let y, z be Element of X1; ( ( for x1, x2, x3, x4, x5 being set st x = [x1,x2,x3,x4,x5] holds
y = x1 ) & ( for x1, x2, x3, x4, x5 being set st x = [x1,x2,x3,x4,x5] holds
z = x1 ) implies y = z )
assume A3:
for x1, x2, x3, x4, x5 being set st x = [x1,x2,x3,x4,x5] holds
y = x1
; ( ex x1, x2, x3, x4, x5 being set st
( x = [x1,x2,x3,x4,x5] & not z = x1 ) or y = z )
assume A4:
for x1, x2, x3, x4, x5 being set st x = [x1,x2,x3,x4,x5] holds
z = x1
; y = z
consider xx1 being Element of X1, xx2 being Element of X2, xx3 being Element of X3, xx4 being Element of X4, xx5 being Element of X5 such that
A5:
x = [xx1,xx2,xx3,xx4,xx5]
by A1, Th17;
y = xx1
by A5, A3;
hence
y = z
by A5, A4; verum