reserve r1,r2,r3 for non negative Real;
reserve n,m1 for Nat;
reserve s for Real;
reserve cn,cd,i1,j1 for Integer;
reserve r for irrational Real;
reserve q for Rational;
reserve c0,c1,c2,u,a0,b0 for Real;
reserve a,b for Real;
reserve n for Integer;

theorem Th32:
  |.a-[\a/].|*|.b-[\a/].| >= |.a-b.|/2 &
  |.a-([\a/]+1).|*|.b-([\a/]+1).| >= |.a-b.|/2 implies
     a is Integer or [\a/]<=b
   proof
     set u = [\a/];
     set v = [\a/]+1;
     assume
A5:  |.a-[\a/].|*|.b-[\a/].| >= |.a-b.|/2 &
       |.a-([\a/]+1).|*|.b-([\a/]+1).| >= |.a-b.|/2;
     assume that
A1:  a is not Integer and
A2:  [\a/] > b;
A3:  a-[\a/] > 0 by A1,INT_1:26, XREAL_1:50;
A4:  a-1 < [\a/] by INT_1:def 6;
S:   [\a/]+1 > b by A2,XREAL_1:40; then
A6:  [\a/]+1 - b > 0 by XREAL_1:50;
A8:  [\a/]-b > 0 by A2,XREAL_1:50;
A9:  [\a/]+1-a > 0 by A4,XREAL_1:19, XREAL_1:50;
S2:  a < [\a/] + 1 by A4,XREAL_1:19; then
a5:  |.a-[\a/].|*|.b-[\a/].|*|.a-[\a/]-1.|*|.b-[\a/]-1.| <= |.a - b.|^2/4
       by A1,A2,INT_1:26,Th25;
     per cases by A5,XXREAL_0:1;
     suppose
A5:  |.a-[\a/].|*|.b-[\a/].| = |.a-b.|/2 &
       |.a-([\a/]+1).|*|.b-([\a/]+1).| > |.a-b.|/2;
A8:  b - [\a/] < 0 by A2,XREAL_1:49;
     (|.a-[\a/].|*|.b-[\a/].|)*(|.a-([\a/]+1).|*|.b-([\a/]+1).|)
       >(|.a-b.|/2)*(|.a-b.|/2) by A3,A8,XREAL_1:98,A5;
       hence contradiction by a5;
     end;
     suppose
A5:  |.a-[\a/].|*|.b-[\a/].| > |.a-b.|/2 &
       |.a-([\a/]+1).|*|.b-([\a/]+1).| = |.a-b.|/2;
A4:  |.a-[\a/].|*|.b-[\a/].|*|.a-[\a/]-1.|*|.b-[\a/]-1.| <= |.a - b.|^2/4
       by A1,A2,INT_1:26,Th25,S2;
a6:  a-([\a/]+1) < 0 by XREAL_1:49,S2;
     b-([\a/]+1) < 0 by XREAL_1:49,S; then
     (|.a-[\a/].|*|.b-[\a/].|)*(|.a-([\a/]+1).|*|.b-([\a/]+1).|)
     > (|.a-b.|/2)*(|.a-b.|/2) by A5,a6,XREAL_1:98;
     hence contradiction by A4;
     end;
     suppose
A5:  |.a-[\a/].|*|.b-[\a/].| > |.a-b.|/2 &
       |.a-([\a/]+1).|*|.b-([\a/]+1).| > |.a-b.|/2;
     |.a-[\a/].|*|.b-[\a/].|*|.a-[\a/]-1.|*|.b-[\a/]-1.| <= |.a - b.|^2/4
       by A1,A2,S2,INT_1:26,Th25;
     hence thesis by Th23,A5;
     end;
     suppose
A5:  |.a-[\a/].|*|.b-[\a/].| = |.a-b.|/2 &
       |.a-([\a/]+1).|*|.b-([\a/]+1).| = |.a-b.|/2;
A10: |.a-u.|*|.u-b.|= |.a-u.|*|.-(u-b).| by COMPLEX1:52
   .=|.a-b.|/2 by A5;
A11: |.a-u.|*|.u-b.|= |.a-u.|*|.-(u-b).| by COMPLEX1:52
   .=|.a-v.|*|.b-v.| by A5;
A13: (a-u)*(u+1-b) = |.(a-u)*(u+1-b).| by A6,A3,ABSVALUE:def 1
   .=|.(a-u).| *|.(u+ 1- b).| by COMPLEX1:65;
a14: (u-b)*(u+1-a) = |.(u-b)*(u+1-a).| by A8,A9,ABSVALUE:def 1
     .= |.u-b.| * |.u+ 1-a.| by COMPLEX1:65; then
A14: |.a-u.|*|.u+1-b.|+|.u-b.|*|.u+1-a.| = |.a-b.| by A13;
A15: |.a-u.|*|.u-b.|*|.a-u-1.|*|.b-u-1.|
    =(|.a-u.|*|.u-b.|)*(|.a-u-1.|*|.b-u-1.|)
   .=((|.a-u.|*|.u+1-b.|+|.u-b.|*|.u+1-a.|)/2)^2 by A13,a14,A10,A11;
     set r1 = |.a-u.|*|.u+1-b.|;
     set r2 = |.u-b.|*|.u+1-a.|;
A17: |.u+1-a.| = |.-(u+1-a).| by COMPLEX1:52 .= |.a-u-1.|;
A18: |.u+1-b.| = |.-(u+1-b).| by COMPLEX1:52 .= |.b-u-1.|;
     r1*r2 = ((r1+r2)/2)^2 by A15,A18,A17; then
A19:  r1 = r2 by Th20;
A23: |.u+1-b.|-|.u-b.|
    =(|.u+1-b.|*|.a-u.|-|.u-b.|*|.a-u.|)/|.a-u.| by A3,XCMPLX_1:129
   .=0 by A10,A14,A19;
     |.u+1-b.|-|.u-b.| = (u+1-b) - |.u-b.| by A6,ABSVALUE:def 1
   .=(u+1-b) - (u-b) by A8,ABSVALUE:def 1
   .= 1;
     hence contradiction by A23;
     end;
    end;
