Scope of Quantifier (Logic Discrete Math)

Scope of Quantifier (Logic Discrete Math)
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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.

Thanks in advance.

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1 Answer

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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.

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Robert Thorne
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Robert Thorne

Robert Thorne covers electric vehicle innovations, autonomous driving systems, global mobility trends, and automotive engineering developments.