Let's discuss Mathematics

A, A>B, D>~C, B>C |- ~D

1.A>B ------------------------------hyp
2.B>C ------------------------------hyp
3.D>~C -----------------------------hyp
4.A ---------------------------------hyp
5.B ---------------------------------MP for. 1, for. 4
6.C ---------------------------------MP for. 2, for. 5
7.C>~D -----------------------------CPI for. 3
8.~D -------------------------------MP for. 7, for. 6
 
Are you tripping on acid or just spamming?
 
That is a mathamatical proof.
 
OK I'll humour you. What is ~ defined as?
 
~= not
>= implies
/\= and
\/= or
<=> = biconditional
MP=Modus Ponus
when from P>Q, and P you get Q
CPI=Contrapositive Inverse
where from P>Q you get ~Q>~P
 
Right, not much discussion though.

I was thinking you were using > as greater than rather than implies and ~ as set complement which would make A>~C silly.
 
I am bored, solve this one, if you can(have know knowelege of your understanding of this, why would I, if not, then someone else do it)

A, A>B, D/\G, D>N, [B/\N]>~[~P\/~Q] |- P\/Q
 
I'll have to bow out of formal logic I'm afraid. Fifty has some knowledge in that area IIRC.
 
A, A>B, D/\G, D>N, [B/\N]>~[~P\/~Q] |- P\/Q

not gonna use your notation cuz i don't feel like it

6. (B&N)->~~(P&Q) 5, DeM
7. (B&N)->(P&Q) 6, DN
8. B 1,2, MP
9. D 3, simp
10. N 4,9,MP
11. B&N 8,10 conj.
12. P&Q 9,11 MP
13. P 12, Simp
14. PvQ 13, add.

Why the crap are we doing logic 101 proofs in thread that is supposed to be about interesting math.
 
Speaking of logic, here's a problem for you gamemaster77:

Consider the following system:

Primitive symbols: infinitely many propositional symbols P, Q, R, S, T, with and without subscripts, parenthesis (or dots), and the single operator symbol "|". The rules for wffs in N may be stated as:

1. Any single letter of N is a wff.
2. If P and Q are wffs, then (P)|(Q) is a wff.
(No formula of N will be regarded as being a wff unless its being so follows from this definition)

Single axiom (in metalanguage... its a pattern for infinite axioms in the object language):

Ax. P.|.Q|R:|::T.|.T|T:.|:.S|Q:|:P|S.|.P|S

Single rule of inference:

Rule. From P and P.|.R|Q to infer Q.

Now:



Prove: Q:.|:.Q|P.|.P:|:Q|P.|.P
 
Sorry fifty I am yet to figure that out.
 
That is fine ParadigmShifter. I only know basic logic fifty
 
No, its best shown by an example.
Say F(z) :-> z^2 = (x^2 - y^2) + i(2xy)
so U(z) = x^2 - y^2 and V(z) = 2xy

so dU/dx = 2x
and dV/dy = 2x

ans similarly

dU/dy = -2y and dV/dx = 2y

Strange huh?

Another example is just confusing since dutchfire already showed a counter-example. Better than simply giving examples would have been to get your hypothesis straightened out.
 
Strange huh?

When you consider the definition of complex differentiable it's pretty straightforward.

Basically, you can also consider a complex function as a function from R^2 to R^2. If it's complex differentiable it's Jacobi Matrix needs to have the form of a rotation times a real factor. That yields the Cauchy-Riemann equations.
 
Hey guys... Given two circles with radii r1 and r2 and and centre-to-centre distance d, how do I find how much area of those circles overlap (if any)?

Basically, I want a function of those three variables (or fewer, somehow), that doesn't involve integrals. That possible? I'll try and work through it myself in the mean time...

EDIT: Nevermind! http://mathworld.wolfram.com/Circle-CircleIntersection.html

Thanks Zelig!
 
Anyone know what this is about?

unambiguous.png
 
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