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--- sav08:herbrand_universe_for_equality [2008/04/02 21:49]
vkuncak
+++ sav08:herbrand_universe_for_equality [2008/04/02 22:58]
vkuncak
@@ Line 17: / Line 17: @@
 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 =====
 **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 ground terms under some congruence.