Mathematical-Reasoning · Unit 5 · Video 4 · Interactive Practice

Statements With Two Quantifiers: Why the Order Matters

IKey Formulas

Statement	Truth value	What decides it
$(\forall x \in \mathbb{Z})(\exists y \in \mathbb{Z})(x + y = 0)$	True	$y$ is chosen after $x$ , so it may depend on $x$ : the rule $y = -x$ works for every $x$
$(\exists x \in \mathbb{Z})(\forall y \in \mathbb{Z})(x + y = 0)$	False	one fixed $x$ must work for every $y$ , and $y = 1 - x$ gives $x + y = 1 \ne 0$
$\neg(\exists x \in U)(\forall y \in U)P(x, y) \equiv (\forall x \in U)(\exists y \in U)\neg P(x, y)$	Negation rule	the negation theorem applied once per quantifier
$\neg(\forall x \in U)(\exists y \in U)P(x, y) \equiv (\exists x \in U)(\forall y \in U)\neg P(x, y)$	Negation rule	flip each quantifier, keep the order $x$ then $y$ , negate $P$

Key Insight: to negate a string of quantifiers, flip every quantifier, keep their order, and negate the predicate. Swapping two quantifiers of the same kind changes nothing, but swapping a $\forall$ with an $\exists$ can change the truth value.

IIThe Truth Grid: Columns Decide

With $x$ as the column, $(\forall x)(\exists y)$ needs a true point in every column; $(\exists x)(\forall y)$ needs one all-true column.

Statement Predicate P(x, y) Outer value x = 5

💡 The grid shows only $-6$ to $6$ ; each verdict rests on a rule, such as $y = 1 - x$ , that works for every integer.

IIIBuild the Negation

A two-quantifier statement is negated by flipping every quantifier, keeping their order, and negating the predicate.

Statement

IVQuiz Questions

Problem 1 · Flip, Keep, Negate

Given: the statement $(\forall x \in U)(\exists y \in U)P(x, y)$ . Which statement is logically equivalent to its negation?

✅ Correct! Both quantifiers flip,

x

still comes before

y

, and the predicate is negated.

❌ Close, but… only the outer quantifier flipped. The negation theorem applies a second time to

(\exists y \in U)P(x, y)

, turning its

\exists

into

\forall

❌ Not quite. The quantifiers flipped, but

y

now comes before

x

. Negation keeps the order, and swapping a

\forall

with an

\exists

changes the meaning: here one

y

would have to work for every

x

❌ Not quite. Only the predicate was negated. Moving the

\neg

inward past each quantifier flips that quantifier, so both

\forall x

and

\exists y

must change.

Show solution

Treat the statement as $(\forall x \in U)Q(x)$ , where $Q(x)$ is $(\exists y \in U)P(x, y)$ . The negation theorem gives

\neg(\forall x \in U)Q(x) \equiv (\exists x \in U)\neg Q(x).

Apply the theorem again to $Q(x)$ :

\neg Q(x) \equiv \neg(\exists y \in U)P(x, y) \equiv (\forall y \in U)\neg P(x, y).

Combining the two results,

\neg(\forall x \in U)(\exists y \in U)P(x, y) \equiv (\exists x \in U)(\forall y \in U)\neg P(x, y).

Problem 2 · Which Order Is True?

Given: the universal set $\mathbb{Z}$ and the predicate $y > x$ . Exactly one of these statements is true. Which one?

✅ Correct! Here

y

is chosen after

x

, and the rule

y = x + 1

gives

y > x

for every integer

x

❌ Not quite. This says one integer

y

is greater than every integer

x

. Whatever

y

is proposed,

x = y

gives

y > y

, which is false. Reading it as “every

x

has a larger

y

” is the order misreading.

❌ Not quite. This says one fixed integer

x

is less than every integer

y

. Whatever

x

is proposed,

y = x

gives

x > x

, which is false.

❌ Not quite. This claims every pair satisfies

y > x

, so one failing pair makes it false:

x = 2

y = 1

gives

1 > 2

, which is false.

Show solution

(a) For each integer $x$ , choose $y = x + 1$ . It is an integer and $y = x + 1 > x$ . So (a) is true.

(b) For any proposed integer $y$ , take $x = y$ ; then $y > x$ reads $y > y$ , which is false. No $y$ works, so (b) is false.

(c) For any proposed integer $x$ , take $y = x$ ; then $y > x$ reads $x > x$ , which is false. No $x$ works, so (c) is false.

(d) The pair $x = 2$ , $y = 1$ is a counterexample, since $1 > 2$ is false. So (d) is false.

Statements (a) and (b) use the same quantifiers on the same predicate in different orders, and only (a) is true.

Problem 3 · Negate, Then Decide

Given: the statement $(\exists x \in \mathbb{Z})(\forall y \in \mathbb{Z})(xy = y)$ . Find its negation, then decide which of the two is true and what evidence settles it.

What is the negation?

Which is true, and why?

✅ Correct! The statement is true with the witness

x = 1

, so its negation is false: for

x = 1

there is no

y

with

1 \cdot y \ne y

❌ Close, but… both quantifiers flipped, but the predicate was not negated:

xy = y

must become

xy \ne y

❌ Not quite. Only the predicate was negated. Each quantifier must flip as well:

\exists x

becomes

\forall x

and

\forall y

becomes

\exists y

❌ Not quite. The quantifiers flipped, but the order changed to

y

before

x

. Negation keeps the order

x

then

y

❌ Not quite. Flip every quantifier, keep the order, and negate the predicate.

❌ Not quite. A single pair can prove a statement of the form

(\exists x)(\exists y)

, but the negation begins with “for every

x

”, so it needs a

y

for each

x

, and

x = 1

has none, since

1 \cdot y = y

for every

y

❌ Not quite.

0 \cdot y = 0

, not

y

; for example

y = 3

gives

0 \cdot 3 = 0 \ne 3

. The value

x = 0

is not a witness.

❌ Not quite. The rule

y = 1

fails at

x = 1

, since

1 \cdot 1 = 1

. In fact no rule works there, because

x = 1

makes

xy = y

for every

y

❌ Not quite. An

(\exists x)(\forall y)

statement is proved by one

x

that works for every

y

Show solution

Negation. Apply the negation theorem once per quantifier, keeping the order:

\neg(\exists x \in \mathbb{Z})(\forall y \in \mathbb{Z})(xy = y) \equiv (\forall x \in \mathbb{Z})\neg(\forall y \in \mathbb{Z})(xy = y)

\equiv (\forall x \in \mathbb{Z})(\exists y \in \mathbb{Z})\neg(xy = y) \equiv (\forall x \in \mathbb{Z})(\exists y \in \mathbb{Z})(xy \ne y).

Truth value. The statement asks for one fixed integer $x$ with $xy = y$ for every integer $y$ . The integer $x = 1$ works, since $1 \cdot y = y$ for every $y$ . So the statement is true, and its negation is false.

The value $x = 1$ also shows directly why the negation fails: it claims that every $x$ has some $y$ with $xy \ne y$ , but for $x = 1$ every $y$ gives $1 \cdot y = y$ .

Problem 4 · Negation in Words

Given: the statement “There exists a real number $x$ such that for every real number $y$ , $x \le y^2$ .” Which sentence is its negation?

✅ Correct! “There exists

x

” became “for every

x

”, “for every

y

” became “there exists

y

”, the order stayed

x

then

y

, and

x \le y^2

became

x > y^2

❌ Not quite. The inequality is negated correctly, but neither quantifier flipped.

❌ Close, but… the quantifiers flipped, but the sentence now chooses

y

first and

x

second. Negation keeps the order:

x

is quantified first.

❌ Close, but… the quantifiers are right, but the negation of

x \le y^2

x > y^2

, not

x < y^2

; the case

x = y^2

belongs to the original inequality.

Show solution

In symbols the statement is $(\exists x \in \mathbb{R})(\forall y \in \mathbb{R})(x \le y^2)$ . Flip each quantifier, keep the order, negate the predicate:

\neg(\exists x \in \mathbb{R})(\forall y \in \mathbb{R})(x \le y^2) \equiv (\forall x \in \mathbb{R})(\exists y \in \mathbb{R})(x > y^2).

In words, following the writing guideline: “For every real number $x$ , there exists a real number $y$ such that $x > y^2$ .”

The original statement is true, since $x = 0$ gives $0 \le y^2$ for every real $y$ . So the negation is false; $x = 0$ shows it directly, because no real $y$ has $y^2 < 0$ .

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