No, it's not. Mathematical elimination only applies to the first ballot.
WTF? I'm just trying to make sense of what you're saying.
I took this:
No. He might not be able to get 1237 on the first ballot (which I believe he's more or less acknowledged for a while now) but that does not mean he's been mathematically eliminated...yet.
...to mean that Cruz could be mathematically eliminated from the first ballot because there wouldn't be enough delegates left for him to get to 1237 on the first vote, but he is not yet mathematically eliminated from wining the nomination because he could, theoretically, win it on the second vote. Phat chance of that, but theoretically.
Now if you meant something more than that, it would have been helpful to say more of what you found inaccurate about what Cruz's team said.
It's not clear from this:
I heard this from some Cruz spokesperson Tuesday night, and it is one of the most absurd things I've ever heard.
The first ballot, and the first ballot only, is what "mathematically eliminated" applies to. It does not apply to any other ballot, because no candidate is "mathematically eliminated" from balloting rounds in which delegates are unbound (which happens, with increasing numbers of delegates, in the second round of balloting).
By this Cruz line of thinking, NO candidate is "mathematically eliminated".
exactly what about Cruz's line of thinking you found so absurd. I thought you were disagreeing that mathematical elimination only applies to the 1st vote.