let E be set ; :: thesis: for A being Subset of (E ^omega )
for n, k being Nat holds A |^.. (n + k) = (A |^.. n) ^^ (A |^ k)
let A be Subset of (E ^omega ); :: thesis: for n, k being Nat holds A |^.. (n + k) = (A |^.. n) ^^ (A |^ k)
let n, k be Nat; :: thesis: A |^.. (n + k) = (A |^.. n) ^^ (A |^ k)
defpred S₁[ Nat] means A |^.. (n + $1) = (A |^.. n) ^^ (A |^ $1);
A1: S₁[ 0 ]

proof

thus A |^.. (n + 0 ) = (A |^.. n) ^^ {(<%> E)} by FLANG_1:14
.= (A |^.. n) ^^ (A |^ 0 ) by FLANG_1:25 ; :: thesis: verum

end;

A2: now

let k be Nat; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A3: S₁[k] ; :: thesis: S₁[k + 1]
A |^.. (n + (k + 1)) = A |^.. ((n + k) + 1)
.= ((A |^.. n) ^^ (A |^ k)) ^^ A by A3, Th16
.= (A |^.. n) ^^ ((A |^ k) ^^ A) by FLANG_1:19
.= (A |^.. n) ^^ (A |^ (k + 1)) by FLANG_1:24 ;
hence S₁[k + 1] ; :: thesis: verum

end;

for k being Nat holds S₁[k] from NAT_1:sch 2(A1, A2);
hence A |^.. (n + k) = (A |^.. n) ^^ (A |^ k) ; :: thesis: verum