 reserve x,y for Element of [.0,1.];
 reserve I for BinOp of [.0,1.];

theorem LemmaAA:
  I is satisfying_(OP) implies I is satisfying_(I3) satisfying_(I4)
    satisfying_(NC) satisfying_(LB) satisfying_(RB) satisfying_(IP)
  proof
    assume
a1: I is satisfying_(OP);
A2: 0 in [.0,1.] by XXREAL_1:1;
A3: 1 in [.0,1.] by XXREAL_1:1;
T3: I is satisfying_(LB)
    proof
      let y be Element of [.0,1.];
      0 <= y by XXREAL_1:1;
      hence thesis by a1,A2;
    end;
    I is satisfying_(RB)
    proof
      let x be Element of [.0,1.];
      x <= 1 by XXREAL_1:1;
      hence thesis by a1,A3;
    end;
    hence thesis by T3,a1;
  end;
