If minimal counterexample method of proof is to assume to opposite of an argument is true and then finding a counterexample for the opposite and then concluding the validity of the original argument, then can't you say that the opposite of the Collatz conjecture is that the process will eventually reach to some number(s) other than 1, and if you put, say, 5 inside the Collatz function you'll reach one, so you found the counterexample, so the Collatz conjecture is valid?!
I know this approach is wrong, because any conjecture that was shown to be wrong with a large counter example will be proved right by this approach, or any theorem in any field that is valid can be easily proved because its opposite must be invalid therefore have counterexamples.
Where is my problem?
Answer
"minimal counterexample method of proof is to assume to opposite of an argument is true and then finding a counterexample for the opposite and then concluding the validity of the original argument"
This is false. The minimal counterexample method is to assume the opposite of a claim is true, and then to examine a minimal (in a sense dependent on context) counterexample to the original claim (a counterexample to the original claim must exist under the assumption of the opposite claim, and if things are nicely ordered, we can talk about a minimal one).
For an example of this type of argument, consider this proof that every number greater than $1$ can be written as a product of primes: Suppose, for sake of contradiction, that the claim is false; suppose that there are numbers greater than $1$ that can't be written as a product of primes. Then there should be a smallest one (a minimal counterexample to the claim "everything is a product of primes"); call it $n$.
Now, if $n$ can only be written as a product in forms like $1\times n$, then $n$ is prime, and hence isn't a counterexample after all. Otherwise, $n=a\times b$ where $a,b>1$. But since $n$ can't be written as a product of primes, at least one of $a$ and $b$ can't either. But since $a,b>1$ they must both be smaller than $n$. And hence $n$ isn't minimal after all. Since the concept of $n$ has problems either way, there must be no counterexample to the original claim after all, and every number greater than $1$ can be written as a product of primes.
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