# How to Prove It - Solutions

## Chapter - 2, Quantificational Logic

### Summary

• Sets can be defined in following ways:
• Listing all elements of the set. Eg: $\{1, 2, 3, 4, 5\}, or \{1, 3, 5, 7, 9, 11, 13, 15, ... \}$.
• Element-hood Test Notation: $\{ x\,\vert\,P(x) \}$. Or sometimes replacing $x$ by an expression eg: $\{ n^2\,\vert\,n \in \mathbb{N} \}$.
• A set can also be defined using indexed family notation. For eg: $P = \{p_i\,\vert\,i ∈ I\}$, where $I = \{ i ∈ N\,\vert\,1 ≤ i ≤ 100 \}$.
• An indexed family $A = \{x_i\,\vert\,i ∈ I\}$ can also be defined as $A = \{x\,\vert\,∃i ∈ I (x = xi )\}$. Thus $x ∈ \{x_i\,\vert\,i ∈ I \}$ means the same thing as $∃i ∈ I(x = x_i)$.
• Sets whose elements are all sets are called families of sets. Eg: $\mathcal{F} = \{ \{1,2\}, \{2,3,4\}, \{6,7,10\} \}$.
• Power Set: $\mathcal{P}(A) = \{x\,\vert\,x ⊆ A\}$.
• If $\mathcal{F}$ is a family of sets, then:
• Intersection of all sets in $\mathcal{F} = \cap \mathcal{F} = \{x\,\vert\,∀A ∈ F(x ∈ A)\} = \{x\,\vert\,∀A(A ∈ F → x ∈ A)\}$.
• Union of all sets in $\mathcal{F} = \cup \mathcal{F} = \{x\,\vert\,∃A ∈ F(x ∈ A)\} = \{x\,\vert\,∃A(A ∈ F \land x ∈ A)\}$.
• Some useful logical forms:
• $x ∈ \{x_i\,\vert\,i ∈ I \}$ is equivalent to $∃i ∈ I(x = x_i)$.
• $x ∈ \mathcal{P}(A)$ is equivalent to $x ⊆ A$ which is equivalent to $∀y(y ∈ x → y ∈ A)$.
• $x ∈ ∩ \mathcal{F}$ is equivalent to $∀A ∈ \mathcal{F}(x ∈ A)$, or equivalently,$∀A(A ∈ \mathcal{F} → x ∈ A)$.
• $x ∈ ∪ \mathcal{F}$ is equivalent to $∃A ∈ \mathcal{F}(x ∈ A)$.
• An alternative notation of union or intersection. If $\mathcal{F} = \{A_i\,\vert\,i ∈ I\}$, where each $A_i$ is a set. Then $∩ \mathcal{F}$ would be the set of all elements common to all the $A_i$’s, for $i ∈ I$, and this can also be written as $∩_{i∈I} A_i$.

Soln1

(a) $\mathcal F \subseteq \mathcal P(A)$
$\quad = \forall x (x \in \mathcal F \to x \in \mathcal P(A)$
$\quad = \forall x (x \in \mathcal F \to x \subseteq A$
$\quad = \forall x (x \in \mathcal F \to \forall y(y \in x \to y \in A))$

(b) $A ⊆ \{ 2n+1\,\vert\,n ∈ \mathbb N \}$
$\quad = \forall x ( x \in A \to x \in \{ 2n+1\,\vert\,n ∈ \mathbb N \} )$
$\quad = \forall x ( x \in A \to \exists n \in \mathbb N( x = 2n+1) )$

(c) $\{n^2 + n + 1\,\vert\,n ∈ N \} ⊆ \{ 2n + 1\,\vert\,n ∈ \mathbb N \}$
$\quad = \forall x ( x \in \{n^2 + n + 1\,\vert\,n ∈ \mathbb N \} \to x \in \{ 2n + 1\,\vert\,n ∈ \mathbb N \} )$
$\quad = \forall x ( \exists n \in \mathbb N(x = n^2 + n + 1) \to \exists m \in \mathbb N (x = 2m + 1) )$
This is equivalent to:
$\quad = \exists n \in \mathbb N \exists m \in \mathbb N(n^2 + n + 1 = 2m + 1)$.

(d) $\mathcal P (∪_{i ∈ I} A_i) \nsubseteq ∪_{i ∈ I} \mathcal P(A_i)$
$\quad = \exists x ( (x \in \mathcal P (∪_{i ∈ I} A_i)) \land \lnot (x \in ∪_{i ∈ I} \mathcal P(A_i)))$
$\quad = \exists x ( (x \subseteq ∪_{i ∈ I} A_i) \land \lnot (\exists i \in I(x \in \mathcal P(A_i))))$
$\quad = \exists x ( \forall y(y \in x \to y \in ∪_{i ∈ I} A_i) \land \lnot (\exists i \in I(x \subseteq A_i)))$
$\quad = \exists x ( \forall y(y \in x \to \exists i \in I(y \in A_i)) \land \forall i \in I \lnot(x \subseteq A_i))$
$\quad = \exists x ( \forall y(y \in x \to \exists i \in I(y \in A_i)) \land \forall i \in I \lnot \forall y( y \in x \to y \in A_i))$
$\quad = \exists x ( \forall y(y \in x \to \exists i \in I(y \in A_i)) \land \forall i \in I \exists y \lnot ( y \notin x \lor y \in A_i))$
$\quad = \exists x ( \forall y(y \in x \to \exists i \in I(y \in A_i)) \land \forall i \in I \exists y ( y \in x \land y \notin A_i))$

Soln2

(a) $x ∈ ∪ \mathcal F \setminus ∪ \mathcal G$
$\quad = x ∈ ∪ \mathcal F \land \lnot x \in ∪ \mathcal G$
$\quad = \exists A \in \mathcal F(x \in A) \land \lnot \exists A \in \mathcal G(x \in A)$
$\quad = \exists A \in \mathcal F(x \in A) \land \forall A \in \mathcal G \lnot(x \in A)$
$\quad = \exists A \in \mathcal F(x \in A) \land \forall A \in \mathcal G(x \notin A)$

(b) $\{ x \in B\,\vert\,x \notin C\} ∈ \mathcal P(A)$
$\quad = \{ x \in B\,\vert\,x \notin C\} \subseteq A$
$\quad = \forall y (y \in \{ x \in B\,\vert\,x \notin C\} \to y \in A )$
$\quad = \forall y ((y \in B \land y \notin C) \to y \in A )$

(c) $x ∈ ∩_{i∈I}(A_i ∪ B_i)$
$\quad = \forall i \in I(x \in (A_i ∪ B_i))$
$\quad = \forall i \in I(x \in A_i \lor x \in B_i)$

(d) $x ∈ (∩_{i ∈ I}A_i) ∪ (∩_{i ∈ I}B_i)$
$\quad = x ∈ (∩_{i ∈ I}A_i) \lor x \in (∩_{i ∈ I}B_i)$
$\quad = \forall i ∈ I(x \in A_i) \lor \forall i ∈ I(x \in B_i)$

Soln3 $\mathcal P({\phi}) = \{ \phi, \{ \phi \} \}$

Soln4 $\cap \mathcal F = \{ \text{red, blue} \}$
$\cup \mathcal F = \{ \text{red, green, blue, orange, purple} \}$

Soln5 $\cap \mathcal F = \phi$
$\cup \mathcal F = \{ 3, 5, 7, 12, 16, 23 \}$

Soln6

Given: $A_i = \{i, i + 1, i −1, 2i \}$ where $I = \{ 2, 3, 4, 5\}$

$A_2 = \{2, 3, 1, 4 \}$
$A_3 = \{3, 4, 2, 6 \}$
$A_4 = \{4, 5, 3, 8 \}$
$A_5 = \{5, 6, 4, 10 \}$

$\cap_{i \in I} A = \{ 4 \}$
$\cup_{i \in I} A = \{ 1, 2, 3, 4, 5, 6, 8, 10 \}$

Soln8

Given: $A_i = \{i,2i\}$ and $B_i = \{i, i+1 \}$ where $I = \{2, 3\}$

(a)

$A_2 = \{ 2, 4\}$
$A_3 = \{ 3, 6\}$

$B_2 = \{ 2, 3\}$
$B_3 = \{ 3, 4\}$

(b)

$∩_{i∈I}(A_i ∪ B_i) = \{3, 4\}$
$(∩_{i ∈ I}A_i) ∪ (∩_{i ∈ I}B_i) = \{3 \}$

(c) No. The given statements are not equivalent.

Soln9

$I = \{1, 2 \}$
$A_1 = \{2, 3 \}, \quad B_1 = \{2, 5\}$
$A_2 = \{3, 5 \}, \quad B_2 = \{7, 11\}$

$A_1 \cup A_2 = \{ 2, 3, 5\}, \quad B_1 \cup B_2 = \{ 2, 5, 7, 11\}$
$(A_1 \cup A_2) \cap (B_1 \cup B_2) = \{ 2, 5 \}$

$A_1 \cap B_1 = \{ 2 \}, \quad A_2 \cap B_2 = \phi$
$(A_1 \cap B_1) \cup (A_2 \cap B_2) = \{ 2 \}$

Thus $(A_1 \cup A_2) \cap (B_1 \cup B_2) \neq (A_1 \cap B_1) \cup (A_2 \cap B_2)$

Soln10

$x ∈ \mathcal P(A) ∩ \mathcal P (B)$
$\quad = x ∈ \mathcal P(A) \land x ∈ \mathcal P (B)$
$\quad = x \subseteq A \land x \subseteq B$
$\quad = \forall y (y \in x \to y \in A ) \land \forall y (y \in x \to y \in B )$
$\quad = \forall y \in x(y \in A) \land \forall y \in x(y \in B )$
$\quad = \forall y \in x(y \in A \land y \in B )$
$\quad = \forall y \in x(y \in (A \cap B))$
$\quad = x \subseteq (A \cap B))$
$\quad = x \to \mathcal P(A \cap B))$ Hence Proved.

Soln11

$Taking A = \{1, 2\}$, and $B = \{2, 3\}$
$A \cup B = \{1, 2, 3\}$

$\mathcal P(A) = \{\{1\}, \{2\}, \{1, 2\}, \phi \}$
$\mathcal P(B) = \{\{2\}, \{3\}, \{2, 3\}, \phi \}$
$\mathcal P(A) \cup \mathcal P(B) = \{\{1\}, \{2\}, \{3\}, \{1, 2\}, \{2, 3\}, \phi \}$

And $\mathcal P(A \cup B) = \{\{1\}, \{2\}, \{3\}, \{1, 2\}, \{2, 3\}, \{1, 3\}, \{1, 2, 3\}, \phi \}$
Clearly $\mathcal P(A \cup B) \neq \mathcal P(A) \cup \mathcal P(B)$

Soln12

(a)

$x \in ∪_{i ∈ I}(A_i ∪ B_i)$
$\quad = \exists i \in I(x \in (A_i ∪ B_i)$
$\quad = \exists i \in I(x \in A_i \lor x \in B_i)$
Existential quantifier distributes over interjection:
$\quad = \exists i \in I(x \in A_i) \lor \exists i \in I(x \in B_i)$
$\quad = x \in ∪_{i∈I}A_i \lor x \in ∪_{i ∈ I}B_i$
$\quad = x \in ((∪_{i∈I}A_i) \cup (∪_{i ∈ I}B_i))$
This is equivalent to RHS.

(b)

$x \in ( ∩ \mathcal F) ∩ ( ∩ \mathcal G)$
$\quad = x \in ( ∩ \mathcal F) \land x \in ( ∩ \mathcal G)$
$\quad = (\forall A \in \mathcal F(x \in A)) \land (\forall A \in \mathcal G(x \in A))$
$\quad = \forall A(A \in \mathcal F \to x \in A) \land \forall A(A \in \mathcal G \to x \in A)$
Using reverse of, Universal Quantifier distributes over conjunction: $\quad = \forall A((A \in \mathcal F \to x \in A) \land (A \in \mathcal G \to x \in A))$
$\quad = \forall A((A \notin \mathcal F \lor x \in A) \land (A \notin \mathcal G \lor x \in A))$
Using inverse of law of distribution:
$\quad = \forall A((A \notin \mathcal F \land A \notin \mathcal G) \lor x \in A)$
Using Demorgan’s Law:
$\quad = \forall A(\lnot(A \in \mathcal F \lor A \in \mathcal G) \lor x \in A)$
$\quad = \forall A((A \in \mathcal F \lor A \in \mathcal G) \to x \in A)$
$\quad = \forall A \in (\mathcal F \cup \mathcal G)(x \in A)$
$\quad = x \in \cap(\mathcal F \cup \mathcal G)$
= RHS.

(c)

$x \in \cap_{i∈I}(A_i \setminus B_i)$
$\quad = \forall i \in I(x \in (A_i \setminus B_i))$
$\quad = \forall i \in I((x \in A_i) \land \lnot (x \in B_i))$
$\quad = \forall i \in I(x \in A_i) \land \forall i \in I (\lnot (x \in B_i))$
$\quad = \forall i \in I(x \in A_i) \land \lnot \exists i \in I (x \in B_i)$
$\quad = (x \in \cap_{i \in I}A_i) \land (\lnot x \in \cup_{i \in I} B_i)$
$\quad = x \in (\cap_{i \in I}A_i \setminus \cup_{i \in I} B_i)$
= RHS.

Soln13

Given: $A_{i,j} = \{i, j,i + j\}$ where $I = \{1, 2\}$ and $J = \{3, 4\}$

(a) $B_j = ∪_{i∈I} A_{i,j} = A_{1,j} ∪ A_{2,j}$
$B_3 = A_{1,3} \cup A_{2,3} = \{1, 3, 4\} \cup \{2, 3, 5\} = \{1, 2, 3, 4, 5\}$
$B_4 = A_{1,4} \cup A_{2,4} = \{1, 4, 5\} \cup \{2, 4, 6\} = \{1, 2, 4, 5, 6\}$

(b)

Putting values from (a) : $B_3 \cap B_4 = \{1, 2, 4, 5\}$

(c)

To find $\cup{i ∈ I}(\cap_{j∈J} A_{i,j})$, lets first compute $C_i = \cap_{j∈J} A_{i,j} = A_{i,3} \cap A_{i,4}$
$C_1 = A_{1,3} \cap A_{1,4} = \{1, 3, 4\} \cap \{1, 4, 5\} = \{1, 4\}$
$C_2 = A_{2,3} \cap A_{2,4} = \{2, 3, 5\} \cap \{2, 4, 6\} = \{2\}$
Thus we have: $\cup{i ∈ I}(\cap_{j∈J} A_{i,j}) = C_1 \cup C_2 = \{1, 2, 4\}$
No, expression of (b) and (c) parts are not equal.

(d)

LHS: $x ∈ \cap_{j \in J} (\cup_{i \in I} A_{i, j} )$
$\quad = \forall j \in J \exists i \in I(x \in A_{i, j})$

RHS: $x ∈ \cup_{i \in I} (\cap_{j \in J} A_{i, j} )$
$\quad = \exists i \in I \forall j \in J(x \in A_{i, j})$

Clearly LHS and RHS are not equal. We already saw this that if order of quantifiers are changed, expressions are also changed.

Soln14

(a)

$x ∈ \cup \mathcal F$
$\quad = \exists A \in \mathcal F (x \in A)$
$\quad = \exists A(A \in \mathcal F \land x \in A)$
If $\mathcal F = \phi$, $A \in \mathcal F$ will always be false.
Thus whole statement will be false. Thus $x ∈ \cup \phi = false$, irrespective of the value of x.
which means $\cup \phi = \phi$.

(b)

$x ∈ \cap \mathcal F$
$\quad = \forall A(A \in \mathcal F \to x \in A)$
If $\mathcal F = \phi$, $A \in \mathcal F$ will always be false.
Thus $A \in \mathcal F \to x \in A$ will always be true.
That means $x ∈ \cap \mathcal \phi = true$, irrespective of the value of x.
That means $\cap \phi = U$.

Soln15

(a)

• According to the definition, R is the set of all sets that does not contains themselves.
• R itself should also be a set that does not contain itself, as if it contains itself it will contradict its own definition.
• Now if R does not contain itself, then again it contradicts with its definition which says it is a set that contains all the sets that does not contain themselves.
• Thus such set R can not exist.

(b) It follows that there is no universal set of sets.