Nested Quantifiers Essentials Cheatsheet

Foundations of Quantifiers

QUANTIFIER SYMBOLS & BASICS

`∀` (Universal Quantifier)	Meaning: “For all”, “For every”, “For any”. Truth Condition: True if the predicate holds for every element in the domain. Example: `∀x P(x)` - “For every x, P(x) is true.” Tip: To prove `∀x P(x)`, you must show `P(x)` is true for every single `x`. To disprove, find just one counterexample.
`∃` (Existential Quantifier)	Meaning: “There exists”, “For some”, “At least one”. Truth Condition: True if the predicate holds for at least one element in the domain. Example: `∃x P(x)` - “There exists an x such that P(x) is true.” Tip: To prove `∃x P(x)`, you only need to find one example. To disprove, you must show `P(x)` is false for every `x`.
Domain of Discourse	The set of all possible values for the variables in a quantified statement. Crucial for truth evaluation. Example: `∀x (x > 0)` If domain is integers `Z`, it’s false (e.g., x=-1). If domain is positive integers `Z+`, it’s true.
Free vs. Bound Variables	`Bound`: A variable is bound if it falls within the scope of a quantifier (e.g., `x` in `∀x P(x)`). `Free`: A variable is free if it is not bound by any quantifier. Example: In `P(x, y) ∧ ∃y Q(y, z)`, `x` and `z` are free, while the first `y` is free and the second `y` is bound.
COMMON MISTAKE	Forgetting the domain of discourse. The truth value of a quantified statement always depends on its domain. Strategy Tip: Always state your domain of discourse clearly before evaluating or constructing proofs.
SCOPE OF QUANTIFIERS	A quantifier’s scope extends to the smallest subformula that immediately follows it. Parentheses `()` are used to explicitly define scope. Example: `∀x (P(x) → Q(x))` means “For all x, if P(x) then Q(x).” `∀x P(x) → Q(x)` means “(For all x, P(x)) implies Q(x).” (Q(x) is outside the scope of ∀x and x here is a free variable in Q(x)).

NESTED QUANTIFIER STRUCTURES

Order Matters! The sequence of quantifiers drastically changes the meaning of a statement, especially when mixing ∀ and ∃.

∀x ∃y P(x, y)

Plain English: “For every x, there exists a y such that P(x, y) is true.”
Interpretation: For each x you pick, you can find some y (which might depend on x) that makes P(x, y) true. Each x gets its own y.
Real-world Example: Let P(x, y) be “x loves y”. Domain is people. “Everybody loves somebody.” (Each person has someone they love, but not necessarily the same person for everyone).

∃y ∀x P(x, y)

Plain English: “There exists a y such that for every x, P(x, y) is true.”
Interpretation: There is a single, specific y that makes P(x, y) true for all x. This y is independent of x.
Real-world Example: Let P(x, y) be “x loves y”. “There is somebody whom everyone loves.” (One specific person is loved by everyone).

∀x ∀y P(x, y) (or ∀y ∀x P(x, y))

Plain English: “For all x and for all y, P(x, y) is true.”
Interpretation: The predicate P(x, y) holds for every possible pair of x and y from the domain.
Order: Order does not matter for two universal quantifiers.

∃x ∃y P(x, y) (or ∃y ∃x P(x, y))

Plain English: “There exists an x and there exists a y such that P(x, y) is true.”
Interpretation: There is at least one pair (x, y) for which P(x, y) holds.
Order: Order does not matter for two existential quantifiers.

STRATEGY TIP: Dependency

Think of nested quantifiers as a game:

If ∀ comes first, the y you pick can depend on the x chosen by your opponent.
If ∃ comes first, you pick a y first, and it must work for all x chosen by your opponent later.

Evaluating & Negating Quantifiers

TRUTH VALUE ANALYSIS

To determine the truth value of a nested quantified statement over a finite domain, systematically check all possibilities.

Example: ∀x ∃y (x + y = 0) over domain D = {-1, 0, 1}.

For x = -1: Is there a y in D such that -1 + y = 0? Yes, y = 1. (True for x=-1)
For x = 0: Is there a y in D such that 0 + y = 0? Yes, y = 0. (True for x=0)
For x = 1: Is there a y in D such that 1 + y = 0? Yes, y = -1. (True for x=1)

Since for every x, we found a y, the statement ∀x ∃y (x + y = 0) is True.

Example: ∃y ∀x (x + y = 0) over domain D = {-1, 0, 1}.

Try y = -1: Is it true for all x that x + (-1) = 0? No, if x = 1, 1 + (-1) = 0 (True), but if x = 0, 0 + (-1) ≠ 0 (False). So, y = -1 doesn’t work for all x.
Try y = 0: Is it true for all x that x + 0 = 0? No, if x = 1, 1 + 0 ≠ 0 (False). So, y = 0 doesn’t work for all x.
Try y = 1: Is it true for all x that x + 1 = 0? No, if x = 0, 0 + 1 ≠ 0 (False). So, y = 1 doesn’t work for all x.

Since no single y worked for all x, the statement ∃y ∀x (x + y = 0) is False.

STRATEGY TIP: Verification vs. Counterexample

To verify ∀x P(x), you need a proof that works for all x.
To verify ∃x P(x), you need one concrete example.
To disprove ∀x P(x), you need one counterexample.
To disprove ∃x P(x), you need a proof that works for all x showing P(x) is always false.

NEGATING NESTED QUANTIFIERS

De Morgan’s Laws for Quantifiers	`¬∀x P(x)` is logically equivalent to `∃x ¬P(x)` `¬∃x P(x)` is logically equivalent to `∀x ¬P(x)`
Rule for Negation	When negating a quantified statement, move the negation `¬` inward past each quantifier, flipping the quantifier type (`∀` to `∃`, `∃` to `∀`) as you pass it. Finally, negate the predicate.
Original Statement	Negated Statement
`¬∀x (P(x))`	`∃x (¬P(x))`
`¬∃x (P(x))`	`∀x (¬P(x))`
`¬∀x ∃y P(x, y)`	Move `¬` past `∀x`: `∃x ¬∃y P(x, y)` Move `¬` past `∃y`: `∃x ∀y ¬P(x, y)` English Example: `¬("Everybody loves somebody")` `"Somebody loves nobody."` (i.e., there exists a person x such that for all people y, x does not love y).
`¬∃x ∀y P(x, y)`	Move `¬` past `∃x`: `∀x ¬∀y P(x, y)` Move `¬` past `∀y`: `∀x ∃y ¬P(x, y)` English Example: `¬("There is somebody whom everyone loves")` `"For every person, there is someone they don't love."` (i.e., for every person x, there exists a person y such that x does not love y).
COMMON MISTAKE	Negating only the predicate and not flipping the quantifiers. Remember to flip each quantifier as the negation passes it.

Quantifiers in Proofs

APPLICATION IN PROOFS

To Prove `∀x P(x)`	Strategy: Universal Generalization. Let `x` be an arbitrary (but specific) element from the domain. Show that `P(x)` holds for this arbitrary `x`. Example: Prove `∀x (x^2 ≥ 0)` for `x ∈ R`. Let `x` be any real number. We know `x^2` is always non-negative. Thus, `x^2 ≥ 0`.
To Prove `∃x P(x)`	Strategy: Existential Generalization (Constructive Proof). Find or construct a specific element `x` in the domain for which `P(x)` is true. Example: Prove `∃x (x^2 = 4)` for `x ∈ Z`. We can pick `x = 2`. `2^2 = 4`. Done.
To Disprove `∀x P(x)` (i.e., Prove `∃x ¬P(x)`)	Strategy: Find a Counterexample. Provide one specific element `x` in the domain for which `P(x)` is false (or `¬P(x)` is true). Example: Disprove `∀x (x > 5)` for `x ∈ Z`. Let `x = 3`. `3 > 5` is false. Counterexample found.
To Disprove `∃x P(x)` (i.e., Prove `∀x ¬P(x)`)	Strategy: Show `¬P(x)` for all `x`. Use Universal Generalization on `¬P(x)`, or proof by contradiction. Example: Disprove `∃x (x^2 < 0)` for `x ∈ R`. Let `x` be any real number. We know `x^2` is always non-negative, so `x^2 < 0` is always false. Thus `∀x (x^2 ≥ 0)`.
Proving `∀x ∃y P(x, y)`	Strategy: For an arbitrary `x` from the domain, you must show how to find or construct a `y` (which typically depends on `x`) such that `P(x, y)` holds. Example: Prove `∀x ∈ R ∃y ∈ R (x + y = 0)`. Let `x` be an arbitrary real number. We need to find a `y` such that `x + y = 0`. Let `y = -x`. Since `x ∈ R`, `y = -x ∈ R`. So, for any `x`, we can find such a `y`.
Proving `∃y ∀x P(x, y)`	Strategy: You must find or construct a specific `y` from the domain, and then prove that for every `x` in the domain, `P(x, y)` holds for that fixed `y`. Example: Prove `∃y ∈ R ∀x ∈ R (x + y = x)`. We need to find a `y` that works for all `x`. Let `y = 0`. Now, for any arbitrary real number `x`, `x + 0 = x` is true. Thus, such a `y` exists.
Constructive vs. Non-Constructive Existence Proofs	Constructive: You explicitly provide an example of the element whose existence is claimed (e.g., finding a `y` for `∃y P(y)`). Non-Constructive: You prove the element must exist without actually finding it (e.g., using proof by contradiction, or showing that if it didn’t exist, a contradiction would arise).
COMMON MISTAKE & STRATEGY TIP	A common mistake is confusing the order or dependency. Always remember: if `∃y` is inside `∀x`, `y` can depend on `x`. If `∃y` is outside `∀x`, `y` must be a single, fixed value for all `x`. Strategy Tip: To approach proofs with nested quantifiers, unpack them layer by layer, starting with the outermost quantifier. The strategy for handling the inner quantifier will often depend on the specific element picked by the outer quantifier.