let A be non trivial set ; for B being set
for Bo1, Bo2 being Subset of B
for yo1 being Function of Bo1,A
for yo2 being Function of Bo2,A st Bo1 <> Bo2 holds
{ y where y is Function of B,A : y | Bo1 = yo1 } <> { y where y is Function of B,A : y | Bo2 = yo2 }
let B be set ; for Bo1, Bo2 being Subset of B
for yo1 being Function of Bo1,A
for yo2 being Function of Bo2,A st Bo1 <> Bo2 holds
{ y where y is Function of B,A : y | Bo1 = yo1 } <> { y where y is Function of B,A : y | Bo2 = yo2 }
let Bo1, Bo2 be Subset of B; for yo1 being Function of Bo1,A
for yo2 being Function of Bo2,A st Bo1 <> Bo2 holds
{ y where y is Function of B,A : y | Bo1 = yo1 } <> { y where y is Function of B,A : y | Bo2 = yo2 }
let yo1 be Function of Bo1,A; for yo2 being Function of Bo2,A st Bo1 <> Bo2 holds
{ y where y is Function of B,A : y | Bo1 = yo1 } <> { y where y is Function of B,A : y | Bo2 = yo2 }
let yo2 be Function of Bo2,A; ( Bo1 <> Bo2 implies { y where y is Function of B,A : y | Bo1 = yo1 } <> { y where y is Function of B,A : y | Bo2 = yo2 } )
assume A1:
Bo1 <> Bo2
; { y where y is Function of B,A : y | Bo1 = yo1 } <> { y where y is Function of B,A : y | Bo2 = yo2 }
per cases
( not Bo2 c= Bo1 or Bo2 c= Bo1 )
;
suppose
not
Bo2 c= Bo1
;
{ y where y is Function of B,A : y | Bo1 = yo1 } <> { y where y is Function of B,A : y | Bo2 = yo2 } then consider f being
Function of
B,
A such that A2:
f | Bo1 = yo1
and A3:
f | Bo2 <> yo2
by Lm2;
not
f in { y where y is Function of B,A : y | Bo2 = yo2 }
proof
assume
f in { y where y is Function of B,A : y | Bo2 = yo2 }
;
contradiction
then
ex
y being
Function of
B,
A st
(
f = y &
y | Bo2 = yo2 )
;
hence
contradiction
by A3;
verum
end; hence
{ y where y is Function of B,A : y | Bo1 = yo1 } <> { y where y is Function of B,A : y | Bo2 = yo2 }
by A2;
verum end; suppose
Bo2 c= Bo1
;
{ y where y is Function of B,A : y | Bo1 = yo1 } <> { y where y is Function of B,A : y | Bo2 = yo2 } then consider f being
Function of
B,
A such that A4:
f | Bo2 = yo2
and A5:
f | Bo1 <> yo1
by A1, Lm3;
not
f in { y where y is Function of B,A : y | Bo1 = yo1 }
proof
assume
f in { y where y is Function of B,A : y | Bo1 = yo1 }
;
contradiction
then
ex
y being
Function of
B,
A st
(
f = y &
y | Bo1 = yo1 )
;
hence
contradiction
by A5;
verum
end; hence
{ y where y is Function of B,A : y | Bo1 = yo1 } <> { y where y is Function of B,A : y | Bo2 = yo2 }
by A4;
verum end; end;