28x = 0n (which implies
x = 0n
wherein x = Dinosaurusgede's words, and n = normal words
I can understand Dinosaurusgede, which implies I use the same words as her. therefor:
x = y
y = Catey's words.
y = x = 0n
but I know for a fact that 28y = 28n .
therefor (28n = 28y = 28x)/28 ≠ 0
it is clear we have made an error. On reinspection in becomes clear that speech(x) ≠ words(x) [that formula lies in the linquistics, which is unique for ever individual.]
For me, speech(y) = (words(y)/ b) = communication(x) .
Therefore if I were to say I am speechless: speech(x) = 
it would also imply that words(y) and communication(y) also = 
It is clear though that
(words(x) / b) = communication(x) ≠ speech(x) .
therefore while speech(x) =  , communication(x) and words(x) ≠ 
in fact, one can specify that communication is running: communication(x) =  and words(x) also: words(x) = [0<] this we know by counting the words and placing that number into b:
(words(x) /28) = communication(x) = 1
words(x) = [0<28]
hence, you were able to commune without using speech.
...so much to the topic that you be speechless.
yes, maybe I should add those to the keywords^^