
theorem Th23:
  for L being non empty RelStr, x being Element of L for F being
non empty NetStr over L holds rng (the mapping of x "/\" F) = {x} "/\" rng the
  mapping of F
proof
  let L be non empty RelStr, x be Element of L, F be non empty NetStr over L;
  set f = the mapping of F, h = the mapping of x "/\" F, A = rng the mapping
  of F;
A1: {x} "/\" A = {x "/\" y where y is Element of L: y in A} by YELLOW_4:42;
A2: the RelStr of x "/\" F = the RelStr of F by Def3;
  then
A3: dom h = the carrier of F by FUNCT_2:def 1;
A4: dom f = the carrier of F by FUNCT_2:def 1;
  thus rng the mapping of x "/\" F c= {x} "/\" A
  proof
    let q be object;
    assume q in rng h;
    then consider a being object such that
A5: a in dom h and
A6: q = h.a by FUNCT_1:def 3;
    reconsider a as Element of x "/\" F by A5;
    consider y being Element of L such that
A7: y = f.a and
A8: h.a = x "/\" y by Def3;
    y in A by A2,A4,A7,FUNCT_1:def 3;
    hence thesis by A1,A6,A8;
  end;
  let q be object;
  assume q in {x} "/\" A;
  then consider y being Element of L such that
A9: q = x "/\" y and
A10: y in A by A1;
  consider z being object such that
A11: z in dom f and
A12: y = f.z by A10,FUNCT_1:def 3;
  reconsider z as Element of x "/\" F by A2,A11;
  ex w being Element of L st w = f.z & h.z = x "/\" w by Def3;
  hence thesis by A3,A9,A11,A12,FUNCT_1:def 3;
end;
