let f be one-to-one Function; for A, B being Subset of (dom f) st A misses B holds
rng (f | A) misses rng (f | B)
let A, B be Subset of (dom f); ( A misses B implies rng (f | A) misses rng (f | B) )
assume ASC0:
A misses B
; rng (f | A) misses rng (f | B)
assume
rng (f | A) meets rng (f | B)
; contradiction
then consider y being set such that
ASC1:
( y in rng (f | A) & y in rng (f | B) )
by XBOOLE_0:3;
consider xa being set such that
ASC2:
( xa in dom (f | A) & y = (f | A) . xa )
by ASC1, FUNCT_1:def 5;
consider xb being set such that
ASC3:
( xb in dom (f | B) & y = (f | B) . xb )
by ASC1, FUNCT_1:def 5;
ASC4:
( xa in dom f & xb in dom f )
by RELAT_1:86, ASC2, ASC3;
( y = f . xa & y = f . xb )
by ASC2, ASC3, FUNCT_1:70;
then ASC5:
xa = xb
by FUNCT_1:def 8, ASC4;
( dom (f | A) c= A & dom (f | B) c= B )
by RELAT_1:87;
then
xa in A /\ B
by XBOOLE_0:def 4, ASC5, ASC2, ASC3;
hence
contradiction
by ASC0, XBOOLE_0:4; verum