The proof divides a single multi-step test into one per command. For example,
if we have G(p,A) = (q, B, L), we used to create a command

c_{p,A}(s_i, s_{i-1})
  if p in A[s_i,s_i] and A in A[s_i, s_i] and own in A[s_{i-1}, s_i]
    delete p from A[s_i,s_i]
    delete A from A[s_i,s_i]
    insert B into A[s_i,s_i]
    insert q into A[s_{i-1},s_{i-1}]

Now we create 4 steps:

c_{p,A}_step1(s_i)
  if p in A[s_i,s_i] and A in A[s_i, s_i] and ready in A[s_{state},s_{state}]
    delete ready from A[s_{state},s_{state}]
    delete p from A[s_i,s_i]
    insert r_{p,A,1} into A[s_{state},s_i]

c_{p,A}_step2(s_i)
  if r_{p,A,1} in A[s_{state},s_i]
    delete r_{p,A,1} from A[s_{state},s_i]
    delete A from A[s_i,s_i]
    insert r_{p,A,2} into A[s_{state},s_i]


c_{p,A}_step3(s_i)
  if r_{p,A,2} in A[s_{state},s_i]
    delete r_{p,A,2} from A[s_{state},s_i]
    insert B into A[s_i,s_i]
    insert r_{p,A,3} into A[s_{state},s_i]

c_{p,A}_step4(s_i)
  if r_{p,A,3} in A[s_{state},s_i] and own in [s_{i-1},s_i]
    delete r_{p,A,3} from A[s_{state},s_i]
    insert q into A[s_{i-1},s_{i-1}]
    insert ready into A[s_{state},s_{state}]

Then you have to prove that exactly one command is applicable at any point in
time, and that this continues to properly execute the Turing Machine. This is
because if ready is in A[s_{state}, s_{state}], then the argument from before
continues to hold; if ready is not in A[s_{state}, s_{state}], then there's
exactly one applicable command based on the r_{p,A,n} rights in the matrix
(be sure to use a different one for "rightmost"). It continues to simulate a
Turing Machine because when steps 1-4 are executed in a row (which they must
be, we proved it earlier), it's the same as the original, and the original
executed a Turing Machine, so we're done.