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sav08:herbrand_universe_for_equality [2008/04/02 21:49] vkuncak |
sav08:herbrand_universe_for_equality [2008/04/02 22:59] vkuncak |
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Let $S$ be a set of formulas in first-order logic with equality and $S'$ result of replacing '=' with 'eq' in $S$. Suppose $S' \cup AxEq$ is satisfiable. | Let $S$ be a set of formulas in first-order logic with equality and $S'$ result of replacing '=' with 'eq' in $S$. Suppose $S' \cup AxEq$ is satisfiable. | ||
- | Let $I_H = (GT,\alpha_H)$ be Herbrand model for $S \cup AxEq$. We construct a new model as [[Interpretation Quotient Under Congruence]] of $(GT,\alpha)$ under congruence 'eq'. Denote quoient structure by $I_Q = ([GT],\alpha_Q)$. By theorem on quotient structures, $S'$ is true in $I_Q$. Moreover, $I_Q(eq)$ is the identity relation, so $S$ is true under $I_Q$ as well. Therefore, $S$ is satisfiable. | + | Let $I_H = (GT,\alpha_H)$ be Herbrand model for $S \cup AxEq$. We construct a new model as [[Interpretation Quotient Under Congruence]] of $(GT,\alpha)$ under congruence 'eq'. Denote quoient structure by $I_Q = ([GT],\alpha_Q)$. By theorem on quotient structures, $S$ is true in $I_Q$. Therefore, $S$ is satisfiable. |
===== Herbrand-Like Theorem for Equality ===== | ===== Herbrand-Like Theorem for Equality ===== | ||
**Theorem:** For every set of formulas with equality $S$ the following are equivalent | **Theorem:** For every set of formulas with equality $S$ the following are equivalent | ||
- | - $S$ has a model | + | - $S$ has a model; |
- | - $S' \cup AxEq$ has a model (where $AxEq$ are [[Axioms for Equality]] and $S'$ is result of replacing '=' with 'eq' in $S$) | + | - $S' \cup AxEq$ has a model (where $AxEq$ are [[Axioms for Equality]] and $S'$ is result of replacing '=' with 'eq' in $S$); |
- | - $S$ has a model whose domain is the quotient $[GT]$ of ground terms under some congruence | + | - $S$ has a model whose domain is the quotient $[GT]$ of the set of ground terms under some congruence. |