reserve c,c1,c2,x,y,z,z1,z2 for set;
reserve C1,C2,C3 for non empty set;

theorem
  for f,g be RMembership_Func of C1,C2 holds f c= g implies converse f
  c= converse g
proof
  let f,g be RMembership_Func of C1,C2;
  assume
A1: f c= g;
  let c being Element of [:C2,C1:];
  ex y,x being object st y in C2 & x in C1 & c = [y,x] by ZFMISC_1:def 2;
  then consider x,y being set such that
A2: x in C1 and
A3: y in C2 and
A4: c = [y,x];
  [x,y] in [:C1,C2:] by A2,A3,ZFMISC_1:87;
  then f. [x,y] <= g. [x,y] by A1;
  hence thesis by A2,A3,A4,Th9;
end;
