2 votes
Accepted

Do the "nothing" and "unique existence" quantifiers respect or at least semi-respect ordered pairs?

To give an example where $NxNyP(x,y) \neq N(x,y)P(x,y)$, consider $P(x,y):=x=y$ where the universe from where $x$ and $y$ come is any non-empty set, say $\mathbb N$. Given an $x$, $NyP(x,y)$ then ...
Ingix's user avatar
  • 14.3k
2 votes

Changing the definition of almost sure convergence

We use the notation in the comments. Define the probability measure $\tilde{\mathbb P}(\cdot) = \mathbb P(\cdot \cap A_1)/\mathbb P(A_1)$. Note $\tilde{\mathbb P}(A_\epsilon) = 1$ for all $\epsilon>...
Andrew's user avatar
  • 1,634
2 votes

$\forall x(P(x) \rightarrow Q(x))\stackrel{?}{\equiv}\neg\exists x(P(x) \rightarrow \neg Q(x))\stackrel{?}{\equiv}\neg\exists x(P(x) \wedge\neg Q(x))$

We can work with the different equivalent transformations of the formulas. We agree that 1 and 3 are equivalent and 3 says that "there is no object $a$ such that $P(a)$ and $\lnot Q(a)$". ...
Mauro ALLEGRANZA's user avatar
2 votes

Is $\exists x [(P(x) \vee Q(x))\rightarrow R(x)]$ logically equivalent to $\exists x [(P(x) \rightarrow R(x)) \vee (Q(x)\rightarrow R(x))]$?

To supplement the answer given by Jean Abou Samra: If $P(x)$ and $R(x)$ are identically false, while $Q(x)$ is identically true, then $(P(x) \lor Q(x)) \to R(x)$ is false, but $(P(x) \to R(x)) \lor (...
Rob Arthan's user avatar
  • 48.1k
2 votes
Accepted

Is $\exists x [(P(x) \vee Q(x))\rightarrow R(x)]$ logically equivalent to $\exists x [(P(x) \rightarrow R(x)) \vee (Q(x)\rightarrow R(x))]$?

No. In fact, $(P(x) ∨ Q(x)) ⇒ R(x)$ is equivalent to $(P(x) ⇒ R(x)) \color{red}{∧} (Q(x) ⇒ R(x))$. Proof: Suppose $(P(x) ∨ Q(x)) ⇒ R(x)$. If $P(x)$, then $P(x) ∨ Q(x)$, and thus $R(x)$, which proves $...
Jean Abou Samra's user avatar
1 vote
Accepted

Definition of non-monotonic sequence

Negate the definition: $\exists m,n, p, q \in\mathbb{N}$ s.t ($ n>m $ but $x_n<x_m$) and ($p>q$ but $x_p >x_q$)
J.Dmaths's user avatar
  • 662
1 vote

Quantifiers and the meaning of the statement

If two formulas seem confusingly similar like $i$ and $ii$, then you can always take their negations and compare to see if they’re actually logically equivalent. In this case, the negation of $i$ is ...
PW_246's user avatar
  • 1,156
1 vote
Accepted

The quantifiers in the definition of almost sure convergence

The following proves $(1) \iff (2)$ and it follows the suggestions in the comments. To see this, note that we have $$\{X_n\to X\}=\bigcap_{k\in \mathbb{N}}\bigcup_{N\in \mathbb{N}}\bigcap_{n\geq N}\{|...
Snoop's user avatar
  • 15k

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