let E be set ; :: thesis: for A being Subset of ()
for k, n being Nat holds (A |^ k) ^^ (A |^.. n) = (A |^.. n) ^^ (A |^ k)

let A be Subset of (); :: thesis: for k, n being Nat holds (A |^ k) ^^ (A |^.. n) = (A |^.. n) ^^ (A |^ k)
let k, n be Nat; :: thesis: (A |^ k) ^^ (A |^.. n) = (A |^.. n) ^^ (A |^ k)
defpred S1[ Nat] means (A |^ \$1) ^^ (A |^.. n) = (A |^.. n) ^^ (A |^ \$1);
A1: now :: thesis: for k being Nat st S1[k] holds
S1[k + 1]
let k be Nat; :: thesis: ( S1[k] implies S1[k + 1] )
assume A2: S1[k] ; :: thesis: S1[k + 1]
(A |^ (k + 1)) ^^ (A |^.. n) = ((A |^ k) ^^ A) ^^ (A |^.. n) by FLANG_1:23
.= (A ^^ (A |^ k)) ^^ (A |^.. n) by FLANG_1:32
.= A ^^ ((A |^.. n) ^^ (A |^ k)) by
.= (A ^^ (A |^.. n)) ^^ (A |^ k) by FLANG_1:18
.= ((A |^.. n) ^^ A) ^^ (A |^ k) by Th23
.= (A |^.. n) ^^ (A ^^ (A |^ k)) by FLANG_1:18
.= (A |^.. n) ^^ ((A |^ k) ^^ A) by FLANG_1:32
.= (A |^.. n) ^^ (A |^ (k + 1)) by FLANG_1:23 ;
hence S1[k + 1] ; :: thesis: verum
end;
(A |^ 0) ^^ (A |^.. n) = {(<%> E)} ^^ (A |^.. n) by FLANG_1:24
.= A |^.. n by FLANG_1:13
.= (A |^.. n) ^^ {(<%> E)} by FLANG_1:13
.= (A |^.. n) ^^ (A |^ 0) by FLANG_1:24 ;
then A3: S1[ 0 ] ;
for k being Nat holds S1[k] from NAT_1:sch 2(A3, A1);
hence (A |^ k) ^^ (A |^.. n) = (A |^.. n) ^^ (A |^ k) ; :: thesis: verum