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consequent
▪ I. consequent, n. (ˈkɒnsɪkwənt) [a. F. conséquent, ad. L. consequens, -ent-, a consequence, subst. use of pr. pple.: see next.] † 1. = consequence 1. Obs. exc. as in b.c 1386 Chaucer Melib. ¶421 (Harl. MS.) Let vs now examyne þe þridde poynt þat Tullius clepeþ consequente. Þou schalt vnderstonde þ...
Oxford English Dictionary
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Affirming the consequent
The name affirming the consequent derives from using the consequent, Q, of , to conclude the antecedent P. Affirming the consequent and denying the antecedent are invalid.
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consequent
consequent/ˈkɔnsɪkwənt; `kɑnsəˌkwɛnt/ adj~ (on/upon sth) (fml 文) following sth as a result or an effect 由某事物引起的; 随之发生的 his resignation and the consequent public uproar 他的辞职以及由此而引起的公众的哗然 the rise in prices consequent upon the failure of the crops 由於农作物歉收而引起的物价上涨.
牛津英汉双解词典
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Finding a guaranteed consequent of a satisfiable sentence ### Let A = ∀x∀y(P(x,y)). A is satisfiable. Find any B such that it isn't a tautology and is a consequent of A. Now if I understand it correctly, then a corre...
How about **reflexivity** \- $\forall x (P(x, x))$? Similarly, **transitivity** $\forall x\forall y\forall z(P(x, y)\wedge P(y, z)\implies P(x, z))$ and **symmetry** $\forall x\forall y(P(x, y)\implies P(y, x))$. And there are lots of others . . .
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If the antecedents are mutually exclusive, then is the consequent true? Suppose I know that the following implications are true: $$P_1 \Longrightarrow (A \land B)$$ $$P_2 \Longrightarrow (A \land B)$$ for some premi...
No; given that $P_1$ and $P_2$ are mutually exclusive conditions, it is still possible that they're both false, and so we cannot deduce anything from $P_1\rightarrow Q,$ nor from $P_2 \rightarrow Q$. Explicitly (note that false implies false), the following is a counterexample to the conjecture: $$P...
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Find M <= N such that sum of consequent numbers from 1 to x is M. I am given a number $N$. I need to find $M\le N$ such that $M$ can be written as a sum of consequent numbers from 1 to x. I know this formula $1+2+\c...
$$ \begin{align} &\, M \,=\, \sum_{k=1}^{n}k = \frac{n(n+1)}{2} \le N \\\\[2mm] &\, \frac{n(n+1)}{2}=N \,\Rightarrow\, n^2+n-2N=0 \,\Rightarrow\, n=\frac{-1\pm\sqrt{1+8N}}{2} \\\\[2mm] &\, n\in{\mathbb N}^{+} \,\Rightarrow\, \color{red}{n=\left\lfloor \frac{\sqrt{8N+1}-1}{2} \right\rfloor} ,\,\, \co...
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How Can This Antecedent Be Evaluated to True? I've a couple of questions. From MIT notes: > $\frac{NOT(P)\;IMPLIES\;NOT(Q)}{P\;IMPLIES\;Q}$ > > is not sound: if P is assigned T and Q is assigned F, then the anteced...
In checking validity of an argument, it's not a good practice to just say "the right conclusion would be this, rather than the given conclusion." What you need to do in showing an argument is not valid, is to come up with truth assignments for the variables which make all premises true but make the ...
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小人物的尊严 La dignidad de los nadies
The degraded socio-economic condition of Argentina leading to the December 2001 rebellions, and its consequent social chaos analyzed by focusing on real The degraded socio-economic condition of Argentina leading to the December 2001 rebellions, and its consequent
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Can I assume a condition in the consequent? Im reading Axler's Linear Algebra Done Right. In an exercise, he ask to prove that $$a\in F,v\in V,av=0 \implies a=0 \lor v=0 $$ where $V$ is a vector space over the field $...
In order to prove $a=0\vee q=0$, yes, it suffices to assume $q\neq0$ and conclude $a=0$. _Alternatively_ , it suffices to assume $a\neq0$ and conclude $q=0$. You don't have to do both. (In fact, you shouldn't do both, because it's confusing.) Feel free to pick whichever tactic seems more convenient....
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Arithmetic progression with common difference 2061 If there are 30 consequent members of an arithmetic progression with CD of 2061, show that among them are at most 20 squares of natural numbers. I wrote out $a_1$ th...
Let $\\{a_k\\}_{k=1}^{30}$ be your arithmetic progression and suppose $a_k=y_k^2,\,\forall k.$ Consider that equation $$y_{k+2}^2-y_k^2=2\times 2061$$ $$(y_{k+2}-y_k)(y_{k+2}+y_k)=2\times 3^2\times 229.$$ If $y_{k+2}, y_k$ are integers, should be same parity. Otherwise $y_{k+2}^2-y_k^2$ become an od...
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Conditional statement with unknown components Can there be a conditional statement where the truth value of both the antecedent $P$ and consequent $Q$ are unknown but the truth value of $P\Rightarrow Q$ is known?
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Clearly, the truth value of the conditional is known (it's false), but we do not a priori know truth values for the antecedent or consequent.
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ZF Extensionality axiom To familiarize myself with axiomatic set theory, I am reading Kenneth Kunen's The Foundations of Mathematics that presents ZF set theory. I haven't gotten really far since I am stuck at the axi...
The reason the converse is not part of the axiom is that it already follows from the axioms of first-order logic: $x = y \implies (\varphi(x) \iff \varphi(y))$ for any formula $\varphi$. This is known as the substitution property of equality.
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Find $N \gt 0$ so that $ n \ge N \implies \lvert \left(1+\frac3n \right)^2 -1\rvert \lt \frac{1}{10}$. Here's the problem: Find $N \gt 0$ so that $$ n \ge N \implies \left\lvert \left(1+\frac3n \right)^2 -1\right\rv...
Let $m = 1+\dfrac{3}{n} \to |m^2-1| \dfrac{3}{\sqrt{10}}$. And $m \dfrac{3\sqrt{10}}{\sqrt{11}-\sqrt{10}}=3\sqrt{110}+30\approx 61.5$. Thus $N = 62$ or higher.
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Is this a valid interpretation that shows this sentence, (∀x)(Mx ⊃ Kx) ⊃ (∃x)(Mx & Kx), is not quantificationally true? Here's the interpretation I thought up. UD: Set of all positive integers Mx: x is less than 1 ...
Here, you have that both the quantified antecedent (call it $p$) is false, and the quantified consequent (call it $q$) is false. Put another way, an implication is False ONLY when the antecedent is true, and the consequent is true.
So, $F\to T, T\to T, F\to F$ are all true.
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Limit of the ratio of consecutive Fibonacci numbers I have read in a book that the limit of the ratio of consequent Fibonacci numbers is the golden ratio. However, it was just mentioned thus not justified. So, my ques...
HINT: Let $\displaystyle\lim_{n\to\infty}\frac{F_{n+1}}{F_n}=u$ Clearly, $u\not<0$ By definition, we have $\displaystyle F_{n+1}=F_n+F_{n-1}$ $\displaystyle\implies \frac{F_{n+1}}{F_n}=1+\frac1{\frac{F_n}{F_{n-1}}}$ Setting $\displaystyle n\to\infty, u=1+\frac1u$
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