deffunc H1( object , object ) -> object = [((L_ $2) \/ (sqrt (x,(L_ $2),(R_ $2)))),(((R_ $2) \/ (sqrt (x,(L_ $2),(L_ $2)))) \/ (sqrt (x,(R_ $2),(R_ $2))))];
consider f being Function such that
A1: ( dom f = NAT & f . 0 = x0 & ( for n being Nat holds f . (n + 1) = H1(n,f . n) ) ) from NAT_1:sch 11();
take f ; :: thesis: ( dom f = NAT & f . 0 = x0 & ( for n being Nat holds
( f . n is pair & (f . (n + 1)) `1 = (L_ (f . n)) \/ (sqrt (x,(L_ (f . n)),(R_ (f . n)))) & (f . (n + 1)) `2 = ((R_ (f . n)) \/ (sqrt (x,(L_ (f . n)),(L_ (f . n))))) \/ (sqrt (x,(R_ (f . n)),(R_ (f . n)))) ) ) )

thus ( dom f = omega & f . 0 = x0 ) by A1; :: thesis: for n being Nat holds
( f . n is pair & (f . (n + 1)) `1 = (L_ (f . n)) \/ (sqrt (x,(L_ (f . n)),(R_ (f . n)))) & (f . (n + 1)) `2 = ((R_ (f . n)) \/ (sqrt (x,(L_ (f . n)),(L_ (f . n))))) \/ (sqrt (x,(R_ (f . n)),(R_ (f . n)))) )

let k be Nat; :: thesis: ( f . k is pair & (f . (k + 1)) `1 = (L_ (f . k)) \/ (sqrt (x,(L_ (f . k)),(R_ (f . k)))) & (f . (k + 1)) `2 = ((R_ (f . k)) \/ (sqrt (x,(L_ (f . k)),(L_ (f . k))))) \/ (sqrt (x,(R_ (f . k)),(R_ (f . k)))) )
thus f . k is pair :: thesis: ( (f . (k + 1)) `1 = (L_ (f . k)) \/ (sqrt (x,(L_ (f . k)),(R_ (f . k)))) & (f . (k + 1)) `2 = ((R_ (f . k)) \/ (sqrt (x,(L_ (f . k)),(L_ (f . k))))) \/ (sqrt (x,(R_ (f . k)),(R_ (f . k)))) )
proof
per cases ( k = 0 or k > 0 ) ;
suppose k = 0 ; :: thesis: f . k is pair
hence f . k is pair by A1; :: thesis: verum
end;
suppose k > 0 ; :: thesis: f . k is pair
then reconsider k1 = k - 1 as Nat by NAT_1:20;
f . (k1 + 1) = H1(k1,f . k1) by A1;
hence f . k is pair ; :: thesis: verum
end;
end;
end;
f . (k + 1) = H1(k,f . k) by A1;
hence ( (f . (k + 1)) `1 = (L_ (f . k)) \/ (sqrt (x,(L_ (f . k)),(R_ (f . k)))) & (f . (k + 1)) `2 = ((R_ (f . k)) \/ (sqrt (x,(L_ (f . k)),(L_ (f . k))))) \/ (sqrt (x,(R_ (f . k)),(R_ (f . k)))) ) ; :: thesis: verum