let n be Nat; :: thesis: ( omega *` (card n) c= omega & (card n) *` omega c= omega )
defpred S1[ Nat] means omega *` (card $1) c= omega ;
A1: S1[ 0 ] ;
A2: for k being Nat st S1[k] holds
S1[k + 1]
proof
let k be Nat; :: thesis: ( S1[k] implies S1[k + 1] )
assume A3: S1[k] ; :: thesis: S1[k + 1]
card (k + 1) = Segm (k + 1)
.= succ (Segm k) by NAT_1:38 ;
then card (k + 1) = card (succ k) ;
then omega *` (card (k + 1)) = card ((succ k) *^ omega) by Th13, CARD_1:47
.= card ((k *^ omega) +^ omega) by ORDINAL2:36
.= (card (k *^ omega)) +` omega by Th12, CARD_1:47
.= (omega *` (card k)) +` omega by Th13, CARD_1:47
.= omega by A3, Th75 ;
hence S1[k + 1] ; :: thesis: verum
end;
A4: for k being Nat holds S1[k] from NAT_1:sch 2(A1, A2);
 hence omega *` (card n) c= omega ; :: thesis: (card n) *` omega c= omega
thus (card n) *` omega c= omega by A4; :: thesis: verum