theorem
  sin(PI*i) = 0
  proof
    per cases;
    suppose i is even;
      then consider j being Integer such that
A1:   i = 2*j by INT_1:def 3,ABIAN:def 1;
      thus sin(PI*i) = sin(2*PI*j+0) by A1
      .= 0 by SIN_COS:31,COMPLEX2:8;
    end;
    suppose i is odd;
      then consider j being Integer such that
A2:   i = 2*j+1 by ABIAN:1;
      thus sin(PI*i) = sin(2*PI*j+PI) by A2
      .= 0 by SIN_COS:77,COMPLEX2:8;
    end;
  end;
