reserve a,b,c,k,m,n for Nat;
reserve i,j for Integer;
reserve p for Prime;

theorem Th45:
  7 divides 2|^(2|^(n+1)) + 2|^(2|^n) + 1
  proof
    per cases;
    suppose
A1:   n = 0;
      2|^(2|^(0+1)) + 2|^(2|^0) + 1 = 2|^2 + 2|^1 + 1 by NEWTON:4
      .= 7 by Lm2;
      hence thesis by A1;
    end;
    suppose
A2:   n > 0;
      set m = 2|^(2|^(n+1)) + 2|^(2|^n) + 1;
A3:   1 mod 7 = 1 by NAT_D:24;
      8 = 1+7*1;
      then
A4:   8 mod 7 = 1 mod 7 by NAT_D:61;
      2|^(n+1) = 2|^n*2|^1 by NEWTON:8;
      then
A5:   2|^(2|^(n+1)) = 2|^(2|^n)|^2 by NEWTON:9
      .= 2|^(2|^n) * 2|^(2|^n) by WSIERP_1:1;
      now per cases;
        suppose n is even;
          then 2|^n mod 3 = 1 by Th42;
          then consider k being Integer such that
A6:       2|^n = k*3+1 by Lm8,NAT_D:64,NAT_6:9;
          now
            assume k < 0;
            then 3*k+1 < 0+1 by XREAL_1:6;
            hence contradiction by A6,NAT_1:14;
          end;
          then reconsider k as Element of NAT by INT_1:3;
A7:       1 mod 7 = 1 by NAT_D:24;
A8:       2|^(2|^n) = 2|^(3*k) * 2|^1 by A6,NEWTON:8
          .= 8|^k * 2 by Lm3,NEWTON:9;
A9:       (8|^k) mod 7 = (8 mod 7)|^k mod 7 by GR_CY_3:30
          .= 1 by A3,A4;
A10:      8|^k * 2 mod 7 = ((8|^k mod 7) * (2 mod 7)) mod 7 by NAT_D:67
          .= 2 by A9,NAT_D:24;
          then 2|^(2|^(n+1)) mod 7 = (2*2) mod 7 by A5,A8,NAT_D:67
          .= 4 by NAT_D:24;
          hence m mod 7 = (4 + 2 + 1) mod 7 by A7,A8,A10,NUMBER05:8
          .= 0;
        end;
        suppose n is odd;
          then 2|^n mod 3 = 2 by Th43;
          then consider k being Integer such that
A11:      2|^n = k*3+2 by Lm9,NAT_D:64,NAT_6:9;
          now
            assume k < 0;
            then 3*k+2 < 0+2 by XREAL_1:6;
            then 2|^n = 1 by A11,NAT_1:23
            .= 2|^0 by NEWTON:4;
            hence contradiction by A2;
          end;
          then reconsider k as Element of NAT by INT_1:3;
A12:      1 mod 7 = 1 by NAT_D:24;
A13:      2|^(2|^n) = 2|^(3*k) * 2|^2 by A11,NEWTON:8
          .= 8|^k * 4 by Lm2,Lm3,NEWTON:9;
A14:      (8|^k) mod 7 = (8 mod 7)|^k mod 7 by GR_CY_3:30
          .= 1 by A3,A4;
A15:      8|^k * 4 mod 7 = ((8|^k mod 7) * (4 mod 7)) mod 7 by NAT_D:67
          .= 4 by A14,NAT_D:24;
          then 2|^(2|^(n+1)) mod 7 = (4*4) mod 7 by A5,A13,NAT_D:67
          .= (2+7*2) mod 7
          .= 2 mod 7 by NAT_D:61
          .= 2 by NAT_D:24;
          hence m mod 7 = (4 + 2 + 1) mod 7 by A12,A13,A15,NUMBER05:8
          .= 0;
        end;
      end;
      hence thesis by INT_1:62;
    end;
  end;
