First, we'll prove that either a proposition P holds, or else it implies everything.

To do so, we will admit that we can quantify over and form sets of propositions and that we have the axiom of choice. We'll let T stand for a true proposition, whichever you like.

Code:

1) Consider the set of propositions Q such that P or Q. This is non-empty since it at least contains T.
2) Consider the set of propositions R such that either P or else (R implies everything). This is non-empty since it at least contains the proposition "everything holds".
3) Pair the above sets and use the axiom of choice to pick out a Q and R respectively.
4) Assume P
5) Then the sets considered above are the same set: the universal set of propositions.
6) Thus choice picked out only one proposition above. In other words, Q = R
7) Assume Q
8) Assume R implies everything
9) Hence Q implies everything (from 6 and 8)
10) Everything (from 7 and 9)
11) If Q and R implies everything, then everything (from 7-10)
12) If P, Q and R implies everything, then everything (from 4-11).
13) (if Q and R implies everything) then P implies everything (reordering the antecedents of 12)
14) (P or Q) and (P or R implies everything) (from 1, 2 and 3)
15) P or (Q and R implies everything) (from 14 by distributivity)
16) P or P implies everything (from 13 and 15)

As a corollary:

Code:

1) Assume P.
2) Assume not-P.
3) Have P or P implies everything (from the proof above)
4) P implies everything (disjunctive syllogism from 2 and 3).
5) Anything. (from 1 and 4).