PEXISTS_UNIQUE_CONVPairRules.PEXISTS_UNIQUE_CONV : convExpands with the definition of paired unique existence.
Given a term of the form "?!p. t[p]", the conversion
PEXISTS_UNIQUE_CONV proves that this assertion is
equivalent to the conjunction of two statements, namely that there
exists at least one pair p such that t[p], and
that there is at most one value p for which
t[p] holds. The theorem returned is:
|- (?!p. t[p]) = (?p. t[p]) /\ (!p p'. t[p] /\ t[p'] ==> (p = p'))where p' is a primed variant of the pair p
none of the components of which appear free in the input term. Note that
the quantified pair p need not in fact appear free in the
body of the input term. For example,
PEXISTS_UNIQUE_CONV "?!(x,y). T" returns the theorem:
|- (?!(x,y). T) =
(?(x,y). T) /\ (!(x,y) (x',y'). T /\ T ==> ((x,y) = (x',y')))PEXISTS_UNIQUE_CONV tm fails if tm does not
have the form "?!p.t".