Logic problem

assuming you have some set of expressions/variables: in this case you can't have a mix of true and false, so all the expressions in your set are logically equivalent

>>8604709
That's a cool math trick

As for your question, isn't this basically the AND operation?

so p1 <=> p2 <=> .... <=> pn

where {p1...pn} is your set of expressions

>>8604729
Or this too
With AND it would be p1 ^ p2 .... ^ pn - truth condition (all of them have to be true otherwise it's false) stays the same

>>8604726
Actually, yes! But I wanted to express it more mathematically instead of using just words. Also it is a set, so I guess like a function...? Like (∀P(x)) & ... Agh, I guess I don't know how to frase it. Hope you guys get the point

>>8604734

they can all be false, in which case the statement still holds, so it isn't AND.

>>8604709
∀x∈X(¬x→∀y∈X(¬y)) should work

>>8604709

If ∃x∈S s.t. x=0, then ∀x∈S x=0

>>8604740

you don't really need quantifiers/first order logic. again, this is just logical equivalence.

>>8604709
[math](\exists x \in S / \neg x) \implies ( \forall x \in S, \neg x ) [/math]

Is one way to put it, but I also thought of longer ways to put it. You can read it as

If there exists x element of S such that not x is true then for all x element of S, not x is true.

>>8604709
All elements are either false or true

You would write

If there exists an element a in X such that x is false, then all elements in X are false.

and could symbolize that like this

[math]\exists a: a \in X, \neg x \rightarrow \forall a: a \in X, \neg a[/math]

>>8604709
>2 digit numbers
>large

Bin Laden Math

>>8604709
jesus people, you don't need quantifiers to deal with this problem. if your domain of discourse is just a set of propositions, then you can use propositional logic

>>8604787
what does : mean. How could I learn more about logic?

>>8604798
>you don't need quantifiers to deal with this problem

But it looks better with quantifiers. >>8604768 looks better than >>8604729

Plus, >>8604729 is incomplete. It would be more like: X={p1,...pn} is such that p1 <=> p2 <=> .... <=> pn

>>8604804

you're misusing the notation of FOL.

if you want a pretty looking expression:

[math]\wedge_{p,q \in X} p \Leftrightarrow q [/math]

>>8604803
I'd read : as "such that". Though I wrote it wrong, as is it reads "If there exists an a such that a in an element of X and a is not ture, then for every a, a is an element of X and a is not true." Some fiddling will make it read "If there exists an a in X such that not a, then every a in x is not true.

The topic you want to study is formal logic, there are a lot of good introductory books. (Though I find the practice of encoding english into formal logic symbols boring and would instead recommend a discrete mathematics textbook)

>>8604821
>FOL

First order logic is for wimps. I never understood what the problem is with higher order logic.

And that expression is pretty good. I approve.

>>8604709
So how does that work for two wildly different numbers, like

65 x 26?

>>8604839

it's not necessarily wrong to use the language of FOL for this, but it indicates some confusion about what quantifiers are usually used for.

>>8604839
>>8604729
Doesnt this assume that every element of the matrix is an expression? Couldnt the matrix include both expressions and numbers?

>>8604860

there is no matrix involved.

>>8604868
set whatever

>>8604844
for two wildly different numbers you could do 50 or 10
for that I would double 65, times that by 10, then 65*6
so that would be initially 1300
390 is 65*6 (To make it easier just double it then 130*3)
so its 1690
solved in like 5 seconds

>>8604875
for 50, just do by 100 then halve it. using 10s, 50s, and 100s, you can get most things quickly

>>8604860
>Doesnt this assume that every element of the matrix is an expression?

Yeah. You can't "not" things that are not propositions so obviously it is a set of propositions.

>>8604883
>If 1 element in X is false, then all elements in X are false
is not equivalent to
>p1 <=> p2, for all p1, p2 in X
because take the set {True, 2}, clearly True <=/=> 2

>>8604872

the OP didn't really say, but you can't call it an unreasonable interpretation that each element is either true or false. otherwise, it's a very different question.

>>8604888

give me a break...

>>8604896
exactly, you cant make assumptions that change the context of the statement

>>8604901

yes, yes you can. what kind of retarded question would it be otherwise?

>>8604888

since the OP didn't provide any formal definition/predicate for equality, i think that i'm in the right.

why can't 2 = True? if you discard my original interpretation, then this is also fair game.

>>8604989
>why can't 2 = True?

This is funny because in programming a constant 2 is actually true. Actually, all numbers except for 0 are true.

For example, if you run this program:
#include <iostream>
#include <string>

int main()
{
if(2) std::cout << "lol";
}

Then it will print "lol".

>>8604745
/thread