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 S₂[ Nat] means A +^ (succ $1) = (succ A) +^ $1;
A1: S₂[ 0 ]

proof

thus A +^ (succ 0 ) = succ A by ORDINAL2:48
.= (succ A) +^ 0 by ORDINAL2:44 ; :: thesis: verum

end;

A2: for k being Nat st S₂[k] holds
S₂[k + 1]

proof

let k be Nat; :: thesis: ( S₂[k] implies S₂[k + 1] )
assume A3: S₂[k] ; :: thesis: S₂[k + 1]
A4: k + 1 = succ k by NAT_1:39;
hence A +^ (succ (k + 1)) = succ ((succ A) +^ k) by A3, ORDINAL2:45
.= ((succ A) +^ k) +^ 1 by ORDINAL2:48
.= (succ A) +^ (k +^ 1) by ORDINAL3:33
.= (succ A) +^ (k + 1) by A4, ORDINAL2:48 ;
:: thesis: verum

end;

for k being Nat holds S₂[k] from NAT_1:sch 2(A1, A2);
hence A +^ (succ n) = (succ A) +^ n ; :: thesis: A +^ (n + 1) = (succ A) +^ n
hence A +^ (n + 1) = (succ A) +^ n by NAT_1:39; :: thesis: verum