theorem Th39:
  for f being Assign of BASSModel(R,BASSIGN), s being Element of S
  holds s|= EG(f) iff s|= Fax(f,EG(f))
proof
  let f be Assign of BASSModel(R,BASSIGN);
  let s be Element of S;
  set g = EG(f);
  thus s|= EG(f) implies s|= Fax(f,EG(f))
  proof
    set g = EG(f);
    assume s|= EG(f);
    then consider pai be inf_path of R such that
A1: pai.0 = s and
A2: for n being Element of NAT holds pai.n |= f by Th15;
    set pai1 = PathShift(pai,1);
A3: for n being Element of NAT holds pai1.n |= f
    proof
      let n be Element of NAT;
      set n1 = n+1;
      pai1.n = pai.n1 by Def67;
      hence thesis by A2;
    end;
    pai.(0+1) = pai1.0 by Def67;
    then pai.1 |= g by A3,Th15;
    then
A4: s |= EX(g) by A1,Th14;
    s |= f by A1,A2;
    hence thesis by A4,Th13;
  end;
  assume
A5: s|= Fax(f,g);
  then s|= EX(g) by Th13;
  then consider pai1 be inf_path of R such that
A6: pai1.0 = s and
A7: (pai1.1) |= g by Th14;
  consider pai2 being inf_path of R such that
A8: pai2.0 = pai1.1 and
A9: for n being Element of NAT holds (pai2.n) |= f by A7,Th15;
  set pai = PathConc(pai1,pai2,1);
  reconsider pai as inf_path of R by A8,Th38;
A10: pai.0 = PathChange(pai1,pai2,1,0) by Def69
    .= s by A6,Def68;
  for n being Element of NAT holds pai.n |= f
  proof
    let n be Element of NAT;
    per cases;
    suppose
A11:  n < 1;
      n=0
      proof
        assume
A12:    n <> 0;
        n<0+1 by A11;
        hence contradiction by A12,NAT_1:22;
      end;
      hence thesis by A5,A10,Th13;
    end;
    suppose
A13:  not n < 1;
      then reconsider m = n-1 as Element of NAT by NAT_1:21;
      pai.n = PathChange(pai1,pai2,1,n) by Def69
        .= pai2.m by A13,Def68;
      hence thesis by A9;
    end;
  end;
  hence thesis by A10,Th15;
end;
