
theorem LMStat0:
  for k be Nat st 1 <= k & k <= 128 ex i,j be Nat st i in Seg 4 & j in Seg 4 &
  1+(i-'1)*8+(j-'1)*32 <= k & k <= 1+(i-'1)*8+(j-'1)*32+7
proof
  let k be Nat;
  assume
A1: 1 <= k & k <= 128;
A3: k = 32*(k div 32)+(k mod 32) by NAT_D:2;
  reconsider m = k div 32 as Nat;
  reconsider n = k mod 32 as Nat;
  k div 32 <= (32*4) div 32 by A1,NAT_2:24;
  then
M1: m <= 4 by NAT_D:18;
  per cases;
  suppose
A4: n = 0;
A5: 1 <= m
    proof
      assume not 1 <= m;
      then m = 0 by NAT_1:14;
      hence contradiction by A1,A3,A4;
    end;
    set j = m;
A8: j in Seg 4 by M1,A5;
    set i = 4;
A10: i in Seg 4;
A11: j-'1 = j-1 by XREAL_1:233,A5;
A13: k = 32*(k div 32)+(k mod 32) by NAT_D:2
      .= 32*(j-'1)+8*((i-1)+1) by A4,A11
      .= 32*(j-'1)+8*((i-'1)+1) by XREAL_1:233
      .= 1+(i-'1)*8+(j-'1)*32+7;
    1+(i-'1)*8+(j-'1)*32+0 <= 1+(i-'1)*8+(j-'1)*32+7 by XREAL_1:7;
    hence ex i,j be Nat st i in Seg 4 & j in Seg 4 &
    1+(i-'1)*8+(j-'1)*32 <= k & k <= 1+(i-'1)*8+(j-'1)*32+7 by A8,A10,A13;
  end;
  suppose
A14: n <> 0;
    then
XX0: 1 <= n by NAT_1:14;
XX1: n <= 32 by NAT_D:1;
    m <> 4
    proof
      assume
U1:   m = 4;
U2:   k = 32*4+n by NAT_D:2,U1
        .= 128+n;
      128+1 <= 128+n by XX0,XREAL_1:7;
      hence contradiction by U2,XXREAL_0:2,A1;
    end;
    then m < 4 by XXREAL_0:1,M1;
    then
A15: m+1 <= 4 by NAT_1:13;
A16: 1 <= m+1 by NAT_1:11;
    set j = m+1;
A18: j in Seg 4 by A15,A16;
A19: j-'1 = j-1 by XREAL_1:233,NAT_1:11
      .=m;
A20: k = 32*(j-'1)+n by NAT_D:2,A19;
A22: n = 8*(n div 8)+(n mod 8) by NAT_D:2;
    reconsider s = n div 8 as Nat;
    reconsider t = n mod 8 as Nat;
    n div 8 <= (8*4) div 8 by XX1,NAT_2:24;
    then
M2: n div 8 <= 4 by NAT_D:18;
    now per cases;
    suppose
A23:  t = 0;
A24:  1 <= s
      proof
        assume not 1 <= s;
        then n = 8*0+0 by NAT_1:14,A22,A23;
        hence contradiction by A14;
      end;
      set i = s;
A28:  i in Seg 4 by M2,A24;
A29:  i-'1 = i-1 by XREAL_1:233,A24;
A30:  n = 8*s+0 by NAT_D:2,A23
        .= 8*(i-'1)+8*1 by A29;
      1+(i-'1)*8+(j-'1)*32+0 <= 1+(i-'1)*8+(j-'1)*32+7 by XREAL_1:7;
      hence ex i,j be Nat st i in Seg 4 & j in Seg 4 &
      1+(i-'1)*8+(j-'1)*32 <= k & k <= 1+(i-'1)*8+(j-'1)*32+7
      by A28,A18,A20,A30;
    end;
    suppose
      t <> 0;
      then
XX0:  1 <= t by NAT_1:14;
XXX1: t <= 8 by NAT_D:1;
      s <> 4
      proof
        assume
U1:     s = 4;
U2:     n = 8*4+t by NAT_D:2,U1
          .= 32+t;
        32+1 <= 32+t by XX0,XREAL_1:7;
        hence contradiction by U2,XXREAL_0:2,XX1;
      end;
      then s < 4 by XXREAL_0:1,M2;
      then
B15:  s+1 <= 4 by NAT_1:13;
B16:  1 <= s+1 by NAT_1:11;
      set i = s+1;
B18:  i in Seg 4 by B15,B16;
B19:  i-'1 = i-1 by XREAL_1:233,NAT_1:11
        .=s;
B20:  n = 8*(i-'1)+t by NAT_D:2,B19;
B220: 32*(j-'1)+8*(i-'1)+1 <= 32*(j-'1)+8*(i-'1)+t by XX0,XREAL_1:7;
      32*(j-'1)+8*(i-'1)+t <= 32*(j-'1)+8*(i-'1)+8 by XXX1,XREAL_1:7;
      then k <= 1+8*(i-'1)+32*(j-'1)+7 by A20,B20;
      hence ex i,j be Nat st i in Seg 4 & j in Seg 4 &
      1+(i-'1)*8+(j-'1)*32 <= k & k <= 1+(i-'1)*8+(j-'1)*32+7
      by B220,A20,B20,B18,A18;
    end;
  end;
  hence ex i,j be Nat st i in Seg 4 & j in Seg 4 & 1+(i-'1)*8+(j-'1)*32 <= k &
  k <= 1+(i-'1)*8+(j-'1)*32+7;
end;
end;
