Page 139 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Conjunctions and Biconditionals 125
Example 3.4.1. Suppose A ⊆ B, and A and C are disjoint. Prove that
A ⊆ B \ C.

Scratch work
Givens Goal
A ⊆ B A ⊆ B \ C
A ∩ C = ∅

Analyzing the logical form of the goal, we see that it has the form ∀x(x ∈
A → x ∈ B \ C), so we let x be arbitrary, assume x ∈ A, and try to prove that
x ∈ B \ C. The new goal x ∈ B \ C means x ∈ B ∧ x /∈ C, so according to
our strategy we should split this into two goals, x ∈ B and x /∈ C, and prove
them separately.
Givens Goals
A ⊆ B x ∈ B
A ∩ C = ∅ x /∈ C
x ∈ A
The final proof will have this form:

Let x be arbitrary.
Suppose x ∈ A.
[Proof of x ∈ B goes here.]
[Proof of x /∈ C goes here.]
Thus, x ∈ B ∧ x /∈ C,so x ∈ B \ C.
Therefore x ∈ A → x ∈ B \ C.
Since x was arbitrary, ∀x(x ∈ A → x ∈ B \ C), so A ⊆ B \ C.
The first goal, x ∈ B, clearly follows from the fact that x ∈ A and A ⊆ B.
The second goal, x /∈ C, follows from x ∈ A and A ∩ C =∅. You can see
this by analyzing the logical form of the statement A ∩ C =∅. It is a negative
statement, but it can be reexpressed as an equivalent positive statement:
A ∩ C =∅ is equivalent to ¬∃y(y ∈ A ∧ y ∈ C) (definitions of ∩ and ∅),
which is equivalent to ∀y¬(y ∈ A ∧ y ∈ C) (quantifier negation law),
which is equivalent to ∀y(y /∈ A ∨ y /∈ C) (DeMorgan’s law),
which is equivalent to ∀y(y ∈ A → y /∈ C) (conditional law).
Plugging in x for y in this last statement, we see that x ∈ A → x /∈ C, and since
we already know x ∈ A, we can conclude that x /∈ C.
Solution

Theorem. Suppose A ⊆ B, and A and C are disjoint. Then A ⊆ B \ C