Question
Write out $\exists ! x P(x),$ where the domain consists of the integers $1,2,$ and $3,$ in terms of negations, conjunctions, and disjunctions.
Step 1
It means "there exists a unique x such that P(x) is true". In other words, only one x in the domain makes P(x) true, and all other x in the domain make P(x) false. Show more…
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Key Concepts
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Suppose the domain of the propositional function $P(x, y)$ consists of pairs $x$ and $y,$ where $x$ is $1,2,$ or 3 and $y$ is $1,2,$ or $3 .$ Write out these propositions using disjunctions and conjunctions. $$ \begin{array}{ll}{\text { a) } \forall x \forall y P(x, y)} & {\text { b) } \exists x \exists y P(x, y)} \\ {\text { c) } \exists x \forall y P(x, y)} & {\text { d) } \forall y \exists x P(x, y)}\end{array} $$
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Suppose the domain of the propositional function $P(x, y)$ consists of pairs $x$ and $y,$ where $x$ is $1,2,$ or 3 and $y$ is $1,2,$ or $3 .$ Write out these propositions using disjunctions and conjunctions. $$ \begin{array}{ll}{\text { a) } \exists x P(x, 3)} & {\text { b) } \forall y P(1, y)} \\ {\text { c) } \exists y \neg P(2, y)} & {\text { d) } \forall x \neg P(x, 2)}\end{array} $$
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Suppose that the domain of $Q(x, y, z)$ consists of triples $x, y, z,$ where $x=0,1,$ or $2, y=0$ or $1,$ and $z=0$ or $1 .$ Write out these propositions using disjunctions and conjunctions. $$ \begin{array}{ll}{\text { a) } \forall y Q(0, y, 0)} & {\text { b) } \exists x Q(x, 1,1)} \\ {\text { c) } \exists z \neg Q(0,0, z)} & {\text { d) } \exists x \neg Q(x, 0,1)}\end{array} $$
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