Um, no.
Your outcomes are as follows with a brutal 2 d12:
Roll a 1 or 2 and reroll. 1/6 chance
Roll a 3 to 12 and keep it. 5/6 chance, equal opp
If you reroll, you get the same chance as above.
Now, to figure out your probabilities after all is said and done.
Rolling a 3 to 12, each outcome has the same probability to occur, 1/n. We don't know what n is yet, we have to solve for that.
Rolling a final result of 1 or 2, however, has a probability of 0/n. This is an impossible event, as if it ever occured, you'd not have reached a final result.
There are no other outcomes.
So you have 10 possible outcomes each with the exact same probability.
10/n = 1. n = 10.
So you have 1/10 chance of each outcome happening.
1d10+2 has 10 possible outcomes, each with 1/10 chance of occuring, and also creates the same series of integers, from 3 to 12.
In other words, the results, statisticly, between 1d12 brutal 2 and 1d10+2 are the same.
However, 1d12 brutal 2, unlike 1d10+2, does not interact with -other- effects that reroll 1's. Also, unlike 1d10+2, there's no confusion/ambiguity/counterintuitivity/hard math with how you deal with x[W] where x > 1.
1d10+2's average damage, by the way, is 7.5. Brutal 2 1d12's average damage is also 7.5 by no small coincidence....