Determine the truth-value of the following propositions: 15) (∀x ∈ R) (x > -2 ∧ x > -1) A) TRUE B) FALSE 16) ~(∀x ∈ R) (x > 1 ∧ 5x) A) TRUE B) FALSE 17) (∃y ∈ R)(∀x ∈ R) (y < x^2) A) TRUE B) FALSE 18) (∃x ∈ R) (x^2 - 2 = 0) A) TRUE B) FALSE 19) (∀x ∈ R)(∃y ∈ R)(y < x) A) TRUE B) FALSE 20) (∃x ∈ R)((∀y ∈ N)(x < y)) A) TRUE B) FALSE
Added by Andr-S P.
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This is false because there are numbers between -2 and -1 that satisfy x > -2 but not x > -1. For example, x = -1.5. So, the answer is $\boxed{\textbf{(B) FALSE}}$. Show more…
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What are the truth values of these statements? $$ \begin{array}{l}{\text { a) } \exists ! x P(x) \rightarrow \exists x P(x)} \\ {\text { b) } \forall x P(x) \rightarrow \exists ! x P(x)} \\ {\text { c) } \exists ! x \neg P(x) \rightarrow \neg \forall x P(x)}\end{array} $$
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Nested Quantifiers
Determine the truth value of each of these statements if the domain for all variables consists of all integers. $$ \begin{array}{ll}{\text { a) } \forall n \exists m\left(n^{2}<m\right)} & {\text { b) } \exists n \forall m\left(n<m^{2}\right)} \\ {\text { c) } \forall n \exists m(n+m=0)} & {\text { d) } \exists n \forall m(n m=m)} \\ {\text { e) } \exists n \exists m\left(n^{2}+m^{2}=5\right)} & {\text { f) } \exists n \exists m\left(n^{2}+m^{2}=6\right)}\end{array} $$ $$ \begin{array}{l}{\text { g) } \exists n \exists m(n+m=4 \wedge n-m=1)} \\ {\text { h) } \exists n \exists m(n+m=4 \wedge n-m=2)} \\ {\text { i) } \forall n \forall m \exists p(p=(m+n) / 2)}\end{array} $$
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