SUBST : (term,thm) subst -> term -> thm -> thm

- STRUCTURE
- SYNOPSIS
- Makes a set of parallel substitutions in a theorem.
- DESCRIPTION
- Implements the following rule of simultaneous substitutionEvaluating
A1 |- t1 = u1 , ... , An |- tn = un , A |- t[t1,...,tn] ------------------------------------------------------------- A u A1 u ... u An |- t[u1,...,un]

returns the theorem A u A1 u ... u An |- t[u1,...,un] (perhaps with fewer assumptions, see paragraph below). The term argument t[x1,...,xn] is a template which should match the conclusion of the theorem being substituted into, with the variables x1, ... , xn marking those places where occurrences of t1, ... , tn are to be replaced by the terms u1, ... , un, respectively. The occurrence of ti at the places marked by xi must be free (i.e. ti must not contain any bound variables). SUBST automatically renames bound variables to prevent free variables in ui becoming bound after substitution.SUBST [x1 |-> (A1 |- t1=u1) ,..., xn |-> (An |- tn=un)] t[x1,...,xn] (A |- t[t1,...,tn])

The assumptions of the returned theorem may not contain all the assumptions A1 u ... u An if some of them are not required. In particular, if the theorem Ak |- tk = uk is unnecessary because xk does not appear in the template, then Ak is not be added to the assumptions of the returned theorem.

- FAILURE
- If the template does not match the conclusion of the hypothesis, or the terms in the conclusion marked by the variables x1, ... , xn in the template are not identical to the left hand sides of the supplied equations (i.e. the terms t1, ... , tn).
- EXAMPLE
- val x = “x:num” and y = “y:num” and th0 = SPEC “0” arithmeticTheory.ADD1 and th1 = SPEC “1” arithmeticTheory.ADD1; (* x = `x` y = `y` th0 = |- SUC 0 = 0 + 1 th1 = |- SUC 1 = 1 + 1 *) - SUBST [x |-> th0, y |-> th1] “(x+y) > SUC 0” (ASSUME “(SUC 0 + SUC 1) > SUC 0”); > val it = [.] |- (0 + 1) + 1 + 1 > SUC 0 : thm - SUBST [x |-> th0, y |-> th1] “(SUC 0 + y) > SUC 0” (ASSUME “(SUC 0 + SUC 1) > SUC 0”); > val it = [.] |- SUC 0 + 1 + 1 > SUC 0 : thm - SUBST [x |-> th0, y |-> th1] “(x+y) > x” (ASSUME “(SUC 0 + SUC 1) > SUC 0”); > val it = [.] |- (0 + 1) + 1 + 1 > 0 + 1 : thm

- COMMENTS
- SUBST is perhaps overly complex for a primitive rule of inference.
- USES
- For substituting at selected occurrences. Often useful for writing special purpose derived inference rules.
- SEEALSO

HOL Kananaskis-14