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.
If you want to go around calling things basic math, maybe you should actually do the math. You're not going to impress anyone by just parroting the idea that you're an idiot.
I can do it for you:
+----------+--------+
| eloquent | clumsy |
+-------+----------+--------+
| true | A | B |
+-------+----------+--------+
| false | C | D |
+-------+----------+--------+
Imagine there are A papers that are eloquent and true, B papers that are badly worded and true, C papers that are eloquent and false, and D papers that are badly worded and false. We'll also assume none of these values are 0.
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)
Observe here that the four values a, b, c, and d are all positive and their sum is 1.
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]
Let's examine whether eloquent papers are more likely to be true. As an algebraic statement, this is:
a / (a + c) > (a + b) / (a + b + c + d)
Same process:
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
We have now proven that eloquent papers are more likely to be true if and only if ad > bc. We've also proven that, if clumsy papers are less likely to be true, bc < ad. I'm really, really hoping you can fill in the rest of the proof.
Why did you insist - repeatedly - that a very basic and obvious fact was false? What were you thinking?
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.