My Math Forum How to rewrite these formulas in the opposite form?

 Applied Math Applied Math Forum

 March 15th, 2019, 07:37 AM #1 Newbie   Joined: Dec 2018 From: Amsterdam Posts: 28 Thanks: 2 How to rewrite these formulas in the opposite form? I want to know how I can write from some predicate formula its negative form. For example ∀x∃y(x R y) then not(∀x∃y(x R y)) is equal to ∃x,not ∃y(x R y)))(not exist y) How do these formula work with conjunction implication and disjunction? Bellow are examples of some formulas but I also want to know how to rewrite the formulas if there are conjunctions and disjunctions involved. $\forall a,b,c \in X: (aRb \wedge bRc) \Rightarrow aRc$ what do we have if we have the negation of this formula? also the opposite of: $\forall a, b \in X(a R b \Rightarrow \lnot(b R a))$ also for the opposite of: $\forall a, b, c: a R b \land b R c \Rightarrow \lnot (a R c)$ also for the opposite of $\forall a, b, c\in X\,(a\,R\, b \land a \,R\, c \to b \,R\, c)$ also for the opposite of: $\forall a, b \in X(a R b \Leftrightarrow b R a)$ Last edited by skipjack; March 15th, 2019 at 01:30 PM.
 March 15th, 2019, 08:22 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 645 Thanks: 92 Simply there is another symbol of relation R which is an opposite of the given relation .
March 15th, 2019, 02:30 PM   #3
Newbie

Joined: Dec 2018
From: Amsterdam

Posts: 28
Thanks: 2

Quote:
 Originally Posted by idontknow Simply there is another symbol of relation R which is an opposite of the given relation .
What do you mean with this?

 March 15th, 2019, 02:54 PM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 645 Thanks: 92 About the example $\displaystyle A : \forall x \exists y (x R y)$ we can have two opposites of A . $\displaystyle \neg A =\forall x \not\exists y (xRy)\;$ , but instead of $\displaystyle \forall x$ the $\displaystyle \exists ! x$ ( exists only one x) also is a type of opposite of A . So the best way is to just simply write the opposite of relation R with a symbol .
March 15th, 2019, 11:49 PM   #5
Newbie

Joined: Dec 2018
From: Amsterdam

Posts: 28
Thanks: 2

Quote:
 Originally Posted by idontknow $\displaystyle \neg A =\forall x \not\exists y (xRy)\;$ , but instead of $\displaystyle \forall x$ the $\displaystyle \exists ! x$ ( exists only one x) also is a type of opposite of A . So the best way is to just simply write the opposite of relation R with a symbol .
How would you rewrite this
∀w1w2w3(w1Rw2 ∧ w1Rw3 → ∃w(w2Rw ∧ w3Rw))

If you want to rewrite it like:
not(∀w1w2w3(w1Rw2 ∧ w1Rw3 → ∃w(w2Rw ∧ w3Rw))))

Can this be rewritten as:
∃w1w2w3(w1Rw2 ∧ w1Rw3 → (not ∃w(not exist w)(w2Rw ∧ w3Rw))))?

 March 16th, 2019, 08:10 AM #6 Senior Member   Joined: Dec 2015 From: somewhere Posts: 645 Thanks: 92 Yes but still $\displaystyle \rightarrow \exists$ can have the opposite like : $\displaystyle \rightarrow \not\exists$ and $\displaystyle \not\rightarrow \exists$ . So you can write the opposite but depending on what relation is .
March 16th, 2019, 08:23 AM   #7
Newbie

Joined: Dec 2018
From: Amsterdam

Posts: 28
Thanks: 2

Quote:
 Originally Posted by idontknow Yes but still $\displaystyle \rightarrow \exists$ can have the opposite like : $\displaystyle \rightarrow \not\exists$ and $\displaystyle \not\rightarrow \exists$ . So you can write the opposite but depending on what relation is .
I think that you and I had it both wrong and that the implication had to be removed.

∃w1w2w3(w1Rw2 ∧ w1Rw3 ∧ (not ∃w(not exist w)(w2Rw ∧ w3Rw))))

This is the correct formula do you agree?

 March 16th, 2019, 08:42 AM #8 Senior Member   Joined: Dec 2015 From: somewhere Posts: 645 Thanks: 92 Yes this way gives the full or true opposite , removing the implication.

 Tags form, formulas, opposite, predicate, rewrite

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post shaharhada Calculus 2 December 16th, 2016 07:29 AM apartin Calculus 0 March 10th, 2016 01:17 PM deepblue Elementary Math 5 June 9th, 2015 06:44 AM mathkid Calculus 2 September 16th, 2012 04:11 PM Scrimski Algebra 5 October 10th, 2008 04:57 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top