defpred S₁[ Nat, set , set ] means for n, m being Element of NAT st $2 = n & $3 = m holds
( n < m & P₁[F₁() . m] & ( for k being Element of NAT st n < k & P₁[F₁() . k] holds
m <= k ) );
defpred S₂[ Nat] means P₁[F₁() . $1];
ex m1 being Element of NAT st
( 0 <= m1 & P₁[F₁() . m1] ) by A1;
then A2: ex m being Nat st S₂[m] ;
consider M being Nat such that
A3: ( S₂[M] & ( for n being Nat st S₂[n] holds
M <= n ) ) from NAT_1:sch 5(A2);
reconsider M9 = M as Element of NAT by ORDINAL1:def 12;

A4: now :: thesis: for n being Element of NAT ex m being Element of NAT st
( n < m & P₁[F₁() . m] )

let n be Element of NAT ; :: thesis: ex m being Element of NAT st
( n < m & P₁[F₁() . m] )
consider m being Element of NAT such that
A5: n + 1 <= m and
A6: P₁[F₁() . m] by A1;
take m = m; :: thesis: ( n < m & P₁[F₁() . m] )
thus ( n < m & P₁[F₁() . m] ) by A5, A6, NAT_1:13; :: thesis: verum

end;

A7: for n being Nat
for x being Element of NAT ex y being Element of NAT st S₁[n,x,y]

proof

let n be Nat; :: thesis: for x being Element of NAT ex y being Element of NAT st S₁[n,x,y]
let x be Element of NAT ; :: thesis: ex y being Element of NAT st S₁[n,x,y]
defpred S₃[ Nat] means ( x < $1 & P₁[F₁() . $1] );
ex m being Element of NAT st S₃[m] by A4;
then A8: ex m being Nat st S₃[m] ;
consider l being Nat such that
A9: ( S₃[l] & ( for k being Nat st S₃[k] holds
l <= k ) ) from NAT_1:sch 5(A8);
take l ; :: thesis: ( l is Element of NAT & S₁[n,x,l] )
l in NAT by ORDINAL1:def 12;
hence ( l is Element of NAT & S₁[n,x,l] ) by A9; :: thesis: verum

end;

consider F being sequence of NAT such that
A10: ( F . 0 = M9 & ( for n being Nat holds S₁[n,F . n,F . (n + 1)] ) ) from RECDEF_1:sch 2(A7);
A11: rng F c= REAL by NUMBERS:19;
A12: rng F c= NAT ;
A13: dom F = NAT by FUNCT_2:def 1;
then reconsider F = F as Real_Sequence by A11, RELSET_1:4;

A14: now :: thesis: for n being Element of NAT holds F . n is Element of NAT

let n be Element of NAT ; :: thesis: F . n is Element of NAT
F . n in rng F by A13, FUNCT_1:def 3;
hence F . n is Element of NAT by A12; :: thesis: verum

end;

now :: thesis: for n being Nat holds F . n < F . (n + 1)

let n be Nat; :: thesis: F . n < F . (n + 1)
A15: n in NAT by ORDINAL1:def 12;
A16: F . (n + 1) is Element of NAT by A14;
F . n is Element of NAT by A14, A15;
hence F . n < F . (n + 1) by A10, A16; :: thesis: verum

end;

then reconsider F = F as increasing sequence of NAT by SEQM_3:def 6;
A17: for n being Element of NAT st P₁[F₁() . n] holds
ex m being Element of NAT st F . m = n

proof

defpred S₃[ Nat] means ( P₁[F₁() . $1] & ( for m being Element of NAT holds F . m <> $1 ) );
assume ex n being Element of NAT st S₃[n] ; :: thesis: contradiction
then A18: ex n being Nat st S₃[n] ;
consider M1 being Nat such that
A19: ( S₃[M1] & ( for n being Nat st S₃[n] holds
M1 <= n ) ) from NAT_1:sch 5(A18);
defpred S₄[ Nat] means ( $1 < M1 & P₁[F₁() . $1] & ex m being Element of NAT st F . m = $1 );
A20: ex n being Nat st S₄[n]

proof

take M ; :: thesis: S₄[M]
A21: M <> M1 by A10, A19;
M <= M1 by A3, A19;
hence M < M1 by A21, XXREAL_0:1; :: thesis: ( P₁[F₁() . M] & ex m being Element of NAT st F . m = M )
thus P₁[F₁() . M] by A3; :: thesis: ex m being Element of NAT st F . m = M
take 0 ; :: thesis: F . 0 = M
thus F . 0 = M by A10; :: thesis: verum

end;

A22: for n being Nat st S₄[n] holds
n <= M1 ;
consider X being Nat such that
A23: ( S₄[X] & ( for n being Nat st S₄[n] holds
n <= X ) ) from NAT_1:sch 6(A22, A20);
A24: for k being Element of NAT st X < k & k < M1 holds
not P₁[F₁() . k]

proof

given k being Element of NAT such that A25: X < k and
A26: k < M1 and
A27: P₁[F₁() . k] ; :: thesis: contradiction

now :: thesis: contradiction

per cases ( ex m being Element of NAT st F . m = k or for m being Element of NAT holds F . m <> k ) ;

suppose ex m being Element of NAT st F . m = k ; :: thesis: contradiction

hence contradiction by A23, A25, A26, A27; :: thesis: verum

end;

suppose for m being Element of NAT holds F . m <> k ; :: thesis: contradiction

hence contradiction by A19, A26, A27; :: thesis: verum

end;

end;

end;

hence contradiction ; :: thesis: verum

end;

consider m being Element of NAT such that
A28: F . m = X by A23;
M1 in NAT by ORDINAL1:def 12;
then A29: F . (m + 1) <= M1 by A10, A19, A23, A28;
A30: P₁[F₁() . (F . (m + 1))] by A10, A28;
A31: X < F . (m + 1) by A10, A28;

now :: thesis: not F . (m + 1) <> M1

assume F . (m + 1) <> M1 ; :: thesis: contradiction
then F . (m + 1) < M1 by A29, XXREAL_0:1;
hence contradiction by A24, A31, A30; :: thesis: verum

end;

hence contradiction by A19; :: thesis: verum

end;

take F ; :: thesis: ( ( for n being Nat holds P₁[(F₁() * F) . n] ) & ( for n being Element of NAT st ( for r being Real st r = F₁() . n holds
P₁[r] ) holds
ex m being Element of NAT st n = F . m ) )
set q = F₁() * F;
defpred S₃[ Nat] means P₁[(F₁() * F) . $1];
A32: for k being Nat st S₃[k] holds
S₃[k + 1]

proof

let k be Nat; :: thesis: ( S₃[k] implies S₃[k + 1] )
assume P₁[(F₁() * F) . k] ; :: thesis: S₃[k + 1]
S₁[k,F . k,F . (k + 1)] by A10;
then P₁[F₁() . (F . (k + 1))] ;
hence S₃[k + 1] by FUNCT_2:15; :: thesis: verum

end;

A33: S₃[ 0 ] by A3, A10, FUNCT_2:15;
thus for n being Nat holds S₃[n] from NAT_1:sch 2(A33, A32); :: thesis: for n being Element of NAT st ( for r being Real st r = F₁() . n holds
P₁[r] ) holds
ex m being Element of NAT st n = F . m
let n be Element of NAT ; :: thesis: ( ( for r being Real st r = F₁() . n holds
P₁[r] ) implies ex m being Element of NAT st n = F . m )
assume for r being Real st r = F₁() . n holds
P₁[r] ; :: thesis: ex m being Element of NAT st n = F . m
then consider m being Element of NAT such that
A34: F . m = n by A17;
take m ; :: thesis: n = F . m
thus n = F . m by A34; :: thesis: verum