I was looking for evidence showing that $ EQ_ {CFG} $ is co-Turing-recognizable. When I search for evidence, I can only find them on the following form:

Build an MT $ M $ which recognizes the complement of $ EQ_ {CFG} $, M =

"On the entrance ⟨G, H":

- For each rope $ x in Sigma ^ * $ in
:lexicographic order- Test if x ∈ L (G) and if x L (H), using the algorithm of $ A_ {CFG} $ .
- If one of the tests accepts and the other rejects, accept; if not continue.

(From: http://cobweb.cs.uga.edu/~potter/theory/6_reducibility.pdf)

This proof is not incorrect?

Say my language is $ Sigma = {a, b } $ and that I have $ L (G_1) = {b } $ and $ L (G_2) = {bb} $. $ M $ should accept $ langle G_1, G_2 rangle $ since $ L (G_1) neq L (G_2) $. But if we run $ M $ with $ langle G_1, G_2 rangle $ the TM will first generate $ x = a $, which is not available in both languages, the machine returns to step 1 and then generates $ x = aa $then $ x = aaa $ and so on forever. The machine will never arrive at $ x = b $. Therefore, the machine will loop over an entry that it should accept. So $ M $ is not a Turing recognition tool.

Would the proof be correct if step 1 indicated "For each chain $ x in Sigma ^ * $, *controlled by the growing size and then* lexicographic order: "This should generate strings in the following order: $ x = a $, $ x = b $, $ x = aa $, $ x = ab $, $ x = ba $, $ x = $ bb, $ x = aaa $etc. So in the case above $ M $ would refuse when $ x = b $.