\DOC SUBS
\TYPE {SUBS : (thm list -> thm -> thm)}
\SYNOPSIS
Makes simple term substitutions in a theorem using a given list of
theorems.
\KEYWORDS
rule.
\DESCRIBE
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
{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
{
- val thm1 = SPECL [Term`m:num`, Term`n:num`] arithmeticTheory.ADD_SYM
val thm2 = CONJUNCT1 arithmeticTheory.ADD_CLAUSES;
> val thm1 = |- m + n = n + m : thm
val thm2 = |- 0 + m = m : thm
}
depending on the occurrence of free subterms
{
- SUBS [thm1, thm2] (ASSUME (Term `(n + 0) + (0 + m) = m + n`));
> val it = [.] |- n + 0 + m = n + m : thm
- SUBS [thm1, thm2] (ASSUME (Term `!n. (n + 0) + (0 + m) = m + n`));
> val it = [.] |- !n. n + 0 + m = m + n : thm
}
\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.
\SEEALSO
Rewrite.ONCE_REWRITE_RULE, Rewrite.PURE_REWRITE_RULE,
Rewrite.REWRITE_RULE, Thm.SUBST, Rewrite.SUBST_MATCH, Drule.SUBS_OCCS.
\ENDDOC