CNOT gate in ZX-calculus

In ZX-calculus, the CNOT gate is represented by this:

Can someone show my why this is true, using just the basic rewriting rules? All books/papers I have seen simply take it without proof, but I can't see why it is true.

I am going to assume that $$|0\rangle$$ is represented by , and $$|1\rangle$$ by .

Then, represents the bit-flip, or NOT gate. What it does is: $$|x\rangle\mapsto |x\oplus1\rangle$$. This can be verified easily in the graphical calculus, using the "spider" rule:

and

Now, by definition, CNOT controls NOT: $$|x_1,x_2\rangle\mapsto|x_1,x_1\oplus x_2\rangle$$. In other words, if $$|0\rangle$$ is fed to the first qubit, the result is $$|0\rangle$$ on the first qubit, and the identity on the second. If it is fed $$|1\rangle$$, the result is $$|1\rangle$$ on the first qubit, and the gate NOT on the second one. Again this can be verified in the graphical calculus:

where the first equality is the "copy" rule, the second is the "spider" rule, and the last one is the fact that a rotation of angle $$0$$ is the identity.

Similarly:

where the first equality is the "$$\pi$$-copy" rule, and the second is the "spider" rule.

• My apologies for accepting this answer so late. It was very helpful. Much appreciated! – Alex Apr 1 at 13:39