let p be FinSequence; for A being set st p is one-to-one holds
len (p - A) = (len p) - (card (A /\ (rng p)))
let A be set ; ( p is one-to-one implies len (p - A) = (len p) - (card (A /\ (rng p))) )
A1:
p " A c= dom p
by RELAT_1:132;
assume A2:
p is one-to-one
; len (p - A) = (len p) - (card (A /\ (rng p)))
p " A,A /\ (rng p) are_equipotent
proof
deffunc H1(
object )
-> set =
p . $1;
consider f being
Function such that A3:
dom f = p " A
and A4:
for
x being
object st
x in p " A holds
f . x = H1(
x)
from FUNCT_1:sch 3();
take
f
;
WELLORD2:def 4 ( f is one-to-one & dom f = p " A & rng f = A /\ (rng p) )
thus
f is
one-to-one
( dom f = p " A & rng f = A /\ (rng p) )proof
let x,
y be
object ;
FUNCT_1:def 4 ( not x in dom f or not y in dom f or not f . x = f . y or x = y )
assume that A5:
x in dom f
and A6:
y in dom f
and A7:
f . x = f . y
;
x = y
A8:
f . y = p . y
by A3, A4, A6;
p . x = f . x
by A3, A4, A5;
hence
x = y
by A2, A1, A3, A5, A6, A7, A8;
verum
end;
thus
dom f = p " A
by A3;
rng f = A /\ (rng p)
thus
rng f c= A /\ (rng p)
XBOOLE_0:def 10 A /\ (rng p) c= rng f
let x be
object ;
TARSKI:def 3 ( not x in A /\ (rng p) or x in rng f )
assume A13:
x in A /\ (rng p)
;
x in rng f
then
x in rng p
by XBOOLE_0:def 4;
then consider y being
object such that A14:
y in dom p
and A15:
p . y = x
by FUNCT_1:def 3;
p . y in A
by A13, A15, XBOOLE_0:def 4;
then A16:
y in p " A
by A14, FUNCT_1:def 7;
then
f . y = x
by A4, A15;
hence
x in rng f
by A3, A16, FUNCT_1:def 3;
verum
end;
then
card (p " A) = card (A /\ (rng p))
by CARD_1:5;
hence
len (p - A) = (len p) - (card (A /\ (rng p)))
by Th57; verum