# Scope of quantifier (LOGIC DISCRETE MATH)

I am finding it quiet understand the scope of a quantifier, and some other concepts of logic.

For example the below the scenarios below:

Scenario 1) (∀y H(y) → ∃z W(z, y)) → ∃z G(z)

Question: Is the "y" in the above formula being scoped by universal quantifier ∀y ?

Scenario 2) ∃y(∀y H(y) → ∃z W(z, y)) → ∃z G(z)

Question: Can you have 2 quantifiers, quantifying the same variable i.e."y" for scenario 2 ?

Scenario 3) (∀y H(y) → ∃z G(z)) → ∃z W(z, y)

Question :

Is the "y" in the "∃z W(z, y)" of the above formula considered to be free ?

Scenario 4) (∀z H(y) → ∃y W(z, y)) → ∃z G(z)

Question: What would be the scope for the scenario above ?

Scenario 5) (∀y H(y) → ∃z∃y W(z, y)) → ∃z G(z)

Question: What would be the scope for the scenario above ?

Scenario 6) (∀y H(y) → ∃z W(z, y)) → ∃x G(z)

Question: What would be the scope for the scenario above ?

Detail for each question will be highly welcomed.

• Too many questions. Regarding the first one, you ask about "the $y$", but there are two of them and the answer depends on which you mean. The answer to the second question is yes: the first $y$ is bound by the universal quantifier, the second by the existential one. The answer to the third one is yes. Concerning questions 4, 5 and 6: scope of what? Dec 24 '16 at 20:21
• Review the basic definition : Scope of a Quantifier. Dec 24 '16 at 20:36

It's often a matter of parentheses:

In a formula like $\forall y P(y)$ the $y$ in $P(y)$ is within the scope of the $\forall y$, but in a formula like $\forall y Q(x) \land P(y)$ it is not, since this formula would be parsed as being a conjunction, whose left conjunct is $\forall y Q(x)$, and right conjunct being $P(y)$, i.e. they get 'separated'. If you want the $y$ to be within the scope, you'd need to use parentheses: $\forall y (Q(x) \land P(y))$, and indeed now you are dealing with a universal statement.

Let me use colors to indicate the scope in each of your sentences:

Scenario 1) $(\color{red}{∀y }H(\color{red}{y}) → \color{blue}{∃z} W(\color{blue}{z}, y)) → \color{green}{∃z} G(\color{green}{z})$

Question: Is the "y" in the above formula being scoped by universal quantifier ∀y ?

=> first $y$ yes, second one not

Scenario 2) $\color{red}{∃y}(\color{blue}{∀y} H(\color{blue}{y}) → \color{green}{∃z} W(\color{green}{z}, \color{red}{y})) → \color{yellow}{∃z} G(\color{yellow}{z})$

Question: Can you have 2 quantifiers, quantifying the same variable i.e."y" for scenario 2 ?

=> You can never have a variable quantified by two quantifiers ... if a variable falls within the scope of multiple quantifiers for the variable, then it is only quantified by the most 'inside' one

Scenario 3) $(\color{red}{∀y} H(\color{red}{y}) → \color{blue}{∃z} G(\color{blue}{z})) → \color{green}{∃z} W(\color{green}{z}, y)$

Question :

Is the "y" in the "∃z W(z, y)" of the above formula considered to be free ?

=> Correct

Scenario 4) $(\color{red}{∀z} H(y) → \color{blue}{∃y} W(z, \color{blue}{y})) → \color{green}{∃z} G(\color{green}{z})$

Question: What would be the scope for the scenario above ?

=> in all cases, the quantifiers merely quantify the atomic formula right after it. Thus, both the first $y$ and the second $z$ are free.

Scenario 5) $(\color{red}{∀y} H(\color{red}{y}) → \color{blue}{∃z}\color{green}{∃y} W(\color{blue}{z}, \color{green}{y})) → \color{yellow}{∃z} G(\color{yellow}{z})$

=> Again, in all cases, the quantifiers merely quantify the atomic formula right after it.

Question: What would be the scope for the scenario above ?

Scenario 6) $(\color{red}{∀y} H(\color{red}{y}) → \color{blue}{∃z} W(\color{blue}{z}, y)) → \color{green}{∃x} G(\color{green}{z})$

Question: What would be the scope for the scenario above ?

=> And once again, in all cases, the quantifiers merely quantify the atomic formula right after it. So the second $y$ is free.

• Thank you so much for your in-depth analysis of the scenario. Literally, have cleared up all mis-conceptions. Dec 25 '16 at 20:27
• Example 1: ∀yH(y) Example 2: ∀y(H(y) v D(y)) Example 3: ∀y(H(x,y) v D(y)) All these "y" are bound to the universal quantifier, the only variable not bound is the "x" in Example 3. Dec 25 '16 at 20:43
• Summary - For a variable to be bound it needs to be quantified by a quantifier immediately before it. The only instance this does not hold would be the case were ∃y(∀yH(y)→∃zW(z,y)) . As the “y” for “H(y)” is being quantified by “∀y” and the “y” for “∃zW(z,y)” being quantified by “∃y” . In essence, couldn’t the formula original formula be re-written like this “ ∃y(∀xH(x)→∃zW(z,y)). Dec 25 '16 at 20:50
• @JackRoberts Yes, all correct! Dec 26 '16 at 8:16