 reserve X,a,b,c,x,y,z,t for set;
 reserve R for Relation;

theorem :: Theorem 6
  for R1, R2 being non empty RelStr st
    the carrier of R1 = the carrier of R2 holds
    LAp R1 = LAp R2 iff
      the InternalRel of R1 = the InternalRel of R2
  proof
    let R1, R2 be non empty RelStr;
    assume
A1: the carrier of R1 = the carrier of R2;
    hence LAp R1 = LAp R2 implies
      the InternalRel of R1 = the InternalRel of R2 by Corr4;
    assume the InternalRel of R1 = the InternalRel of R2; then
A2: the RelStr of R1 = the RelStr of R2 by A1;
    for x being Subset of R1 holds (LAp R1).x = (LAp R2).x
    proof
      let x be Subset of R1;
      reconsider xx = x as Subset of R2 by A1;
      (LAp R1).x = LAp x by ROUGHS_2:def 10
                .= LAp xx by A2,Pom2
                .= (LAp R2).x by ROUGHS_2:def 10;
      hence thesis;
    end;
    hence thesis by FUNCT_2:63,A1;
  end;
