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sav08:a_simple_sound_combination_method [2009/05/13 10:38]
vkuncak
sav08:a_simple_sound_combination_method [2015/04/21 17:30] (current)
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Improvement of precision: if $F$ is equivalent to disjunction $\vee_{j=1}^m D_j$ then to prove $F$ unsatisfiable,​ for each $1 \le j \le m$ check that $D_j$ is unsatisfiable by applying each of the provers. Improvement of precision: if $F$ is equivalent to disjunction $\vee_{j=1}^m D_j$ then to prove $F$ unsatisfiable,​ for each $1 \le j \le m$ check that $D_j$ is unsatisfiable by applying each of the provers.
+

===== Defining Approximations ===== ===== Defining Approximations =====
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$\alpha^p(\forall x.F) = \forall x. \alpha^p(F)$ $\alpha^p(\forall x.F) = \forall x. \alpha^p(F)$

-$\alpha^p(\exists x.F) = \forall ​x. \alpha^p(F)$+$\alpha^p(\exists x.F) = \exists ​x. \alpha^p(F)$

$\alpha^p(A) = A$ for any supported atomic formula $\alpha^p(A) = A$ for any supported atomic formula
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**Example** (from [[Calculus of Computation Textbook]], page 210, Example 10.1): Let $F$ be **Example** (from [[Calculus of Computation Textbook]], page 210, Example 10.1): Let $F$ be
-$+\begin{equation*} 1 \le x \land x \le 2 \land f(x) \neq f(1) \land f(x) \neq f(2) 1 \le x \land x \le 2 \land f(x) \neq f(1) \land f(x) \neq f(2) -$+\end{equation*}
This formula is unsatisfiable. ​ However, This formula is unsatisfiable. ​ However,
-$+\begin{equation*} ​\alpha_U(F) = true ​\alpha_U(F) = true -$ +\end{equation*}
-$+\begin{equation*} ​\alpha_A(F) = 1 \le x \land x \le 2 ​\alpha_A(F) = 1 \le x \land x \le 2 -$+\end{equation*}
and both approximations are satisfiable in resulting theories. and both approximations are satisfiable in resulting theories.

Instead of doing approximation directly, let us transform the original formula into formula $F'$ where function application and arithmetic are separated: Instead of doing approximation directly, let us transform the original formula into formula $F'$ where function application and arithmetic are separated:
-$+\begin{equation*} 1 \le x \land x \le 2 \land y_1=1 \land f(x) \neq f(y_1) \land y_2=2 \land f(x) \neq f(y_2) 1 \le x \land x \le 2 \land y_1=1 \land f(x) \neq f(y_1) \land y_2=2 \land f(x) \neq f(y_2) -$+\end{equation*}
Variables $x$ and $y_1$ both appear in the formula, so let us convert $F'$ into disjunction of $(F' \land x=y_1) \lor (F' \land x \neq y_1)$. ​ We then check satisfiability for each of the disjuncts. ​ Consider, for example, $F' \land x = y_1$, which is formula $F''​$ Variables $x$ and $y_1$ both appear in the formula, so let us convert $F'$ into disjunction of $(F' \land x=y_1) \lor (F' \land x \neq y_1)$. ​ We then check satisfiability for each of the disjuncts. ​ Consider, for example, $F' \land x = y_1$, which is formula $F''​$
-$+\begin{equation*} 1 \le x \land x \le 2 \land y_1=1 \land f(x) \neq f(y_1) \land y_2=2 \land f(x) \neq f(y_2) \land x = y_1 1 \le x \land x \le 2 \land y_1=1 \land f(x) \neq f(y_1) \land y_2=2 \land f(x) \neq f(y_2) \land x = y_1 -$+\end{equation*}
We have We have
-$+\begin{equation*} ​\alpha_U(F''​) = f(x) \neq f(y_1) \land f(x) \neq f(y_2) \land x=y_1 ​\alpha_U(F''​) = f(x) \neq f(y_1) \land f(x) \neq f(y_2) \land x=y_1 -$+\end{equation*}
which is unsatisfiable. ​ Then, considering formula $F' \land x=y_1$, we can do further case analysis on equality of two variables, say $x=y_2$. ​ For $F' \land x \neq y_1 \land x = y_2$ we similarly obtain unsatisfiability of $\alpha_U$-approximation. ​ The remaining case is $F' \land x \neq y_1 \land x \neq y_2$.  For this formula, denoted $F'''​$,​ we have which is unsatisfiable. ​ Then, considering formula $F' \land x=y_1$, we can do further case analysis on equality of two variables, say $x=y_2$. ​ For $F' \land x \neq y_1 \land x = y_2$ we similarly obtain unsatisfiability of $\alpha_U$-approximation. ​ The remaining case is $F' \land x \neq y_1 \land x \neq y_2$.  For this formula, denoted $F'''​$,​ we have
-$+\begin{equation*} ​\alpha_A(F'''​) = 1 \le x \land x \le 2 \land y_1=1 \land y_2=2 \land x \neq y_1 \land x \neq y_2 ​\alpha_A(F'''​) = 1 \le x \land x \le 2 \land y_1=1 \land y_2=2 \land x \neq y_1 \land x \neq y_2 -$+\end{equation*}
which is unsatisfiable. ​ Therefore, by transforming formula into disjunction of formulas, we were able to prove unsatisfiability. ​ Two things helped precision which is unsatisfiable. ​ Therefore, by transforming formula into disjunction of formulas, we were able to prove unsatisfiability. ​ Two things helped precision
* separating literals into literals understood by individual theories (unlike mixed literals $f(x) \neq f(1)$)   * separating literals into literals understood by individual theories (unlike mixed literals $f(x) \neq f(1)$)