Artificial intelligent assistant

prenex

answer Answers

ProphetesAI is thinking...

MindMap

Sources

1
prenex
prenex, a. Logic. (ˈpriːnɛks) [ad. late L. prae-nex(us tied or bound up in front: see pre- A. 1 and nexus.] Of or relating to a quantifier placed initially in a formula whose scope affects the whole formula; spec. in phr. prenex normal form (see quot. 1944).1944 A. Church in Ann. Math. Stud. xiii. 6... Oxford English Dictionary
prophetes.ai 0.0 3.0 0.0
2
Prenex normal form
but not in prenex normal form. Use of prenex form Some proof calculi will only deal with a theory whose formulae are written in prenex normal form. wikipedia.org
en.wikipedia.org 0.0 1.5 0.0
3
Find Prenex form and skolem form We have a formula: $(\exists x)R(x,y)\iff(\forall y)P(x,y)$ Find a prenex form of the formula and convert it into skolem variant. My solution![enter image description]( Is it correct?
, otherwise you would need to use functions when doing skolemization, and fortunately you can do that by first bringing out the $a$ and $d$ using the Prenex
prophetes.ai 0.0 0.90000004 0.0
4
Is this formula in normal form prenex $(\forall x)(\exists y)(P(x) \land (\exists z)Q(y, z))$? In my course notes, this $(\forall x)(\exists)z(P(x) \land (\exists z)Q(y, z))$ formula is said to be in `normal form pren...
Yes prenex normal form must have all quantifiers at the front so what you said is correct.
prophetes.ai 0.0 0.6 0.0
5
Herbrand's theorem
converted to prenex form and their universal quantifiers can be removed by Herbrandization. Conversion to prenex form can be avoided, if structural Herbrandization is performed. wikipedia.org
en.wikipedia.org 0.0 0.6 0.0
6
Finding prenex normal form of a formula > Find prenex normal form of the formula $(\exists x)S(x,y)\rightarrow (R(x)\rightarrow \neg(\exists u)S(x,u))$ My attempt: * $(\exists x)S(x,y)\rightarrow (R(x)\rightarrow ...
x,u))) \overset{\text{Replace variables}}\Leftrightarrow$ $(\forall u) ((\exists w) S(w,y) \rightarrow (R(x) \rightarrow \neg S(x,y)))\overset{\text{Prenex
prophetes.ai 0.0 0.3 0.0
7
prenex normal equivalence challenges in math consider !enter image description here these two following formula are prenex normal equivalence with the above formula? i think yes, but didn't have any idea to explain...
Now I give an answer without using provable equivalences on Enderton, page 121 and page 130. Main formula is equivalent to $$(\neg\exists x\varphi(x))\vee(\forall x\exists y\psi(x,y))$$ and this is also equivalent to $$(\forall x\neg\varphi(x))\vee(\forall x\exists y\psi(x,y))$$ but we have: $$\fora...
prophetes.ai 0.0 0.3 0.0
8
prenex normal form with two same variables How would I put the following statement into prenex normal form? $∀x P(x) ∨ ∀x Q(x)$ Initially I thought you can use subsumption to do it as follows: $∀x( P(x) ∨ Q(x))$ bu...
\forall y Q(y)$ is equivalent with $\forall x (P(x) \vee \forall y Q(y))$ which is equivalent with $\forall x\forall y( P(x) \vee Q(y))$ which is in prenex
prophetes.ai 0.0 0.3 0.0
9
Quantifier order for prenex normal form $\exists x R(x) \land \forall y S(y)$ is this equivalent to $\exists x \forall y (R(x) \land S(y))$ as well as $\forall y \exists x (R(x) \land S(y))$? Is there a rule for the...
Formally: **Prenex Laws** Where $\varphi$ is any formula and where $x$ is _not_ a free variable in $\psi$: $ \forall x \ \varphi \land \psi \Leftrightarrow
prophetes.ai 0.0 0.3 0.0
10
How can I put this sentence into prenex form I have this sentence, and want to put it in Prenex form: $$ \lnot(( \exists y)( \forall z) Q(y,z)) $$ Is it this?: $$ ( \forall y)( \exists z) \lnot Q(y,z) $$
Yes ... basically the relevant rule is: > When you push/pull a negation sign past an _adjacent_ quantifier, the quantifier flips into its dual. So pushing inwards, $\neg\forall x\varphi$ is equivalent to $\exists x\neg\varphi$. Pulling outwards $\forall x\neg\varphi$ is equivalent to $\neg\exists x\...
prophetes.ai 0.0 0.3 0.0
11
Order of quantifiers in prenex normal form I was wondering while doing some transformation into prenex form whether there is a situation when it does matter in what order you pull out the quantifiers to the front. If...
In converting a wff $\varphi$ by stages into an equivalent wff in some normal form, at each step we go from a wff to something equivalent. So whatever order we apply steps in, if there is a choice in your normalization procedure, then starting with $\varphi$ we must end up with something still equiv...
prophetes.ai 0.0 0.0 0.0
12
Prenex normal form question right? Just to see if I'm on the right track here, I did this sample problem in my book which unfortunately has no answers. It is asking to find an equivalent formula in prenex normal form ...
Your answer is correct, cheers! But, from the looks of it, it is important to realise that you did the following: \begin{align} \forall x(x =0 \lor (\neg\exists z(x+z=x))) &\iff \forall x(x=0\lor (\forall z \neg (x+z=x)))\\\ &\iff \forall x(\forall z(x = 0\lor \neg(x+z=x)))\\\ &\iff \forall z(\foral...
prophetes.ai 0.0 0.0 0.0
13
Convert predicate to prenex normal form $$\color{red}{(}\neg A(x)\lor \exists x\text{ }\color{green}{(}\forall y\text{ } B(x,y) \to \exists y \text{ }C(x,y)\color{green}{)}\color{red}{)}$$ So far I could only perform...
You are right that you cannot move the $\exists x$ over the $\neg A(x)$. But you can rename variables! So, you could replace the $x$ in $A(x)$ with a $z$, and now bring the $\exists x$ out. Remember, variables are just dummy placeholders.
prophetes.ai 0.0 0.0 0.0
14
Transform a wff to a prenex normal form Could you help solve this problem. I need to convert the following wff to a PNF ∀x(L(x)->∃x M(x,y)) I do not understand how universal quantifier and existential quantifier on s...
They are not working on the same variable. In $∀x(L(x)->∃x M(x,y))$, the outer variable $x$ is bounded to the all-quantifier. In $\exists x[M(x,y)]$, $x$ is bounded by the existential quantifier. You could rename it such as in $∀x(L(x)->∃z M(z,y))$. On the other hand, $y$ is a free variable.
prophetes.ai 0.0 0.0 0.0
15
Predicate Calculus - Resolution A question came up at the our schools logic club this week which involves using resolution to prove an argument in predicate calculus. I am slightly aware of how to find prenex normal f...
**Hint** The prenex normal forms are trivial; thus we can go directly to the "elimination" of $\exists$.
prophetes.ai 0.0 0.0 0.0