theorem
  EqR1 "\/" (EqR1 /\ EqR2) = EqR1
proof
  EqR1 = EqR1 \/ (EqR1 /\ EqR2) & for EqR st EqR1 \/ (EqR1 /\ EqR2) c= EqR
  holds EqR1 c= EqR by XBOOLE_1:22;
  hence thesis by Def2;
end;
