is an extension by definition of a the
For function definition, it is common in logic textbooks to only require the existence of $y$ and not its uniqueness. The axiom one would then obtain would only be $\phi[f(x_1,...,x_n)/y]$ This also leads to conservative extension, but it turns out not to be enough in the presence of axiom schemas (axioms containing schematic symbols).
\begin{lemma}
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In ZF, an extension by definition without uniqueness doesn't necessarily yields a conservative extension if the use of the new symbol is allowed in axiom schemas.
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In ZF, an extension by definition without uniqueness doesn't necessarily yield a conservative extension if the use of the new symbol is allowed in axiom schemas.
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\end{lemma}
\begin{proof}
In ZF, consider the formula $\phi_c := \forall x. \exists y. (x \neq \emptyset) \implies y \in x$ expressing that nonempty sets contain an element, which is provable in ZFC.
......@@ -66,13 +58,9 @@ For the definition with uniqueness, there is a stronger result than only conserv
\begin{defin}
A theory $\mathcal{T}_2$ is a fully conservative extension over a theory $\mathcal{T}_1$ if:
\begin{itemize}
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\item It is conservative
\item For any formula $\phi_2$ with free variables $x_1, ..., x_k$ in the language of $\mathcal{T}_2$, there exists a formula $\phi_1$ in the language of $\mathcal{T}_1$ with free variables among $x_1, ..., x_k$ such that
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\item it is conservative, and
\item for any formula $\phi_2$ with free variables $x_1, ..., x_k$ in the language of $\mathcal{T}_2$, there exists a formula $\phi_1$ in the language of $\mathcal{T}_1$ with free variables among $x_1, ..., x_k$ such that
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$$\mathcal{T}_2 \vdash \forall x_1...x_k. (\phi_1 \leftrightarrow \phi_2)$$
\end{itemize}
\end{defin}
......@@ -81,11 +69,8 @@ An extension by definition with uniqueness is fully conservative.
\end{thm}
The proof is done by induction on the height of the formula and isn't difficult, but fairly tedious.
\begin{thm}
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If an extension $\mathcal{T}_2$ of a theory $\mathcal{T}_1$ with axiom schemas is fully conservative, then for any instance of the axiom schemas of an axiom schemas $\alpha$ containing a new symbol, $\Gamma \vdash \alpha$ where $\Gamma$ contains no axiom schema instantiated with new symbols.
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If an extension $\mathcal{T}_2$ of a theory $\mathcal{T}_1$ with axiom schemas is fully conservative, then for any instance of the axiom schemas containing a new symbol $\alpha$, $\Gamma \vdash \alpha$ where $\Gamma$ contains no axiom schema instantiated with new symbols.
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\end{thm}
\begin{proof}
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