let E be set ; :: thesis: for A being Subset of (E ^omega)

for m, n being Nat holds (A |^.. m) ^^ (A |^.. n) = A |^.. (m + n)

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

let m, n be Nat; :: thesis: (A |^.. m) ^^ (A |^.. n) = A |^.. (m + n)

defpred S_{1}[ Nat] means (A |^.. m) ^^ (A |^.. $1) = A |^.. (m + $1);

.= A |^.. (m + 0) by Th17 ;

then A3: S_{1}[ 0 ]
;

for n being Nat holds S_{1}[n]
from NAT_1:sch 2(A3, A1);

hence (A |^.. m) ^^ (A |^.. n) = A |^.. (m + n) ; :: thesis: verum

for m, n being Nat holds (A |^.. m) ^^ (A |^.. n) = A |^.. (m + n)

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

let m, n be Nat; :: thesis: (A |^.. m) ^^ (A |^.. n) = A |^.. (m + n)

defpred S

A1: now :: thesis: for n being Nat st S_{1}[n] holds

S_{1}[n + 1]

(A |^.. m) ^^ (A |^.. 0) =
(A |^.. m) ^^ (A *)
by Th11
S

let n be Nat; :: thesis: ( S_{1}[n] implies S_{1}[n + 1] )

assume A2: S_{1}[n]
; :: thesis: S_{1}[n + 1]

(A |^.. m) ^^ (A |^.. (n + 1)) = (A |^.. m) ^^ ((A |^.. n) ^^ A) by Th16

.= (A |^.. (m + n)) ^^ A by A2, FLANG_1:18

.= A |^.. ((m + n) + 1) by Th16 ;

hence S_{1}[n + 1]
; :: thesis: verum

end;assume A2: S

(A |^.. m) ^^ (A |^.. (n + 1)) = (A |^.. m) ^^ ((A |^.. n) ^^ A) by Th16

.= (A |^.. (m + n)) ^^ A by A2, FLANG_1:18

.= A |^.. ((m + n) + 1) by Th16 ;

hence S

.= A |^.. (m + 0) by Th17 ;

then A3: S

for n being Nat holds S

hence (A |^.. m) ^^ (A |^.. n) = A |^.. (m + n) ; :: thesis: verum