for t1, t2 being sequence of S st ( for n being Nat holds t1 . n = t . (n + k) ) & ( for n being Nat holds t2 . n = t . (n + k) ) holds
t1 = t2
proof
let t1,
t2 be
sequence of
S;
( ( for n being Nat holds t1 . n = t . (n + k) ) & ( for n being Nat holds t2 . n = t . (n + k) ) implies t1 = t2 )
assume that A3:
for
n being
Nat holds
t1 . n = t . (n + k)
and A4:
for
n being
Nat holds
t2 . n = t . (n + k)
;
t1 = t2
for
x being
set st
x in NAT holds
t1 . x = t2 . x
proof
let x be
set ;
( x in NAT implies t1 . x = t2 . x )
assume
x in NAT
;
t1 . x = t2 . x
then reconsider x =
x as
Nat ;
t1 . x =
t . (x + k)
by A3
.=
t2 . x
by A4
;
hence
t1 . x = t2 . x
;
verum
end;
hence
t1 = t2
by FUNCT_2:18;
verum
end;
hence
for b1, b2 being sequence of S st ( for n being Nat holds b1 . n = t . (n + k) ) & ( for n being Nat holds b2 . n = t . (n + k) ) holds
b1 = b2
; verum