reserve x,y for object,X,Y,A,B,C,M for set;
reserve P,Q,R,R1,R2 for Relation;
reserve X,X1,X2 for Subset of A;
reserve Y for Subset of B;
reserve R,R1,R2 for Subset of [:A,B:];
reserve FR for Subset-Family of [:A,B:];
reserve R for Relation of A,B;
reserve S for Relation of B,C;

theorem Th49: :: (10.1)
  (R.:X)` = (R`).:^X
proof
  thus (R.:X)` c= (R`).:^X
  proof
    let a be object;
    assume
A1: a in (R.:X)`;
    then reconsider B as non empty set;
    reconsider a as Element of B by A1;
    assume not thesis;
    then consider x being set such that
A2: x in X and
A3: not a in Im(R`,x) by Th25;
    [x,a] in R by A2,A3,Th41;
    then a in R.:X by A2,RELAT_1:def 13;
    hence contradiction by A1,XBOOLE_0:def 5;
  end;
  let a be object;
  assume
A4: a in (R`).:^X;
A5: a in (R`).:^X implies for x being set st x in X holds not [x,a] in R
  proof
    assume a in (R`).:^X;
    let x be set;
    assume
A6: x in X;
    assume
A7: not thesis;
    a in Im(R`,x) by A4,A6,Th24;
    then ex b being object st ( [b,a] in R`)&( b in {x}) by RELAT_1:def 13;
    then
A8: [x,a] in R` by TARSKI:def 1;
    R` misses R by XBOOLE_1:79;
    hence contradiction by A7,A8,XBOOLE_0:3;
  end;
  assume
A9: not thesis;
  per cases by A9,XBOOLE_0:def 5;
  suppose not a in B;
    hence contradiction by A4;
  end;
  suppose a in R.:X;
    then ex y being object st ( [y,a] in R)&( y in X) by RELAT_1:def 13;
    hence contradiction by A4,A5;
  end;
end;
