let X be set ; for C being non empty set
for c being Element of C holds
( (chi (X,C)) . c <> 1 iff (chi (X,C)) . c = 0 )
let C be non empty set ; for c being Element of C holds
( (chi (X,C)) . c <> 1 iff (chi (X,C)) . c = 0 )
let c be Element of C; ( (chi (X,C)) . c <> 1 iff (chi (X,C)) . c = 0 )
thus
( (chi (X,C)) . c <> 1 implies (chi (X,C)) . c = 0 )
( (chi (X,C)) . c = 0 implies (chi (X,C)) . c <> 1 )proof
c in C
;
then
c in dom (chi (X,C))
by Th77;
then A1:
(chi (X,C)) . c in rng (chi (X,C))
by FUNCT_1:def 3;
A2:
rng (chi (X,C)) c= {0,1}
by FUNCT_3:39;
assume
(chi (X,C)) . c <> 1
;
(chi (X,C)) . c = 0
hence
(chi (X,C)) . c = 0
by A1, A2, TARSKI:def 2;
verum
end;
assume that
A3:
(chi (X,C)) . c = 0
and
A4:
(chi (X,C)) . c = 1
; contradiction
thus
contradiction
by A3, A4; verum