let F, G be XFinSequence; ( F ^ G is one-to-one implies rng F misses rng G )
assume A1:
F ^ G is one-to-one
; rng F misses rng G
assume
rng F meets rng G
; contradiction
then consider y being object such that
A2:
( y in rng F & y in rng G )
by XBOOLE_0:3;
consider n1 being object such that
A3:
( n1 in dom F & F . n1 = y )
by A2, FUNCT_1:def 3;
consider n2 being object such that
A4:
( n2 in dom G & G . n2 = y )
by A2, FUNCT_1:def 3;
reconsider n1 = n1, n2 = n2 as Nat by A3, A4;
A5:
( (F ^ G) . n1 = F . n1 & (F ^ G) . ((len F) + n2) = G . n2 )
by A3, A4, AFINSQ_1:def 3;
dom F c= dom (F ^ G)
by AFINSQ_1:21;
then
( n1 in dom (F ^ G) & (len F) + n2 in dom (F ^ G) )
by A3, A4, AFINSQ_1:23;
then
( n1 = n2 + (len F) & n1 < len F )
by A5, A1, FUNCT_1:def 4, A3, A4, AFINSQ_1:86;
hence
contradiction
by NAT_1:11; verum