reconsider x9 = x, y9 = y, w9 = w, z9 = z as Element of REAL by XREAL_0:def 1;
thus
( w = 0 & z = 0 implies ex t being Element of QUATERNION ex x9, y9 being Element of REAL st
( x9 = x & y9 = y & t = [*x9,y9*] ) )
( ( not w = 0 or not z = 0 ) implies ex b1 being Element of QUATERNION st b1 = (0,1,2,3) --> (x,y,w,z) )
set e = (0,1,2,3) --> (x9,y9,w9,z9);
thus
( ( w <> 0 or z <> 0 ) implies ex t being Element of QUATERNION st t = (0,1,2,3) --> (x,y,w,z) )
verumproof
assume A1:
(
w <> 0 or
z <> 0 )
;
ex t being Element of QUATERNION st t = (0,1,2,3) --> (x,y,w,z)
A2:
(
0,1,2,3)
--> (
x9,
y9,
w9,
z9)
in Funcs (4,
REAL)
by CARD_1:52, FUNCT_2:8;
now not (0,1,2,3) --> (x9,y9,w9,z9) in { r where r is Element of Funcs (4,REAL) : ( r . 2 = 0 & r . 3 = 0 ) } assume
(
0,1,2,3)
--> (
x9,
y9,
w9,
z9)
in { r where r is Element of Funcs (4,REAL) : ( r . 2 = 0 & r . 3 = 0 ) }
;
contradictionthen
ex
r being
Element of
Funcs (4,
REAL) st
( (
0,1,2,3)
--> (
x9,
y9,
w9,
z9)
= r &
r . 2
= 0 &
r . 3
= 0 )
;
hence
contradiction
by A1, FUNCT_4:139, FUNCT_4:140;
verum end;
then
(
0,1,2,3)
--> (
x9,
y9,
w9,
z9)
in (Funcs (4,REAL)) \ { r where r is Element of Funcs (4,REAL) : ( r . 2 = 0 & r . 3 = 0 ) }
by A2, XBOOLE_0:def 5;
then reconsider e = (
0,1,2,3)
--> (
x9,
y9,
w9,
z9) as
Element of
QUATERNION by XBOOLE_0:def 3;
take
e
;
e = (0,1,2,3) --> (x,y,w,z)
thus
e = (
0,1,2,3)
--> (
x,
y,
w,
z)
;
verum
end;