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sav08:substitutions_for_first-order_logic [2008/03/19 16:04]
vkuncak
sav08:substitutions_for_first-order_logic [2015/04/21 17:30] (current)
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It is important to be precise about substitutions in first-order logic. ​ For example, we would like to derive from formula $\forall x.F(x)$ formula $F(t)$, denoted $subst(\{x \mapsto t\})(F)$ that results from substituting a term $t$ instead of $x$.  For example, from $\forall x. x < x + 5$ we would like to derive $y - 1 < (y - 1) + 5$.  Consider, however formula It is important to be precise about substitutions in first-order logic. ​ For example, we would like to derive from formula $\forall x.F(x)$ formula $F(t)$, denoted $subst(\{x \mapsto t\})(F)$ that results from substituting a term $t$ instead of $x$.  For example, from $\forall x. x < x + 5$ we would like to derive $y - 1 < (y - 1) + 5$.  Consider, however formula
-$+\begin{equation*} \forall x. \exists y. x < y \forall x. \exists y. x < y -$+\end{equation*}
Consider an interpretation in integers. ​ This formula is true in this domain. ​ Now substitute instead of x the term y+1.  We obtain Consider an interpretation in integers. ​ This formula is true in this domain. ​ Now substitute instead of x the term y+1.  We obtain
-$+\begin{equation*} \exists y. y + 1 < y \exists y. y + 1 < y -$+\end{equation*}
This formula is false. ​ We say that the variable $y$ in term $y+1$ was captured during substitution. ​ When doing substitution in first-order logic we must avoid variable capture. ​ One way to do this is to rename bound variables. ​ Suppose we want to instantiate the formula ​ $\forall x. \exists y. x<y$ with $y+1$. ​ Then we first rename variables in the formula, obtaining This formula is false. ​ We say that the variable $y$ in term $y+1$ was captured during substitution. ​ When doing substitution in first-order logic we must avoid variable capture. ​ One way to do this is to rename bound variables. ​ Suppose we want to instantiate the formula ​ $\forall x. \exists y. x<y$ with $y+1$. ​ Then we first rename variables in the formula, obtaining
-$+\begin{equation*} \forall x_1. \exists y_1. x_1 < y_1 \forall x_1. \exists y_1. x_1 < y_1 -$+\end{equation*}
and then after substitution $\{x_1 \mapsto y+1\}$ we obtain $\exists y_1. y + 1 < y_1$, which is a correct consequence of $\forall x. \exists y. x < y$. and then after substitution $\{x_1 \mapsto y+1\}$ we obtain $\exists y_1. y + 1 < y_1$, which is a correct consequence of $\forall x. \exists y. x < y$.
+

===== Naive and Safe Substitutions ===== ===== Naive and Safe Substitutions =====
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We define naive substitution recursively,​ first for terms: We define naive substitution recursively,​ first for terms:
-$\begin{array}{rcl}+\begin{equation*}\begin{array}{rcl} subst(\sigma)( x ) &=& \sigma( x ),\ \sigma \mbox{ d{}efined at } x \\ subst(\sigma)( x ) &=& \sigma( x ),\ \sigma \mbox{ d{}efined at } x \\ subst(\sigma)( x ) &=& x,\ \sigma \mbox{ not d{}efined at } x \\ subst(\sigma)( x ) &=& x,\ \sigma \mbox{ not d{}efined at } x \\ subst(\sigma)(f(t_1,​\ldots,​t_n)) &=& f(subst(\sigma)(t_1),​\ldots,​subst(\sigma)(t_n)) subst(\sigma)(f(t_1,​\ldots,​t_n)) &=& f(subst(\sigma)(t_1),​\ldots,​subst(\sigma)(t_n)) -\end{array}$+\end{array}\end{equation*}

and then for formulas: and then for formulas:
-$\begin{array}{rcl}+\begin{equation*}\begin{array}{rcl} nsubst(\sigma)(R(t_1,​\ldots,​t_n)) &=& R(nsubst(\sigma)(t_1),​\ldots,​nsubst(\sigma)(t_n)) \\ nsubst(\sigma)(R(t_1,​\ldots,​t_n)) &=& R(nsubst(\sigma)(t_1),​\ldots,​nsubst(\sigma)(t_n)) \\ nsubst(\sigma)(t_1 = t_2) &=& (nsubst(\sigma)(t_1) = nsubst(\sigma)(t_n)) \\ nsubst(\sigma)(t_1 = t_2) &=& (nsubst(\sigma)(t_1) = nsubst(\sigma)(t_n)) \\ - nsubst(\sigma)(\lnot F) &=& \\ + nsubst(\sigma)(\lnot F) &​=& ​\neg nsubst(\sigma)(F) ​\\ - nsubst(\sigma)(F_1 \land F_2) &=& \\ + nsubst(\sigma)(F_1 \land F_2) &​=& ​nsubst(\sigma)(F_1) \wedge nsubst(\sigma)(F_2) \\ - nsubst(\sigma)(\forall x.F) &​=& ​ \\ + nsubst(\sigma)(F_1 \lor F_2) &=& nsubst(\sigma)(F_1) \vee nsubst(\sigma)(F_2) ​\\ - nsubst(\sigma)(\exists x.F) &​=& ​+ nsubst(\sigma)(\forall x.F) &​=& ​\forall x.\ nsubst(\sigma)(F)\\ + nsubst(\sigma)(\exists x.F) &​=& ​\exists x.\ nsubst(\sigma)(F) \end{array} \end{array} -$+\end{equation*}

**Lemma:** Let $\sigma = \{x_1 \mapsto t_1,​\ldots,​x_n \mapsto t_n\}$ be a variable substitution and $t$ a term.  Then for every interpretation $I$, **Lemma:** Let $\sigma = \{x_1 \mapsto t_1,​\ldots,​x_n \mapsto t_n\}$ be a variable substitution and $t$ a term.  Then for every interpretation $I$,
-$+\begin{equation*} e_T(nsubst(\sigma)(t))(I) = e_T(t)(I[x_1 \mapsto e_T(t_1)(I),​\ldots,​ x_n \mapsto e_T(t_n)(I)]) e_T(nsubst(\sigma)(t))(I) = e_T(t)(I[x_1 \mapsto e_T(t_1)(I),​\ldots,​ x_n \mapsto e_T(t_n)(I)]) -$+\end{equation*}

-To avoid variable capture, we introduce in addition to $subst$ a safe substitution,​ $ssubst$. +To avoid variable capture, we introduce in addition to $subst$ a safe substitution,​ $sfsubst$.
-$+\begin{equation*} \begin{array}{rcl} \begin{array}{rcl} sfsubst(\sigma)(R(t_1,​\ldots,​t_n)) &=& R(nsubst(\sigma)(t_1),​\ldots,​nsubst(\sigma)(t_n)) \\ sfsubst(\sigma)(R(t_1,​\ldots,​t_n)) &=& R(nsubst(\sigma)(t_1),​\ldots,​nsubst(\sigma)(t_n)) \\ sfsubst(\sigma)(t_1 = t_2) &=& (nsubst(\sigma)(t_1) = nsubst(\sigma)(t_n)) \\ sfsubst(\sigma)(t_1 = t_2) &=& (nsubst(\sigma)(t_1) = nsubst(\sigma)(t_n)) \\ - sfsubst(\sigma)(\lnot F) &=& \\ + sfsubst(\sigma)(\lnot F) &​=& ​\neg sfsubst(\sigma)(F) ​\\ - sfsubst(\sigma)(F_1 \land F_2) &=& \\ + sfsubst(\sigma)(F_1 \land F_2) &​=& ​ ​sfsubst(\sigma)(F_1) \wedge sfsubst(\sigma)(F_2) \\ - sfsubst(\sigma)(\forall x.F) &=& \\ + sfsubst(\sigma)(F_1 \lor F_2) &=& sfsubst(\sigma)(F_1) \vee sfsubst(\sigma)(F_2) ​\\ - sfsubst(\sigma)(\exists x.F) &​=& ​+ sfsubst(\sigma)(\forall x.F) &​=& ​\left\{ \begin{array}{ll} + \forall x. sfsubst(\sigma)(F) & \text{if} ~ x \notin \text{domain}(\sigma) \wedge x \notin \bigcup_{v \in \text{domain}(\sigma)} FV(v) \\ + sfsubst(\sigma)(\forall x^\prime. sfsubst(\{x \mapsto x^\prime \})(F)) & \text{else}} + \end{array} \right.\\ + sfsubst(\sigma)(\exists x.F) &​=& ​\left\{ \begin{array}{ll} + \exists x. sfsubst(\sigma)(F) & \text{if} ~ x \notin \text{domain}(\sigma) \wedge x \notin \bigcup_{v \in \text{domain}(\sigma)} FV(v) \\ + sfsubst(\sigma)(\exists x^\prime. sfsubst(\{x \mapsto x^\prime \})(F)) & \text{else}} + \end{array} \right. \end{array} \end{array} -$+\end{equation*}

**Lemma:** Let $\sigma = \{x_1 \mapsto t_1,​\ldots,​x_n \mapsto t_n\}$ be a variable substitution and $t$ a term.  Then for every interpretation $I$, **Lemma:** Let $\sigma = \{x_1 \mapsto t_1,​\ldots,​x_n \mapsto t_n\}$ be a variable substitution and $t$ a term.  Then for every interpretation $I$,
-$+\begin{equation*} e_F(sfsubst(\sigma)(F))(I) = e_F(F)(I[x_1 \mapsto e_T(t_1)(I),​\ldots,​ x_n \mapsto e_T(t_n)(I)]) e_F(sfsubst(\sigma)(F))(I) = e_F(F)(I[x_1 \mapsto e_T(t_1)(I),​\ldots,​ x_n \mapsto e_T(t_n)(I)]) -$+\end{equation*}

**Lemma:** $(\forall x.F) \models sfsubst(\{x \mapsto t\}(F)$. **Lemma:** $(\forall x.F) \models sfsubst(\{x \mapsto t\}(F)$.