We start with a macroscopic system ##\mathcal{S}## in a state ##|\psi\rangle## and we have an apparatus that measures the basis ##\{|\sigma_j\rangle\}##. There is also an environment ##\mathcal{E}##.
The combined system-environment state is then assumed to evolve into Schmidt form after measurement:
$$\Psi_{\mathcal{S}\cdot\mathcal{E}} = \sum_i \alpha_i |\sigma_i\rangle |\eta_i\rangle$$
with ##|\eta_i\rangle## a orthonormal basis for the environment.
What we want to show is that the probability of measuring a particular outcome ##j## of the basis ##|\sigma_j\rangle## from the microstate ##\psi##, denoted ##p\left(j,\{\sigma_j\},\psi\right)## is derivable from some property of the final state ##\Psi_{\mathcal{S}\cdot\mathcal{E}}##. That is:
$$p\left(j,\{\sigma_j\},\psi\right) = F\left(\sigma_i,\eta_i,\Psi_{\mathcal{S}\cdot\mathcal{E}}\right)$$
and that this probability agrees with the Born rule.
Assumption 1, Environmental Noncontextuality, EN:
$$p\left(j,\{\sigma_j\},\psi\right) = F\left(\sigma_i,\Psi_{\mathcal{S}\cdot\mathcal{E}}\right)$$
The probabilities do not depend on the environmental states after measurement.
Envariance is really just a consequence of this assumption. For if some unitary on the microscopic system can be undone by the environment and probabilities do not depend on the environment, then that unitary can have no effect on the probabilities.
Assumption 2, Perfect correlation, PC:
The chance of observing an environmental state ##\eta_i##, denoted ##G\left(\eta_i,\Psi_{\mathcal{S}\cdot\mathcal{E}}\right)##, obeys:
$$G\left(\eta_i,\Psi_{\mathcal{S}\cdot\mathcal{E}}\right) = F\left(\sigma_i,\Psi_{\mathcal{S}\cdot\mathcal{E}}\right)$$
That is, in a Schmidt state the environment and the system are perfectly correlated.
Zurek's proof in the equal amplitude case then be seen more easily via looking at the state:
$$\psi = \sqrt{\frac{1}{2}}\left(|0\rangle + |1\rangle\right)$$
for which the post-measurement Schmidt state is:
$$\Psi_{\mathcal{S}\cdot\mathcal{E}} = \sqrt{\frac{1}{2}}\left(|00\rangle + |11\rangle\right)$$
So:
$$p\left(0,\{|0\rangle,|1\rangle\},\sqrt{\frac{1}{2}}\left(|0\rangle + |1\rangle\right)\right) = F\left(|0\rangle,\sqrt{\frac{1}{2}}\left(|00\rangle + |11\rangle\right)\right)$$
Swapping the microscopic states can be undone by swapping the environment states, so by assumption 1 it has no effect on the system, hence:
$$F\left(|0\rangle,\sqrt{\frac{1}{2}}\left(|00\rangle + |11\rangle\right)\right) = F\left(|0\rangle,\sqrt{\frac{1}{2}}\left(|10\rangle + |01\rangle\right)\right)$$
In this new state the chance to measure the system in ##|0\rangle## is the same as the environment in ##|1\rangle##, hence:
$$F\left(|0\rangle,\sqrt{\frac{1}{2}}\left(|10\rangle + |01\rangle\right)\right) = G\left(|1\rangle,\sqrt{\frac{1}{2}}\left(|10\rangle + |01\rangle\right)\right)$$
We can then use the observation that swaps have no effect once again, this time on the environment:
$$G\left(|1\rangle,\sqrt{\frac{1}{2}}\left(|10\rangle + |01\rangle\right)\right) = G\left(|1\rangle,\sqrt{\frac{1}{2}}\left(|11\rangle + |00\rangle\right)\right)$$
And once more use the correlation between system and environment:
$$G\left(|1\rangle,\sqrt{\frac{1}{2}}\left(|11\rangle + |00\rangle\right)\right) = F\left(|1\rangle,\sqrt{\frac{1}{2}}\left(|11\rangle + |00\rangle\right)\right)$$
However this is just ##p\left(1,\{|0\rangle,|1\rangle\},\sqrt{\frac{1}{2}}\left(|0\rangle + |1\rangle\right)\right)##, hence:
$$p\left(0,\{|0\rangle,|1\rangle\},\sqrt{\frac{1}{2}}\left(|0\rangle + |1\rangle\right)\right) = p\left(1,\{|0\rangle,|1\rangle\},\sqrt{\frac{1}{2}}\left(|0\rangle + |1\rangle\right)\right)$$
Note how this works. We use EN to swap environments, PC then let's us convert this to a probability for environment observations. The latter step is what allows us to alter it from a statement about ##|0\rangle## states to one about ##|1\rangle## states, we just have to use EN once more to reattach the system states in their original order.
So envariance alone only tells us we can swap environments in certain scenarios, but it will never allow us to convert a statement about one element of a basis to another, we need PC for that (which is B3 in my list above, you could also use B1 or B2).
Okay, onto problems next.