
theorem Theorem72V1:
  for f being increasing natural-valued Arithmetic_Progression
    st (for i being Nat st i < 10 holds f.i is odd Prime) &
      difference f = 210 holds
        f.0 >= 199
  proof
    let f be increasing natural-valued Arithmetic_Progression;
    assume
A1: for i being Nat st i < 10 holds f.i is odd Prime; then
L1: f.0 is odd Prime;
    assume
KK: difference f = 210;
AA: f = ArProg (f.0,difference f) by NUMBER06:6;
    f.0 is odd by A1; then
T1: f.0 mod 2 = 1 by NAT_2:22;
wv: f.0 mod 11 = 1 by A1,KK,Diff210Mod11; then
WV: f.0 mod 22 = 1 by T1,Mod22;
h3: f.0 div 22 = 0 implies f.0 = 1
    proof
      assume
H1:   f.0 div 22 = 0;
      f.0 = 22 * (f.0 div 22) + (f.0 mod 22) by NAT_D:2
         .= 1 by wv,H1,T1,Mod22;
      hence thesis;
    end;
h2: f.0 div 22 = 1 implies f.0 = 23
    proof
      assume
H1:   f.0 div 22 = 1;
      f.0 = 22 * (f.0 div 22) + (f.0 mod 22) by NAT_D:2
         .= 23 by WV,H1;
      hence thesis;
    end;
vw: 5 divides 5 * 9;
h1: f.0 div 22 = 2 implies f.0 = 45
    proof
      assume
H1:   f.0 div 22 = 2;
      f.0 = 22 * (f.0 div 22) + (f.0 mod 22) by NAT_D:2
         .= 45 by WV,H1;
      hence thesis;
    end;
h4: f.0 div 22 = 3 implies f.0 = 67
    proof
      assume
H1:   f.0 div 22 = 3;
      f.0 = 22 * (f.0 div 22) + (f.0 mod 22) by NAT_D:2
         .= 67 by WV,H1;
      hence thesis;
    end;
h5: f.0 div 22 = 4 implies f.0 = 89
    proof
      assume
H1:   f.0 div 22 = 4;
      f.0 = 22 * (f.0 div 22) + (f.0 mod 22) by NAT_D:2
         .= 89 by WV,H1;
      hence thesis;
    end;
kc: 3 divides 3 * 37;
h6: f.0 div 22 = 5 implies f.0 = 111
    proof
      assume
H1:   f.0 div 22 = 5;
      f.0 = 22 * (f.0 div 22) + (f.0 mod 22) by NAT_D:2
         .= 111 by WV,H1;
      hence thesis;
    end;
kd: 7 divides 7 * 19;
h7: f.0 div 22 = 6 implies f.0 = 133
    proof
      assume
H1:   f.0 div 22 = 6;
      f.0 = 22 * (f.0 div 22) + (f.0 mod 22) by NAT_D:2
         .= 133 by WV,H1;
      hence thesis;
    end;
kb: 5 divides 5 * 31;
h8: f.0 div 22 = 7 implies f.0 = 155
    proof
      assume
H1:   f.0 div 22 = 7;
      f.0 = 22 * (f.0 div 22) + (f.0 mod 22) by NAT_D:2
         .= 155 by WV,H1;
      hence thesis;
    end;
ka: 3 divides 3 * 59;
h9: f.0 div 22 = 8 implies f.0 = 177
    proof
      assume
H1:   f.0 div 22 = 8;
      f.0 = 22 * (f.0 div 22) + (f.0 mod 22) by NAT_D:2
         .= 177 by WV,H1;
      hence thesis;
    end;
hh: f.0 div 22 > 4 implies f.0 >= 199
    proof
      assume f.0 div 22 > 4; then
      f.0 div 22 >= 4 + 1 by NAT_1:13; then
      (f.0 div 22 = 5 + 0 or ... or f.0 div 22 = 5+3) or f.0 div 22 > 8
        by NAT_1:62; then
      f.0 div 22 >= 8+1
        by NAT_1:13,h7,kd,h6,kc,h8,kb,ka,h9,L1,INT_2:def 4; then
      22 * (f.0 div 22) >= 22 * 9 by XREAL_1:64; then
      22 * (f.0 div 22) + 1 >= 198 + 1 by XREAL_1:6;
      hence thesis by WV,NAT_D:2;
    end;
W1: f.0 <> 23
    proof
      assume
L1:   f.0 = 23;
O2:   f.5 = f.0 + 5 * difference f by AA,NUMBER06:7
         .= 29 * 37 by L1,KK; then
O1:   29 divides f.5;
      f.5 is Prime by A1;
      hence thesis by O2,O1,INT_2:def 4;
    end;
xx: f.0 div 22 <= 4 implies
      f.0 div 22 = 0 or ... or f.0 div 22 = 4;
W2: f.0 <> 67
    proof
      assume
L1:   f.0 = 67;
O2:   f.3 = f.0 + 3 * difference f by AA,NUMBER06:7
         .= 17 * 41 by L1,KK; then
O1:   17 divides f.3;
      f.3 is Prime by A1;
      hence thesis by O2,O1,INT_2:def 4;
    end;
    f.0 <> 89
    proof
      assume
L1:   f.0 = 89;
O2:   f.1 = f.0 + 1 * difference f by AA,NUMBER06:7
         .= 13 * 23 by L1,KK; then
O1:   13 divides f.1;
      f.1 is Prime by A1;
      hence thesis by O1,O2,INT_2:def 4;
    end;
    hence thesis
      by W2,xx,h1,h2,h3,h4,h5,hh,W1,vw,L1,INT_2:def 4;
  end;
