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

proof

thus (A |^ 0 ) ^^ (A |^.. n) = {(<%> E)} ^^ (A |^.. n) by FLANG_1:25
.= A |^.. n by FLANG_1:14
.= (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 |^ (k + 1)) ^^ (A |^.. n) = ((A |^ k) ^^ A) ^^ (A |^.. n) by FLANG_1:24
.= (A ^^ (A |^ k)) ^^ (A |^.. n) by FLANG_1:33
.= A ^^ ((A |^.. n) ^^ (A |^ k)) by A3, FLANG_1:19
.= (A ^^ (A |^.. n)) ^^ (A |^ k) by FLANG_1:19
.= ((A |^.. n) ^^ A) ^^ (A |^ k) by Th23
.= (A |^.. n) ^^ (A ^^ (A |^ k)) by FLANG_1:19
.= (A |^.. n) ^^ ((A |^ k) ^^ A) by FLANG_1:33
.= (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 |^ k) ^^ (A |^.. n) = (A |^.. n) ^^ (A |^ k) ; :: thesis: verum