Drule.SUBST

STRUCTURE

SYNOPSIS

Makes substitutions in a term at selected occurrences of subterms, using a list of theorems.

DESCRIPTION

SUBST_CONV implements the following rule of simultaneous substitution

                    A1 |- t1 = v1 ... An |- tn = vn
   ------------------------------------------------------------------
    A1 u ... u An |- t[t1,...,tn/x1,...,xn] = t[v1,...,vn/x1,...,xn]

The first argument to SUBST_CONV is a list [{redex=x1, residue = A1|-t1=v1},...,{redex = xn, residue = An|-tn=vn}]. The second argument is a template term t[x1,...,xn], in which the variables x1,...,xn are used to mark those places where occurrences of t1,...,tn are to be replaced with the terms v1,...,vn, respectively. Thus, evaluating

   SUBST_CONV [{redex = x1, residue = A1|-t1=v1},...,
               {redex = xn, residue = An|-tn=vn}]
              t[x1,...,xn]
              t[t1,...,tn/x1,...,xn]

returns the theorem

   A1 u ... u An |- t[t1,...,tn/x1,...,xn] = t[v1,...,vn/x1,...,xn]

The occurrence of ti at the places marked by the variable xi must be free (i.e. ti must not contain any bound variables). SUBST_CONV automatically renames bound variables to prevent free variables in vi becoming bound after substitution.

FAILURE

SUBST_CONV [{redex=x1,residue=th1},...,{redex=xn,residue=thn}] t[x1,...,xn] t' fails if the conclusion of any theorem thi in the list is not an equation; or if the template t[x1,...,xn] does not match the term t'; or if and term ti in t' marked by the variable xi in the template, is not identical to the left-hand side of the conclusion of the theorem thi.

EXAMPLE

The values

   - val thm0 = SPEC (Term`0`) ADD1
     and thm1 = SPEC (Term`1`) ADD1
     and x = Term`x:num` and y = Term`y:num`;

   > val thm0 = |- SUC 0 = 0 + 1 : thm
     val thm1 = |- SUC 1 = 1 + 1 : thm
     val x = `x` : term
     val y = `y` : term

can be used to substitute selected occurrences of the terms SUC 0 and SUC 1

- SUBST_CONV [{redex=x, residue=thm0},{redex=y,residue=thm1}]
             (Term`(x + y) > SUC 1`)
             (Term`(SUC 0 + SUC 1) > SUC 1`);
> val it = |- SUC 0 + SUC 1 > SUC 1 = (0 + 1) + 1 + 1 > SUC 1 : thm

The |-> syntax can also be used:

- SUBST_CONV [x |-> thm0, y |-> thm1]
             (Term`(x + y) > SUC 1`)
             (Term`(SUC 0 + SUC 1) > SUC 1`);

USES

SUBST_CONV is used when substituting at selected occurrences of terms and using rewriting rules/conversions is too extensive.

SEEALSO

REWR_CONV, SUBS, SUBST, SUBS_OCCS, |->