let A be Ordinal; :: thesis: for n being Nat holds
( A +^ (succ n) = (succ A) +^ n & A +^ (n + 1) = (succ A) +^ n )
let n be Nat; :: thesis: ( A +^ (succ n) = (succ A) +^ n & A +^ (n + 1) = (succ A) +^ n )
defpred S₁[ Nat] means A +^ (succ $1) = (succ A) +^ $1;
A +^ (succ 0) = succ A by ORDINAL2:31
.= (succ A) +^ 0 by ORDINAL2:27 ;
then 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]
A4: Segm (k + 1) = succ (Segm k) by NAT_1:38;
hence A +^ (succ (k + 1)) = succ ((succ A) +^ k) by A3, ORDINAL2:28
.= ((succ A) +^ k) +^ 1 by ORDINAL2:31
.= (succ A) +^ (k +^ 1) by ORDINAL3:30
.= (succ A) +^ (k + 1) by A4, ORDINAL2:31 ;
:: thesis: verum

end;

A5: for k being Nat holds S₁[k] from NAT_1:sch 2(A1, A2);
thus A6: A +^ (succ n) = (succ A) +^ n by A5; :: thesis: A +^ (n + 1) = (succ A) +^ n
Segm (n + 1) = succ (Segm n) by NAT_1:38;
hence A +^ (n + 1) = (succ A) +^ n by A6; :: thesis: verum