1549901369-Elements_of_Real_Analysis__Denlinger_

610 Appendix A • Logic and P roofs

Logicians call this tautology "constructive dilemma." It can be proved by

a truth table. In words, it says

[

If we know that at least one of two possibilities l

("cases") must occur, and if each of these cases

implies R , then R must be true.

PROOF BY CASES:

To prove the theorem

we can first prove

and then prove both

Case 1: [(H1 /\ H 2 /\ · · · /\ Hn) /\ Pi] :::::> C, and

Case 2: [(H1 /\ H2 /\ · · · /\ Hn ) /\ P2] :::::> C.

Of course, proof by cases can be extended to more than two cases. That is, we

can apply this procedure when we have (P 1 V P2 V · · · V Pn) rather than just

(P 1 V P2). Then our proof would break down into Case 1, Case 2, · · · , Case n.

(PS-6') PROOF BY CASES-Using Constructive Dilemma:

To prove the theorem

we need only prove

Case 1: [(H1 /\ H2 /\ · · · /\ Hn) /\ P] :::::> C, and

Case 2: [(H1 /\ H 2 /\ · · · /\ Hn) /\ ,..., P] :::::> C,

where P is any proposition of your choosing.

(PS-7) PROVING THAT SEVERAL STATEMENTS ARE EQUIV-

ALENT:

To prove the

Theorem: P 1 , P2, · · ·, Pn, are all equivalent,

we prove the equivalent

Theorem: P 1 :::::> P2 :::::> · · · :::::> Pn :::::> P1.
Proving the latter establishes a "circular" relationship, showing that any one
of these propositions will imply all of the others.