Update README.md

README.md

@@ -132,6 +132,46 @@ tokenizer_source: model:cognitivecomputations/dolphin-2.2-70b # keep exact token
 
 # Example Output
 
+**Prove P != NP**
+
+In the pursuit to prove P != NP I want to try to prove that no polynomial-time algorithm could possibly solve all problems in NP, even though we may not know of any specific such problem that defies polynomial-time solution.  Can you please make a lemma that would be helpful?
+
+> Certainly! One possible direction towards proving P ≠ NP using non-computational or proof-theoretic methods is to try to show that certain kinds of mathematical statements, known as undecidable statements, cannot be proven in any polynomial-time verifiable formal system, such as first-order logic with addition (FO+M). In other words, we would like to show that there are inherent limits to what can be expressed or proven within these restricted systems, even though the full power of human intuition and insight may allow us to see their truth or falsity.
+>
+> Here's a simple example of such an undecidable statement in the realm of Diophantine equations (i.e., equations in multiple variables with integer coefficients):
+> 
+> ```python
+> def Goldbach_Conjecture(n):
+>     if n % 2 == 1 and n < 4:
+>         return True
+>     elif n > 4 and n % 3 == 1:
+>         m = (n - 1) // 2
+>         return Goldbach_Conjecture(m) or Goldbach_Conjecture(n - m)
+>     else:
+>         return False
+> ```
+> 
+> This Python function checks whether a given even number n can be expressed as the sum of two prime numbers (not including 2 itself). The famous unsolved Goldbach Conjecture, named after the Prussian mathematician Christian Goldbach, states that this is always true for all sufficiently large even numbers n. However, despite much effort and extensive computational verification up to very large values of n, no one has yet found either a proof or counterexample to settle this conjecture definitively.
+> 
+> Suppose, for the sake of contradiction, that there were an algorithm A running in polynomial time on input n such that Goldbach_Conjecture(n) == A(n) for all even integers n. Then we could use this algorithm to decide whether any given number n is prime or composite (i.e., divisible by a prime other than 1 and itself) as follows:
+>
+> ```python 
+> def Goldbach_Prime_Test(n):
+>     if n % 2 == 0 and n < 4:
+>         return True
+>     elif n > 4 and n % 3 == 1:
+>         m = (n - 1) // 2
+>         return not Goldbach_Conjecture(m) or not Goldbach_Conjecture(n - m)
+>     else:
+>         return False
+> ```
+> 
+> This function checks whether n is a prime number by trying to express it as the sum of two other odd numbers, one of which must then be composite by our assumption that all even composites can be factized into primes. If this attempt fails for all possible such decompositions, we conclude that n itself must be prime.
+> 
+> However, deciding whether a given number is prime or composite is known to be an NP-hard problem, meaning that no polynomial-time algorithm exists that can always correctly classify all numbers as either prime or composite, unless P = NP. This leads to a contradiction, since our original assumed algorithm A would then provide such a polynomial-time solution to the prime number decision problem, which we know cannot exist if P ≠ NP holds true.
+> 
+> Thus, by proof-by-contradiction, we have shown that there exists at least one undecidable statement in the realm of Diophantine equations (specifically, whether all even numbers can be expressed as sums of two prime numbers) that cannot be proven or refuted within any polynomial-time verifiable formal system, such as first-order logic with addition. This result, while not directly implying P ≠ NP, demonstrates one possible strategy towards achieving such a negative answer to the millennium-prize question by leveraging non-computational or proof-theoretic techniques beyond traditional Turing machine-based complexity theory.
+
 **Explain principia mathematica**
 
 Please walk me through bertrand russell's proof that 1 + 1 = 2