`SELECT_ELIM : thm -> term * thm -> thm`
STRUCTURE
SYNOPSIS
Eliminates an epsilon term, using deduction from a particular instance.
DESCRIPTION
SELECT_ELIM expects two arguments, a theorem th1, and a pair (v,th2): term * thm. The conclusion of th1 should have the form P(\$@ P), which asserts that the epsilon term \$@ P denotes some value at which P holds. In th2, the variable v appears only in the assumption P v. The conclusion of the resulting theorem matches that of th2, and the hypotheses include the union of all hypotheses of the premises excepting P v.
```    A1 |- P(\$@ P)     A2 u {P v} |- t
-----------------------------------  SELECT_ELIM th1 (v,th2)
A1 u A2 |- t
```
where v is not free in A2. The argument to P in the conclusion of th1 may actually be any term x. If v appears in the conclusion of th2, this argument x (usually the epsilon term) will NOT be eliminated, and the conclusion will be t[x/v].
FAILURE
Fails if the first theorem is not of the form A1 |- P x, or if the variable v occurs free in any other assumption of th2.
EXAMPLE
If a property of functions is defined by:
```   INCR = |- !f. INCR f = (!t1 t2. t1 < t2 ==> (f t1) < (f t2))
```
The following theorem can be proved.
```   th1 = |- INCR(@f. INCR f)
```
Additionally, if such a function is assumed to exist, then one can prove that there also exists a function which is injective (one-to-one) but not surjective (onto).
```   th2 = [ INCR g ] |- ?h. ONE_ONE h /\ ~ONTO h
```
These two results may be combined using SELECT_ELIM to give a new theorem:
```   - SELECT_ELIM th1 (``g:num->num``, th2);
val it = |- ?h. ONE_ONE h /\ ~ONTO h : thm
```

USES
This rule is rarely used. The equivalence of P(\$@ P) and \$? P makes this rule fundamentally similar to the ?-elimination rule CHOOSE.
SEEALSO