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reductio

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reductio
‖ reductio (rɪˈdʌktɪəʊ) Pl. reductiones (rɪdʌktɪˈəʊniːz). [L., = reduction.] Used in various Latin phrases: 1. reductio ad impossibile, reduction to the impossible: a method of proving a proposition by drawing an absurd or impossible conclusion from its contradictory.1552 T. Wilson Rule of Reason (e... Oxford English Dictionary
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Reductio
Reductio and QuickCheck utilise a testing technique called Automated Specification-based Testing. website Reductio User Manual Reductio RequalsHashCode A Case for Automated Testing Tests As Documentation Software testing tools Free software programmed wikipedia.org
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Reductio ad Hitlerum
The phrase was derived from the logical argument termed Reductio ad absurdum. Claims that allegations of antisemitism are reductio ad Hitlerum have also been employed by David Irving, a British Holocaust denier. wikipedia.org
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Why do people say RAA(Reductio Ad Absurdum) is the same as $(\bot E)$? $(\bot E)$ is > $\bot\vdash\psi$. RAA(Reductio Ad Absurdum) says > If $\\{\Gamma,\neg\psi\\}\vdash\bot$, then $\\{\Gamma\\}\vdash\psi$. Yet, o...
See: * Dirk van Dalen, Logic and Structure (5th ed - 2013), page 30 and page 156. Where the presentation consider both _intuitionistic_ and _classical_ logics, the distinction between: > ($\bot$E): $\dfrac \bot \varphi$ and: > (RAA): $\dfrac { [\lnot \varphi] \ldots \bot } \varphi$ has to be maintai...
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Reductio ad absurdum
Greek philosophy Reductio ad absurdum was used throughout Greek philosophy. Greek mathematicians proved fundamental propositions using reductio ad absurdum. wikipedia.org
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The textbook's natural deduction proof for $\vdash(\neg(\phi\to\psi)\to\phi)$ seems to be wrong with regard to RAA(Reductio Ad Absurdum). As you can see below, $\psi$ pops out of nowhere due to RAA(reductio ad absurdu...
The proof is correct. From $\bot$ you can infer anything, which is why the $\psi$ is allowed to "pop out of nowhere" in the first use of RAA in the proof. Note that this use of RAA does not discharge any hypotheses (there is no number next to the line that infers $\psi$ from $\bot$). In the second u...
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Tertium non datur and Inductive Subset of a Well Ordered Set Theorem: *Let (S,) be a well-ordered set. Let T⊆S be an inductive subset of S, that is such that: ∀s∈S:(∀t∈S:t≺st∈T)s∈T (induction hypothesis) Then T=S...
Essentially you're proving the contrapositive: if $T\subsetneq S$ then $T$ is not inductive. Suppose $T\subsetneq S$. Then $S\setminus T\ne \emptyset$, so it has a least element $x$. For any $s\in S$, if $s<x$ then $s\in T$ by definition of $x$; however, $x\notin T$. Therefore, $$ \exists s\in S\,((...
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I'm finding difficult to understand deduction theorem with inference rules I'm finding difficult to understand deduction theorem with inference rules. Now i'm stuck with this question because of lack of understanding ...
Or formally: $$\begin{align}\phi&\vdash \psi\\\ &\vdash \phi\Rightarrow \psi\end{align}$$ This is not that central to the problem, instead it's reductio
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How can I prove the following with natural deduction rules? ¬∀x∃yP(x,y) ⊢ ∃x∀y¬P(x,y) I have been stuck with this problem for a long time, I tried reductio ad absurdum and I got the hypothesys [¬∃x∀y¬P(x,y)], then I t...
You are doing this exactly right! You just have to derive $\forall y \neg P(x,y)$ from $\neg \exists y P(x,y)$ Now, I am not sure how your proof system defines the rule for $\forall$ Introduction ... in the system that I use you designate a 'fresh' constant to take the role of the arbitrary object f...
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In the infinite product topology, rigorously prove a given set is not open For example, how do we prove that $(-1, 1)^\omega$ is not open in $\mathbb{R}^\omega$? I understand the general form of an open set in this to...
You can directly show its complement is not closed. Let $x_n = (0,0,\dots,0,0,2,2,2,2,2,\dots)$ where there are $n$ $0$'s. Then each $x_n \in \mathbb{R}^\omega\setminus (-1,1)^\omega$ but $x_n \to (0,0,0,\dots) \in (-1,1)^\omega$.
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Monotonicity finding based on given releationships I have a function $f:\mathbb{R}\rightarrow\mathbb{R}$ with $f''(x)>0\text{ }\forall x\in\mathbb{R}$ and $\lim_{x\to+\infty}{f(x)}=0$. Now I want to prove that $f$ is ...
We want to prove that $f'(x)\leq 0$ for all $x\in \mathbb R$. Fix $x\in \mathbb R$. By Taylor formula we have $$f(x+h) = f(x)+f'(x)h + h^2\int_0^1(1-t)f''(x+th)dt \geq f(x)+f'(x)h,$$ because $f''(x+th)> 0$. Suppose $f'(x)>0$. We have $$f(x+h) \geq f(x)+f'(x)h,$$ and therefore $\lim_{h\to \infty} f(x...
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Proof that $f(x)=4x^4-2x+1$ has no real roots. My thought was to: 1) hypothesis there are 2 real roots for this equation, 2) apply Rolle's theorem and come to a reductio ad absurdum and then if there aren't 2 re...
* $f'(x)=2(8x^3-1)$, so there's a single critical point: $\; x=\frac12$. * $f''(x)=48x^2\ge 0$, so by the second derivative test, this critical point is a _minimum_ , and this minimum is an absolute minimum. * $f(\frac12)=\frac14>0$.
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Different ways to prove $\sqrt p$ irrational for $p$ prime. I know this fact can be proved by contradiction(reductio ad absurdum) but please give proofs by different methods.
By Eisenstein's criterion, the polynomial $x^2-p$ is irreducible. (To answer WimC's complaint below: Consider the splitting field of $x^2-p$ over $\mathbf Q_p$. It contains an element of valuation $1/2$, so it is a proper over-field of $\mathbf Q_p$, and therefore $x^2-p$ is irreducible over $\mathb...
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Let $a$, $b$, and $c$ be integers satisfying $a^2 + b^2 = c^2$. Prove: $abc$ must be even. I'm pretty sure that this can be proved by reductio ad absurdum, and have a proof for that. However, I'm not sure how to prove...
Hint: Think about even and odd numbers. Can all 3 variables be odd?
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Prove that the diagonals are the longest lines in a square. I just pondered this question and have tried out several methods to solve it(mainly using trigonometry). However, I am not satisfied with my trigonometrical ...
It is sufficient to consider the two cases (since others are repetetive): without loss of generality, take two arbitrary points on two sides such that $0\le x\le y\le a$: ![enter image description here]( Case 1: From the Pythagoras: $$\sqrt{x^2+y^2}\le \sqrt{a^2+a^2}=d.$$ Case 2: From the Pythagoras...
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