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sav08:substitution_theorems_for_propositional_logic [2008/03/06 13:26]
vkuncak
sav08:substitution_theorems_for_propositional_logic [2008/03/11 14:57]
vkuncak
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 ++++ ++++
  
-Corollary: if $\models ​F$, then $\models ​(subst(\sigma)(F))$ for every substitution $\sigma$.+Corrollary: if $I_1(p) = I_2(p)$ for every $p \in FV(F)$, then $e(F)(I_1)=e(F)(I_2)$.
  
-We say that two formulas are equivalent ​if $\models ​(F \leftrightarrow G)$.+Corollary: ​if $\models F$, then $\models (subst(\sigma)(F))$ for every variable substitution $\sigma$.
  
-**Lemma**: If $\models (F \leftrightarrow G)$ then for every interpretation $I$ we have $e(F)(I) = e(G)(I)$. +Corollary: if $\models (F \leftrightarrow G)$ and $\sigma$ is a variable substitution,​ then
- +
-**Theorem on instantiating equivalences**: if $\models (F \leftrightarrow G)$ and $\sigma$ is a variable substitution,​ then+
 \[ \[
     \models (subst(\sigma)(F) \leftrightarrow subst(\sigma)(G)     \models (subst(\sigma)(F) \leftrightarrow subst(\sigma)(G)
 \] \]
-Does the theorem hold if $\sigma$ is not variable substitution?​+ 
 +++Does the theorem hold if $\sigma$ is not variable substitution?​|No.++ 
 + 
 +We say that two formulas are equivalent if $\models (F \leftrightarrow G)$. 
 + 
 +**Lemma**: If $\models (F \leftrightarrow G)$ then for every interpretation $I$ we have $e(F)(I) = e(G)(I)$. 
  
 This theorem was about transforming formulas from the outside, ignoring the structure of certain subformulas. This theorem was about transforming formulas from the outside, ignoring the structure of certain subformulas.
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 **Theorem on Substituting Equivalent Subformulas**:​ if $\sigma$ is equivalence-preserving,​ then for every formula $F$ we have $\models (F \leftrightarrow (subst(\sigma)(F))$. **Theorem on Substituting Equivalent Subformulas**:​ if $\sigma$ is equivalence-preserving,​ then for every formula $F$ we have $\models (F \leftrightarrow (subst(\sigma)(F))$.
 +