Lemma (Impl). (P ==> Q) <==> (P \/ Q) <==> Q.

Proof. We parse the formula to be proved as ((P ==> Q) <==> (P \/ Q)) <==> Q, and start from the left hand side, since it is more complex.

             P ==> Q <==> P \/ Q
  <==> {by def. of implication}
             (!P \/ Q) <==> P \/ Q
 <==> {by disjunction distributes over the equivalence, to pull Q out of both sides}
            !P \/P <==> Q
 <==> {by law of the excluded middle}
            true <==> Q
 <==> {by predicate calculus}
            Q


Lemma (weakening) X ==> X \/ Y.

Proof: We tried this in different ways, but what worked was the suggestion to reduce the entire formula to true.

     X ==> X \/ Y
 = {by def. of implication}
     !X \/ X \/ Y
  = {by !A \/ A = true}
     true \/ Y
  = {by true is the zero of disjunction}
     true