Lamport's Byzantine Generals Algorithm
----I'm not sure if this should be my answer. Many thanks to hengxin. It does explain my confusion, and may be helpful to others seeing the question----

I failed to realise that there are 2 types of unreliable messages,
- some are different for each lieutenant, (eg 'eX' in round 1)
- others are the same for all lieutenants (I will denote these as 'S')

the amended example is,

&lt;(b1) ,c0   ,d0   ,eX   ,f1   ,g1   &gt;  

&lt;     ,(cb1),(db1),(ebX),(fb1),(gb1)&gt;   1  
&lt;(bc0),     ,dc0  ,ecX  ,fc0  ,gc0  &gt;   0  
&lt;(bd0),cd0  ,     ,edX  ,fd0  ,gd0  &gt;   0  
&lt;(beS),ceS  ,deS  ,     ,feS  ,geS  &gt;   S  
&lt;(bf1),cf1  ,df1  ,efX  ,     ,gf1  &gt;   1  
&lt;(bg1),cg1  ,dg1  ,egX  ,fg1  ,     &gt;   1  


now the result from 'round 2' is different to 'round 1', and allows agreement.
however, we could have acheived the same result with:

&lt;(b1) ,c0   ,d0   ,eX   &gt;  

&lt;     ,(cb1),(db1),(ebX)&gt;   1  
&lt;(bc0),     ,dc0  ,ecX  &gt;   0  
&lt;(bd0),cd0  ,     ,edX  &gt;   0  
&lt;(beS),ceS  ,deS  ,     &gt;   S  


Remember we have a traitor commander, but it seems more than coincidence that there is 1 traitor lieutenant out of 4 lieutenants (3t+1)?