SUBS : thm list -> thm -> thm

SYNOPSIS

Makes simple term substitutions in a theorem using a given list of theorems.

DESCRIPTION

Term substitution in HOL is performed by replacing free subterms according to the transformations specified by a list of equational theorems. Given a list of theorems A1|-t1=v1,...,An|-tn=vn and a theorem A|-t, SUBS simultaneously replaces each free occurrence of ti in t with vi:

          A1|-t1=v1 ... An|-tn=vn    A|-t
   ---------------------------------------------  SUBS[A1|-t1=v1;...;An|-tn=vn]
    A1 u ... u An u A |- t[v1,...,vn/t1,...,tn]       (A|-t)

No matching is involved; the occurrence of each ti being substituted for must be a free in t (see SUBST_MATCH). An occurrence which is not free can be substituted by using rewriting rules such as REWRITE_RULE, PURE_REWRITE_RULE and ONCE_REWRITE_RULE.

FAILURE CONDITIONS

SUBS [th1;...;thn] (A|-t) fails if the conclusion of each theorem in the list is not an equation. No change is made to the theorem A |- t if no occurrence of any left-hand side of the supplied equations appears in t.

EXAMPLE

Substitutions are made with the theorems

 # let thm1 = SPEC_ALL ADD_SYM
   and thm2 = SPEC_ALL(CONJUNCT1 ADD_CLAUSES);;
  val thm1 : thm = |- m + n = n + m
  val thm2 : thm = |- 0 + n = n

depending on the occurrence of free subterms

  # SUBS [thm1; thm2] (ASSUME `(n + 0) + (0 + m) = m + n`);;
  val it : thm = (n + 0) + 0 + m = m + n |- (n + 0) + 0 + m = n + m

  # SUBS [thm1; thm2] (ASSUME `!n. (n + 0) + (0 + m) = m + n`);;
  val it : thm = !n. (n + 0) + 0 + m = m + n |- !n. (n + 0) + 0 + m = m + n

USES

SUBS can sometimes be used when rewriting (for example, with REWRITE_RULE) would diverge and term instantiation is not needed. Moreover, applying the substitution rules is often much faster than using the rewriting rules.