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\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)
}
\noindent 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
{
# 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
}
\noindent 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.
\SEEALSO
ONCE_REWRITE_RULE, PURE_REWRITE_RULE, REWRITE_RULE, SUBS_CONV.
\ENDDOC
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