# Without context – Does this evidence show that \$ EQ_ {CFG} \$ is incorrect?

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":

1. For each rope $$x in Sigma ^ *$$ in lexicographic order:
2. Test if x ∈ L (G) and if x L (H), using the algorithm of $$A_ {CFG}$$ .
3. 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 $$x = b$$.