# How To Prove It, Ch-3 Sec-3.2, Proofs Involving Negations and Conditionals

## Chapter - 3, Proofs

### Summary

• There can be following ways to prove a goal of the form $\lnot P$ :
• Convert or re-express the goal to some other form and then use one of the proof strategies for this other goal form. This is generally possible when the original goal is complex goal(containing many components).
• Proof by contradiction: by assuming the goal $P$ is true and try to reach a contradiction. On a contradiction, it can be concluded that P must be false.
• Proof by contradiction is vague as it requires to produce a contradiction by proving something that is known to be false. One approach can be:
• To use a given of the form $\lnot P$: Try making $P$ as goal. Now, If $P$ can be proved, then the proof will be complete, because $P$ contradicts the given $\lnot P$.
• If not using the strategy to prove by contradiction, One approach can be:
• To re-express/convert a given of the form of $\lnot P$, to some other form.
• In previous section, strategy to prove goal of the form of $P \to Q$ was described.
• Apart from all the above strategies, One more strategy can be:
• To use a given of the form of $P \to Q$. Many strategies for using givens suggests ways for drawing inferences from the givens. These strategies are called rules of inference. There can be following two Rules of inference for using given of the form of $P \to Q$:
• Rule modus ponens : If both $P$ and $P \to Q$ are true, then $Q$ must also be true.
• Rule modus tollens : If $P → Q$ is true and $Q$ is false, then $P$ must also be false.
• Till now following strategies are covered:
• To prove a goal of the form $\lnot P$.
• To use a given of the form of $\lnot P$.
• To prove a goal of the form of $P \to Q$.
• To use a given of the form of $P \to Q$.

Soln1

(a)

Suppose $P$. Since $P \to Q$, it follows $Q$. Similarly, since $Q \to R$, it follows $R$. Thus if $P$ then $R$, or $P \to R$.

(b) Suppose $P$. For $Q \to R$, using contra-positive for proof. Suppose $\lnot R$, it follows $P \to \lnot Q$. Since $P$, it follows $\lnot Q$. Thus $\lnot R \to \lnot Q$, or $Q \to R$. It follows that $P \to (Q \to R)$.

Soln2

(a)

Suppose $P$. Since $P \to Q$, it follows $Q$. Since $R \to \lnot Q$, using contrapositive $Q \to \lnot R$, it follows $\lnot R$. Thus, $P \to \lnot R$.

(b)

Simplifying $Q \to \lnot(Q \to ¬P)$, gives $Q \to (Q \land P)$. Since P, it follows $Q \to Q$, which is always true.

Soln3

Suppose $x \in A$. Since $A \subseteq C$, it follows that $x \in C$. Now since $B \cap C = \phi$, it follows that $x \notin B$. Thus $x \in A \to x \notin B$.

Soln4

Suppose $x \in C$. Since $(A \setminus B) \cap C = \phi$, it follows that $x \notin A \setminus B$. And $x \notin A \setminus B$ is equivalent to $x \notin A \lor x \in B$. Since $x \in A$, it follows that $x \in B$. Thus $x \in C \to x \in B$.

Soln5

Suppose $a \in A \setminus B$, which means $a \in A$ and $a \notin B$. Since $a \in C$, it follows that $a \in (A \cap C)$. Since $a \in (A \cap C)$ and $a \notin B$, it follows that $A \cap C \nsubseteq B$. This contradicts the given: $A \cap C \subseteq B$. Thus $a \notin A \setminus B$.

Soln6

Since $a \in A$ and $A \subseteq B$, it follows that $a \in B$. Suppose $a \notin C$, then it follows that $a \in B \setminus C$. But it contradicts with the given $a \notin B \setminus C$. Thus $a \in C$.

Soln7

Suppose $y = 0$. Since $y + x = 2y - x$, it follows that $x = 0$. But this contradicts the given that $\lnot (x = 0 \land y = 0)$, or, equivalantly, $x \neq 0 \lor y \neq 0$. Thus $y \neq 0$.

Soln8

Suppose $a < \frac 1 a < b < \frac 1 b$. Since $a < \frac 1 a$, it follows that $a \in (-\infty, -1) \lor (0, 1)$. Similarly, $b \in (-\infty, -1) \lor (0, 1)$. Thus there are four possible cases:

• Case 1: $a \in (-\infty, -1)$ and $b \in (-\infty, -1)$. (Earlier I missed this case. As pointed out in comments this case is not possible either). Since $\, a \text{ and } b \,$ are both negative(same sign) and $\, \frac 1 a < \frac 1 b \,$, it follows that $\, a > b \,$ which contradicts with $\, a < b \,$.

• Case 2: $a \in (-\infty, -1)$ and $b \in (0, 1)$. This case is possible. For example, when $\, a = -2 \,$ and $\, b = 0.5 \,$, the inequality holds.

• Case 3: $a \in (0, 1)$ and $b \in (-\infty, -1)$. This is not possible because $\frac 1 a > 1$ and $\frac 1 a < b$. But in this case $b < -1$.

• Case 4: $a \in (0, 1)$ and $b \in (0, 1)$. This is also not possible because $\frac 1 a > 1$ and $\frac 1 a < b$. But in this case $0 < b < 1$.

Thus only Case 2 is possible. It follows that $a < -1$.

Soln9

Suppose $x^2y = 2x + y$. We will prove $(y \neq 0 \to x \neq 0)$, using contra-positive. Suppose $x = 0$, it follows that $y = 0$. Thus we have $y \neq 0 \to x \neq 0$.

Soln10

Simplifying $y = \frac {3x^2 + 2y} {x^2 + 2}$, gives $(y - 3)x^2 = 0$. Suppose $x \neq 0$. Suppose $(y - 3)x^2 = 0$, it follows that $y - 3 = 0$, or $y = 3$. Thus we have $x \neq 0 \to (y = \frac {3x^2 + 2y} {x^2 + 2} \to y = 3)$.

Soln11

(a) Reverse of $x \neq 3 \land y \neq 8$, is $x = 3 \lor x = 8$. Thus reverse of the conclusion used in proof is wrong.

(b) Using $x = 2$, and $y = 8$, gives $x + y = 10$. Thus conclusion is not correct for the given $x$, and $y$.

Soln12

(a) Statement:
Since $x \notin B$ and $B ⊆ C$, then $x \notin C$.
is not correct.

(b) Lets say $A = \{1,2\}$, $B = \{3, 4\}$ and $C = \{1, 2, 3, 4\}$. Now if $x = 1$, the theorem is not correct.

Leaving Problems 13 to 16, as all involve truth tables and such problems were already solved in earlier chapters.

Soln17

Suppose $x^2 + y = 13$. Suppose $y = 4$, it follows that $x^2 + 4 = 13$, or $x = 3$ or $x = -3$. Thus for $y = 4$ and $x = -3$, there is no contradiction. Thus theorem is not correct.