theorem
  <%>E ^+ p = p & p +^ <%>E = p
proof
A1: now
    let k be Nat;
    assume k in dom p;
    hence (<%>E ^+ p).k = {}^(p.k) by Def2
      .= p.k;
  end;
  dom (<%>E ^+ p) = dom p by Def2;
  hence <%>E ^+ p = p by A1,FINSEQ_1:13;
A2: now
    let k be Nat;
    assume k in dom p;
    hence (p +^ <%>E).k = (p.k)^{} by Def3
      .= p.k;
  end;
  dom (p +^ <%>E) = dom p by Def3;
  hence thesis by A2,FINSEQ_1:13;
end;
