Online math tool for equation visualisation I am looking for a math editor that is able to present equations and relations between them. I am not searching for a traditional Latex / MathML editor (I know about various online services that allow typing in equations in various formats then outputting a pretty picture). I know about online sage notebook, about mathjax and that like. I also know about Lurch and Wolfram Alpha (thanks to tp1 below).
What I'm looking for is something like pearltrees in interface with pearls being equations and edges indicating the source of the concepts (other equations or axioms). The equations should be editable and each may have an additional page that explains the equation, maybe from external source like Wikipedia.
Here is a sample tree with a direct proof directly from Wikipedia. 

Consider two even integers x and y. Since they are even, they can be written as x = 2a and y = 2b respectively for integers a and b. Then the sum x + y = 2a + 2b = 2(a + b). From this it is clear x + y has 2 as a factor and therefore is even, so the sum of any two even integers is even.

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 A: Trees for some equational proofs
Mathematica (which is Online) recently added some functionality to find proofs for some equational logic theorems and present them in various ways in both text and a tree.
There are a lot of examples/pictures at the reference page for FindEquationalProof. One particularly striking picture is a tree for a proof that one axiom implies a collection of 3 for NAND. If $(ab)$ denotes $a\,\operatorname{NAND}\,b$, then the three axioms

*

*$((aa)(ab))=a$

*$(a(ab))=(a(bb))$

*$(a(a(bc)))=(b(b(ac)))$
all follow from $(((ab)c)(a((ac)a)))=c$. Mathematica proves this in 250 steps with a tree like:


Trees for many theorems
This doesn't cover equational proofs yet because it's only been done for the first 1000 proofs, but Antony is working on automatically converting metamath proofs to trees at this github repository. For example, this page shows the following tree for the theorem "ccase", which roughly says that if we have $P\land R\to T$ and $Q\land R\to T$ and $P\land S\to T$ and $Q\land S\to T$, then we can derive $(P\lor Q)\land (R\lor S)\to T$:

