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Correct the at most degrees 

`L,R,O` are of degree at most `d-1` .
 `T` is of degree at most `d`.
So `P = LR-O=HT` is of degree at most `2(d-1)` and `H` is of degree at most `d-2`.
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_posts/2017-05-10-snark-explain6.md
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@ -23,7 +23,7 @@ As we saw in Part V, Alice will typically want to prove she has a satisfying as
If Alice has a satisfying assignment it means that, defining :math:`L,R,O,P` as above, there exists a polynomial :math:`H` such that :math:`P=H\cdot T`. In particular, for any :math:`s\in\mathbb{F}_p` we have :math:`P(s)=H(s)\cdot T(s)`.
Suppose now that Alice <em>doesn't</em> have a satisfying assignment, but she still constructs :math:`L,R,O,P` as above from some unsatisfying assignment :math:`(c_1,\ldots,c_m)`. Then we are guaranteed that :math:`T` does not divide :math:`P`. This means that for any polynomial :math:`H` of degree at most :math:`d`, :math:`P` and :math:`L,R,O,H` will be different polynomials. Note that :math:`P` and :math:`L,R,O,H` here are both of degree at most :math:`2d`.
Suppose now that Alice <em>doesn't</em> have a satisfying assignment, but she still constructs :math:`L,R,O,P` as above from some unsatisfying assignment :math:`(c_1,\ldots,c_m)`. Then we are guaranteed that :math:`T` does not divide :math:`P`. This means that for any polynomial :math:`H` of degree at most :math:`d-2`, :math:`P` and :math:`L,R,O,H` will be different polynomials. Note that :math:`P` here is of degree at most :math:`2(d-1)`, :math:`L,R,O` here are of degree at most :math:`d-1` and :math:`H` here is degree at most :math:`d-2`.
Now we can use the famous <a href="https://en.wikipedia.org/wiki/Schwartz%E2%80%93Zippel_lemma">Schwartz-Zippel Lemma</a> that tells us that two different polynomials of degree at most :math:`2d` can agree on at most :math:`2d` points :math:`s\in\mathbb{F}_p`. Thus, if :math:`p` is much larger than :math:`2d` the probability that :math:`P(s)=H(s)\cdot T(s)` for a randomly chosen :math:`s\in\mathbb{F}_p` is very small.
@ -93,4 +93,4 @@ We presented a sketch of the Pinocchio Protocol in which Alice can convince Bob
Both these issues can be resolved by the use of pairings of elliptic curves, which we will discuss in the next and final part.
<p><a class="reference external" href="https://z.cash/blog/snark-explain7.html">&gt;&gt; Part VII</a></p>
<p><a class="reference external" href="https://z.cash/blog/snark-explain7.html">&gt;&gt; Part VII</a></p>