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Substitutions for First-Order Logic

Motivating Example

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 \[

  \forall x. \exists y. x < y

\] Consider an interpretation in integers. This formula is true in this domain. Now substitute instead of x the term y+1. We obtain \[

  \exists y. y + 1 < y

\] 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 \[

  \forall x_1. \exists y_1. x_1 < y_1 

\] 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

Let $U$ be a set of variables (that is, $U \subseteq V$). A variable substitution is a function $\sigma : U \to T$ where $T$ is the set of terms first-order logic.

We define naive substitution recursively, first for terms: \[\begin{array}{rcl} subst(\sigma)( x ) &=& \sigma( x ),\ \sigma \mbox{ d{}efined at } x
subst(\sigma)( x ) &=& x,\ \sigma \mbox{ not d{}efined at } x
subst(\sigma)(f(t_1,\ldots,t_n)) &=& f(subst(\sigma)(t_1),\ldots,subst(\sigma)(t_n)) \end{array}\]

and then for formulas: \[\begin{array}{rcl}

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)(\lnot F) &=& \\
nsubst(\sigma)(F_1 \land F_2) &=& \\
nsubst(\sigma)(\forall x.F) &=&  \\
nsubst(\sigma)(\exists x.F) &=& 

\end{array} \]

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 $I$, \[

  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)])

\]

To avoid variable capture, we introduce in addition to $subst$ a safe substitution, $ssubst$. \[ \begin{array}{rcl}

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)(\lnot F) &=& \\
sfsubst(\sigma)(F_1 \land F_2) &=& \\
sfsubst(\sigma)(\forall x.F) &=& \\
sfsubst(\sigma)(\exists x.F) &=& 

\end{array} \]

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 $I$, \[

  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)])

\]

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