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--- sav08:herbrand_universe_for_equality [2008/04/02 22:58]
vkuncak
+++ sav08:herbrand_universe_for_equality [2015/04/21 17:30]
@@ Line 1: / Line 1: @@
-====== Herbrand Universe for Equality ======
-Recall example of formula $F$:
-\[
-    P(a) \land (\forall x.\ P(x) \rightarrow \lnot P(f(x))) \land (\forall x. x=a \lor x=b)
-\]
-Replace '=' with 'eq':
-\[
-    P(a) \land (\forall x.\ P(x) \rightarrow \lnot P(f(x))) \land (\forall x. eq(x,a) \lor eq(x,b))
-\]
-call the resulting formula $F'$.
-Consider Herbrand model $I_H = (GT,\alpha_H)$ for the set $\{F'\} \cup AxEq$.  The relation $\alpha_H(eq)$ splits $GT$ into two sets: the set of terms eq with $a$, and the set of terms eq with $b$.  The idea is to consider these two partitions as domain of new interpretation, denoted $([GT], \alpha_Q)$.
-===== Constructing Model for Formulas with Equality =====
-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 $I_Q$.  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;
-  - $S' \cup AxEq$ 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.