It's a logical tautology, at least if we make the implications definite (i.e., "sounds bad" necessarily implies "not right", and therefore "right" necessarily implies "does not sound bad"). In other words, "A implies B" is logically equivalent to "Not B implies Not A". There's no need to give any further proof.
> you should be well aware that exactly the same proof will quickly show that writing that doesn't sound bad is more likely to be right.
No, it won't. You really need to learn some basic logic.
> They are the same statement
No, they're not. Again, please learn some basic logic. "A implies B" is not logically equivalent to "B implies A". You are claiming that it is. Any basic textbook on logic will tell you that you are wrong.
I can do it for you:
+----------+--------+
| eloquent | clumsy |
+-------+----------+--------+
| true | A | B |
+-------+----------+--------+
| false | C | D |
+-------+----------+--------+
For convenience, I'll use the lowercase letters to refer to probabilities rather than counts: a = A / (A + B + C + D), b = B / (A + B + C + D), etc.
We are given the postulate "papers that sound bad [in our terminology, "clumsy"] are less likely to be true". There are two possible interpretations of this:
1. Clumsy papers are less likely to be true than the average of all papers.
2. Clumsy papers are less likely to be true than eloquent papers.
Fortunately for us, each of these implies the other, so I'll use interpretation (1), again for convenience.
We can now define our postulate precisely:
b / (b + d) < (a + b) / (a + b + c + d)
We can use basic algebra to rearrange the postulate:
b / (b + d) < (a + b) [because a + b + c + d = 1]
b < (a + b)(b + d) [because (b + d) is positive]
b < (ab + ad + b^2 + bd) [algebra]
b < b(a + b + d) + ad [algebra]
b < b(1 - c) + ad [because a + b + c + d = 1]
b < b - bc + ad [algebra]
0 < 0 - bc + ad [algebra]
bc < ad [algebra]
a / (a + c) > (a + b) / (a + b + c + d)
a / (a + c) > (a + b)
a > (a + b)(a + c)
a > a^2 + ac + ab + bc
a > a(a + c + b) + bc
a > a(1 - d) + bc
a > a - ad + bc
0 > 0 - ad + bc
ad > bc
Why did you insist - repeatedly - that a very basic and obvious fact was false? What were you thinking?
> What is necessarily the case is that, if writing that sounds bad is less likely to be right, writing that is right is less likely to sound bad.
That's true. You're 100% right about this.
But if you're able to prove it, you should be well aware that exactly the same proof will quickly show that writing that doesn't sound bad is more likely to be right.
> Which, as above, is not logically equivalent to "writing that doesn't sound bad is more likely to be right".
And that, obviously, is false. They are the same statement; either is sufficient to prove the other. I don't know what happened in your comment.