let IT1, IT2 be Element of NAT ; ( f . IT1 = b & ( for i being Element of NAT st f . i = b holds
IT1 <= i ) & f . IT2 = b & ( for i being Element of NAT st f . i = b holds
IT2 <= i ) implies IT1 = IT2 )
assume that
A6:
f . IT1 = b
and
A7:
for i being Element of NAT st f . i = b holds
IT1 <= i
and
A8:
f . IT2 = b
and
A9:
for i being Element of NAT st f . i = b holds
IT2 <= i
; IT1 = IT2
assume A10:
IT1 <> IT2
; contradiction