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vacuous
vacuous, a. (ˈvækjuːəs) [f. L. vacu-us empty, void, free, clear, etc. (cf. vacuum n.) + -ous.] † 1. Not properly filled out or developed. Obs.—11651 Smallwood Commend. Verses to W. Cartwright's Wks., False Vacuous Births in every street we see: But seldome, true and ripen'd, such as He. 2. Empty of ...
Oxford English Dictionary
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Vacuous truth
Vacuous truths most commonly appear in classical logic with two truth values. yield a vacuous truth under the strict conditional.
wikipedia.org
en.wikipedia.org
vacuous
vacuous/ˈvækjuəs; `vækjʊəs/ adj(fml 文) showing or suggesting absence of thought or intelligence; inane 无思想的; 无智慧的; 空洞的; 无意义的 a vacuous stare, remark, laugh, expression 呆视、 空洞的话、 傻笑、 茫然的表情.
牛津英汉双解词典
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Vacuous truth - Wikipedia
Vacuous truth. In mathematics and logic, a vacuous truth is a conditional or universal statement (a universal statement that can be converted to a conditional statement) that is true because the antecedent cannot be satisfied. [1] It is sometimes said that a statement is vacuously true because it does not really say anything. [2]
en.wikipedia.org
Look up the definition of the word "vacuous".
Vacuous is an adjective that means having or showing a lack of thought or intelligence; empty; void of intelligence; without mental content; stupid.
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Aren't vacuous statements True and False simultaneously? Wikipedia states "a vacuous truth is a statement that asserts that all members of the empty set have a certain property". Clearly the statement: 'all elements o...
You did not negate the statement "all elements of a set S have property X" correctly. The opposite of "all elements of a set S have property X" is _not_ "no elements of set S have property X". The opposite of "all elements of a set S have property X" is "some element of S does not have property X". ...
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Contradictory vacuous truths in consistent formal system Can 2 contradictory vacuously true statements be proved in a consistent formal system?
Indeed, this is exactly what characterizes vacuity: $\forall x(P(x)\implies Q(x))$ is a vacuous truth (with respect to our theory) iff $\forall x(P(x)\
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Vacuous proof of $a+b+c=0$ _"Prove that if $a$, $b$, and $c$ are odd integers such that $a+b+c=0$, then $abc < 0$"_ My attempt: I figured that since you can't make an odd by adding two other odd numbers there was no ...
The standard way to phrase an argument like this is to use modular arithmetic - as used in the comments, if something is even, we write $a\equiv 0 \mod 2$. Because these statements are so 'obvious' e.g. the sum of three odd numbers is odd, you typically wouldn't bother proving it, you'd just take it...
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question about vacuous truth and function I'm confusing about vacuous truth. Let $f: \mathbb{N} \rightarrow \mathbb{N}$ where $f(n)=2n$. we can calculate function values if $n$ belongs to domain. but what if it does ...
It is false that $f(1.5)=3$. As you have pointed out, $(\forall y)(1.5,y)\notin f$, so $(\forall y)y\neq f(1.5)$. In particular $3\neq f(1.5)$. The fact that $1.5$ is not in the domain of $f$ does not stop the proposition "$f(1.5)=3$" from being false.
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What would happen if we just made vacuous truths false instead? It's well known that vacuous truths are a concept, i.e. an implication being true even if the premise is false. What would be the problem with simply re...
Notice that 3=5 is false. but if 3=5 we can prove 8=8 which is true. $$ 3=5$$ therefore $$ 5=3$$ Add both sides, $$8=8$$ We can also prove that $$ 8=10$$ which is false. $$ 3=5$$ Add $5$ to both sides, we get $$8=10$$ The point is that if we assume a false assumption, then we can claim whatever we l...
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Vacuous Domain Mixing For all and There Exists. May be this is a stupid question but I was thinking we know that suppose $D = \varnothing$ then $\forall\, x \in D \,P(x)$ is true vacuously and $\exists y\in D\, P(y)$ ...
$\bigl(\forall x \in D\bigr) \bigl(\exists y \in D\bigr) Q(x,y)$ has the form $\bigl(\forall x \in D\bigr) P(x)$, where $P(x)$ expands to $\bigl(\exists y \in D\bigr) Q(x,y)$. So it is vacuously true. If you switch the order of the quantors, it becomes vacuously false, since you get a proposition of...
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Intuition behind Vacuous proofs My book says we can quickly prove the conditional statement $P \implies Q $ when we know $P $ is false. This much I'm fine with as I can show it with a truth table. But then I'm asked t...
It looks like $P(n)$ is supposed to be the statement: > $n>1\implies n^2>n$. When $n\le1,$ we have that $n>1$ is _false_. So, for example, $P(0)$ is true.
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Strong induction and vacuous truth I was pondering a bit more about this question regarding being able to "omit" the base case in a proof by strong induction due to vacuous truth. The post states: > Strong induction ...
Strong (or : complete) induction_induction) is : > $(∀n)[(∀m)(m $(∀m)(m < 0 \to P(m)) \to P(0)$. But $(m < 0 \to P(m))$ is _vacuously true_ (there are no $m < 0$). Thus, the conditional amounts to : $\text T \to P(0)$ and there is only one possibility to satisfies it : when $P(0)$ is true.
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Can there be a vacuous tautological consequence $F\vDash F$? Can there be a vacuous tautological consequence $F\vDash F$? Since $α⊨φ \iff ⊨α→φ$ then is: $(k∧¬k)⊨(p∧¬p)$ for example considered a tautological consequence?
Yes. This is a consequence of the definition of $\models$ between two formulas: > $\phi \models \psi$ iff $M \models \phi$ implies $M \models \psi$ for all models $M$. (In the case of propositional logic, the models are the lines of a truth table.) Now since $M \models F$ can never occur, it vacuous...
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If each $f_n$ is continuous on a set $S$, does $f_n$ converge pointwise to a function $f$ on $S$? If each $f_n$ is continuous on a set $S$, does $f_n$ converge pointwise to a function $f$ on $S$?I feel I am seriously ...
Vacuous, I'm not sure. But the answer, of course, is no. Even if your set $S$ is a closed interval in $\Bbb R$, consider $f_n(x)=nx$.
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