Question

Rewrite each of these statements so that negations appear only within predicates(that is, so that no negation is outside a quantifier or an expression involving logical connectives).

$$\begin{gathered} \mathbf{a}\mathbf{)}\mathbf{}\mathbf{\neg }\exists \mathbf{y}\exists \mathbf{xP}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{b}\mathbf{)}\mathbf{}\mathbf{\neg }\mathbf{\forall }\mathbf{x}\exists \mathbf{yP}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)} \\ \mathbf{c}\mathbf{)}\mathbf{}\mathbf{\neg }\exists \mathbf{y}\mathbf{(}\mathbf{Q}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{\wedge }\mathbf{\forall }\mathbf{x}\mathbf{\neg }\mathbf{R}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{d}\mathbf{)}\mathbf{}\mathbf{\neg }\exists \mathbf{y}\mathbf{(}\exists \mathbf{xR}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{\vee }\mathbf{\forall }\mathbf{xS}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\left)}\mathbf{\right)} \\ \mathbf{e}\mathbf{)}\mathbf{}\mathbf{\neg }\exists \mathbf{y}\mathbf{(}\mathbf{\forall }\mathbf{x}\exists \mathbf{zT}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{\vee }\exists \mathbf{x}\mathbf{\forall }\mathbf{zU}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{)} \end{gathered}$$

Short answer

The Statements can be rewrite.

Step-by-step solution

Finding the Truth values for ¬∃x∃yP(x,y)

The Statement “$\neg \exists \mathrm{y}\exists \mathrm{xP}(\mathrm{x},\mathrm{y})$”

As it is known that$\neg \exists \mathrm{y}=\forall \mathrm{y},\neg \exists \mathrm{x}=\forall \mathrm{x}$

Then,$\neg \exists \mathrm{y}\exists \mathrm{xP}(\mathrm{x},\mathrm{y})\Rightarrow "\forall \mathrm{y}\forall \mathrm{x}\neg \mathrm{P}(\mathrm{x},\mathrm{y})$

As a result, The Statement “$\neg \exists \mathrm{y}\exists \mathrm{xP}(\mathrm{x},\mathrm{y})$” is rewrite as “$\forall \mathrm{y}\forall \mathrm{x}\neg \mathrm{P}(\mathrm{x},\mathrm{y})$”.

Finding the Truth values for ¬∀x∃yP(x,y)

The Statement “$\neg \forall \mathrm{x}\exists \mathrm{yP}(\mathrm{x},\mathrm{y})$”

As it is known that$\neg \exists \mathrm{y}=\forall \mathrm{y},\neg \forall \mathrm{x}=\exists \mathrm{x}$

Then,$\neg \forall \mathrm{x}\exists \mathrm{yP}(\mathrm{x},\mathrm{y})\Rightarrow "\exists \mathrm{x}\forall \mathrm{y}\neg \mathrm{P}(\mathrm{x},\mathrm{y})"$

As a result, The Statement “$\neg \forall \mathrm{x}\exists \mathrm{yP}(\mathrm{x},\mathrm{y})$” is rewrite as “$\exists \mathrm{x}\forall \mathrm{y}\neg \mathrm{P}(\mathrm{x},\mathrm{y})$”.

Finding the Truth values for ¬∃y(Q(y)∧∀x¬R(x,y))

The Statement “$\neg \exists \mathrm{y}\left(\mathrm{Q}\right(\mathrm{y})\wedge \forall \mathrm{x}\neg \mathrm{R}(\mathrm{x},\mathrm{y}\left)\right)$”

As it is known that role="math" localid="1668600741300" $\neg \exists \mathrm{y}=\forall \mathrm{y},\neg \forall \mathrm{x}=\exists \mathrm{x},\neg (\vee )=\wedge$

Then,$\neg \exists \mathrm{y}\left(\mathrm{Q}\right(\mathrm{y})\wedge \forall \mathrm{x}\neg \mathrm{R}(\mathrm{x},\mathrm{y}\left)\right)\Rightarrow "\forall \mathrm{y}(\neg \mathrm{Q}(\mathrm{y})\vee \exists \mathrm{xR}(\mathrm{x},\mathrm{y}\left)\right)"$

As a result, The Statement “$\neg \exists \mathrm{y}\left(\mathrm{Q}\right(\mathrm{y})\wedge \forall \mathrm{x}\neg \mathrm{R}(\mathrm{x},\mathrm{y}\left)\right)$” is rewrite as

“$\forall \mathrm{y}(\neg \mathrm{Q}(\mathrm{y})\vee \exists \mathrm{xR}(\mathrm{x},\mathrm{y}\left)\right)$”.

Finding the Truth values for ¬∃y(∃xR(x,y)∨∀xS(x,y))

The Statement “$\neg \exists \mathrm{y}(\exists \mathrm{xR}(\mathrm{x},\mathrm{y})\vee \forall \mathrm{xS}(\mathrm{x},\mathrm{y}\left)\right)$”

As it is known that$\neg \exists \mathrm{y}(\exists \mathrm{xR}(\mathrm{x},\mathrm{y})\vee \forall \mathrm{xS}(\mathrm{x},\mathrm{y}\left)\right)$

Then,$\neg \exists \mathrm{y}(\exists \mathrm{xR}(\mathrm{x},\mathrm{y})\vee \forall \mathrm{xS}(\mathrm{x},\mathrm{y}\left)\right)\Rightarrow "\forall \mathrm{y}(\forall \mathrm{x}\neg \mathrm{R}(\mathrm{x},\mathrm{y})\wedge \exists \mathrm{x}\neg \mathrm{S}(\mathrm{x},\mathrm{y}\left)\right)"$

As a result, The Statement “$\neg \exists \mathrm{y}\left(\mathrm{Q}\right(\mathrm{y})\wedge \forall \mathrm{x}\neg \mathrm{R}(\mathrm{x},\mathrm{y}\left)\right)$” is rewrite as

“$\forall \mathrm{y}(\forall \mathrm{x}\neg \mathrm{R}(\mathrm{x},\mathrm{y})\wedge \exists \mathrm{x}\neg \mathrm{S}(\mathrm{x},\mathrm{y}\left)\right)$”.

Finding the Truth values for ¬∃y(∀x∃zT(x,y,z)∨∃x∀zU(x,y,z))

The Statement “$\neg \exists \mathrm{y}(\forall \mathrm{x}\exists \mathrm{zT}(\mathrm{x},\mathrm{y},\mathrm{z})\vee \exists \mathrm{x}\forall \mathrm{zU}(\mathrm{x},\mathrm{y},\mathrm{z}\left)\right)$”

As it is known that$\neg \exists \mathrm{y}=\forall \mathrm{y},\neg \forall \mathrm{x}=\exists \mathrm{x},\neg (\vee )=\wedge$

Then, role="math" localid="1668603146505" $\neg \exists \mathrm{y}(\forall \mathrm{x}\exists \mathrm{zT}(\mathrm{x},\mathrm{y},\mathrm{z})\vee \exists \mathrm{x}\forall \mathrm{z}\cup (\mathrm{x},\mathrm{y},\mathrm{z}\left)\right)\Rightarrow \forall \mathrm{y}(\exists \mathrm{x}\forall \mathrm{z}\neg \mathrm{T}(\mathrm{x},\mathrm{y},\mathrm{z})\wedge \forall \mathrm{x}\exists \mathrm{z}\neg \cup (\mathrm{x},\mathrm{y}\left)\right)$

As a result, The Statement “$\neg \exists \mathrm{y}(\forall \mathrm{x}\exists \mathrm{zT}(\mathrm{x},\mathrm{y},\mathrm{z})\vee \exists \mathrm{x}\forall \mathrm{z}\cup (\mathrm{x},\mathrm{y},\mathrm{z}\left)\right)$” is rewrite as

“$\forall \mathrm{y}(\exists \mathrm{x}\forall \mathrm{z}\neg \mathrm{T}(\mathrm{x},\mathrm{y},\mathrm{z})\wedge \forall \mathrm{x}\exists \mathrm{z}\neg \cup (\mathrm{x},\mathrm{y}\left)\right)$”.