theorem Th158:
  G1 '&' G2 = (H1 '&' H2)/(x,y) iff G1 = H1/(x,y) & G2 = H2/(x,y)
proof
  thus G1 '&' G2 = (H1 '&' H2)/(x,y) implies G1 = H1/(x,y) & G2 = H2/(x,y)
  proof
    assume G1 '&' G2 = (H1 '&' H2)/(x,y);
    then G1 '&' G2 = (H1/(x,y)) '&' (H2/(x,y)) by Lm1;
    hence thesis by ZF_LANG:30;
  end;
  thus thesis by Lm1;
end;
