Page 144 - HOW TO PROVE IT: A Structured Approach, Second Edition
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                                   130                         Proofs
                                   some other statement R, and then using R to deduce Q; and suppose that the
                                   same steps could be used, in reverse order, to prove that Q → P. In other
                                   words, you could assume Q, use this assumption to conclude that R was true,
                                   and then use R to prove P. Since you would be asserting both P → R and
                                   R → P, you could sum up these two steps by saying P ↔ R. Similarly, the
                                   other two steps of the proof tell you that R ↔ Q. These two statements imply
                                   the goal P ↔ Q. Mathematicians sometimes present this kind of proof by
                                   simply writing the string of equivalences

                                                             P iff R iff Q.
                                   You can think of this as an abbreviation for “P iff R and R iff Q (and therefore
                                   P iff Q).” This is illustrated in the next example.

                                   Example 3.4.4. Suppose A, B, and C are sets. Prove that A ∩ (B \ C) =
                                   (A ∩ B) \ C.

                                   Scratch work
                                   As we saw in Chapter 2, the equation A ∩ (B \ C) = (A ∩ B) \ C means
                                   ∀x(x ∈ A ∩ (B \ C) ↔ x ∈ (A ∩ B) \ C), but it is also equivalent to
                                   the statement [A ∩ (B \ C) ⊆ (A ∩ B) \ C] ∧ [(A ∩ B) \ C ⊆ A ∩ (B \ C)].
                                   This suggests two approaches to the proof. We could let x be arbitrary and then
                                   prove x ∈ A ∩ (B \ C) ↔ x ∈ (A ∩ B) \ C, or we could prove the two state-
                                   ments A ∩ (B \ C) ⊆ (A ∩ B) \ C and (A ∩ B) \ C ⊆ A ∩ (B \ C). In fact,
                                   almost every proof that two sets are equal will involve one of these two ap-
                                   proaches. In this case we will use the first approach, so once we have introduced
                                   our arbitrary x, we will have an iff goal.
                                     For the (→) half of the proof we assume x ∈ A ∩ (B \ C) and try to prove
                                   x ∈ (A ∩ B) \ C:

                                                Givens                         Goal
                                            x ∈ A ∩ (B \ C)               x ∈ (A ∩ B) \ C

                                     To see the logical forms of the given and goal, we write out their definitions
                                   as follows:

                                      x ∈ A ∩ (B \ C)iff x ∈ A ∧ x ∈ B \ C iff x ∈ A ∧ x ∈ B ∧ x /∈ C;
                                      x ∈ (A ∩ B) \ C iff x ∈ A ∩ B ∧ x /∈ C iff x ∈ A ∧ x ∈ B ∧ x /∈ C .
                                     At this point it is clear that the given implies the goal, since the last steps
                                   in both strings of equivalences turned out to be identical. In fact, it is also
                                   clear that the reasoning involved in the (←) direction of the proof will be
                                   exactly the same, but with the given and goal columns reversed. Thus, we
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