let s be FinSequence of A; :: thesis: ( s is one-to-one & s is weakly-ascending implies s is ascending )
assume that
A1: s is one-to-one and
A2: s is weakly-ascending ; :: thesis: s is ascending
for n, m being Nat st n in dom s & m in dom s & n < m holds
s /. n <~ s /. m

proof

let n, m be Nat; :: thesis: ( n in dom s & m in dom s & n < m implies s /. n <~ s /. m )
assume that
A3: ( n in dom s & m in dom s ) and
A4: n < m ; :: thesis: s /. n <~ s /. m
A5: s /. n <= s /. m by A3, A4, A2;
not s /. m <= s /. n

proof

assume s /. m <= s /. n ; :: thesis: contradiction
then s /. n = s /. m by A5, ORDERS_2:2;
then s . n = s /. m by A3, PARTFUN1:def 6;
then s . n = s . m by A3, PARTFUN1:def 6;
hence contradiction by A3, A4, A1, FUNCT_1:def 4; :: thesis: verum

end;

hence s /. n <~ s /. m by A5; :: thesis: verum

end;

hence s is ascending ; :: thesis: verum