experiments-gpt3.5.txt

Query to 3.5: 

> Let 0 < a < 1 and k = ceiling(a/(1-a)).  How can I prove that a <= k/(k+1)?

Response:

---------------


To prove that a ≤ k/(k+1) given that 0 < a < 1 and k = ceiling(a/(1-a)), we can proceed as follows:

First, note that ceiling(a/(1-a)) means the smallest integer k such that k ≥ a/(1-a). Since 0 < a < 1, we know that 1-a > 0, and so a/(1-a) > 0 as well.

Therefore, we have k ≥ a/(1-a) > 0.

Multiplying both sides of this inequality by (k+1)(1-a), we get:

k(k+1)(1-a) ≥ a(k+1)

Expanding the left-hand side gives:

k(k+1)(1-a) = k(k+1) - ka

Substituting k = ceiling(a/(1-a)) gives:

k(k+1)(1-a) = k(k+1) - ka ≥ a(k+1)

So:

k/(k+1) - a/(k+1) ≥ 0

Multiplying both sides by (k+1), we get:

k - a ≥ 0

But we already know that k ≥ a/(1-a), so:

k - a ≥ 0 ≥ a(1/(1-a)) - a = a/(1-a)^2

Multiplying both sides by (1-a)^2, we get:

a(1-a) ≤ k(1-a)^2

Dividing both sides by k(1-a)^2, we get:

a/(k+1) ≤ 1/(k(1-a))

Since k ≥ a/(1-a), we have 1-a ≥ k/(k+1), so:

1/(k(1-a)) ≤ k/(k+1)

Therefore, a/(k+1) ≤ k/(k+1), which implies:

a ≤ k/(k+1)

This completes the proof.
-----------------------------

Query to GPT 3.5: 

> I'm sorry but this proof is incorrect.  You wrote, "k(k+1)(1-a) = k(k+1) - ka".  But in reality, k(k+1)(1-a) = k(k+1) - k(k+1)a.  Can you please fix the proof?

Response:

----------------------------


You are correct, and I apologize for the mistake in my previous answer. Here is a corrected proof:

Given 0 < a < 1 and k = ceiling(a/(1-a)), we want to prove that a ≤ k/(k+1).

Since k = ceiling(a/(1-a)), we have k-1 < a/(1-a) ≤ k, which implies (k-1)(1-a) < a ≤ k(1-a).

Multiplying both sides of the inequality (k-1)(1-a) < a by k+1, we get:

(k-1)(k+1)(1-a) < a(k+1)

Simplifying the left-hand side gives:

(k^2-1)(1-a) < a(k+1)

Expanding the left-hand side further gives:

k^2(1-a) - (1-a) < a(k+1)

Substituting k = ceiling(a/(1-a)) and noting that k > a/(1-a), we get:

(k^2 - 1)(1-a) < a(k+1)

(k^2 - 1) - ka < a(k+1)

k^2 - ka - 1 < a(k+1)

Since k = ceiling(a/(1-a)), we have:

k ≤ a/(1-a) + 1

ka ≤ a + a/(1-a)

ka ≤ a(2-a)/(1-a)

Substituting this inequality into the previous inequality, we get:

k^2 - a(2-a)/(1-a) - 1 < a(k+1)

Multiplying both sides by (1-a), we get:

k^2(1-a) - a(2-a) - (1-a) < a(k+1)(1-a)

Simplifying the left-hand side gives:

k^2(1-a) - 2a + a^2 < a(k+1)(1-a)

Rearranging and factoring, we get:

a(k+1)(1-a) > k(k+1)(1-a) - a^2 - 2a

a(k+1) > k(1-a) - a - 2a/(k+1)

Since 0 < a < 1, we have 1-a > 0, so we can divide both sides by 1-a to get:

a/(k+1) > k/(k+1) - a/(1-a) - 2a/(k+1)(1-a)

We can simplify the right-hand side as follows:

k/(k+1) - a/(1-a) = (k(1-a) - a)/(k+1)(1-a) < (k-1)/(k+1)

2a/(k+1)(1-a) < 2a/k(k+1) < 2a/k^2

Substituting these inequalities into the previous inequality, we get:

a/(k+1) > k/(k+1) - (k-1)/(k+1) - 2a/k^2

a/(k+1) > 1/(k+1) - 2a/k^2

Adding 2a/k^2 to both sides, we get:

a/(k+1) + 2a/k^2 > 1/(k+1)

Multiplying both sides by k(k+1), we get:

ak + 2a(k+1) > k

ak + 2a > k(1-a)

Since k = ceiling(a/(1-a)), we have k-1 < a/(1-a) ≤ k, which implies 1-a > k/(
--------------------------------------

Here is the version with GPT4:

Me: Let 0 < a < 1 and k = ceiling(a/(1-a)).  How can I prove that a <= k/(k+1)?

GPT4:
--------------------------------------
To prove that a <= k/(k+1) for 0 < a < 1 and k = ceiling(a/(1-a)), let's first understand what the ceiling function does. The ceiling function, written as ceiling(x), returns the smallest integer that is greater than or equal to x.

Now, let's break down the problem step by step.

1. Since k = ceiling(a/(1-a)), we know that k is the smallest integer greater than or equal to a/(1-a).
2. This implies that k >= a/(1-a) and k-1 < a/(1-a) (because k is the smallest integer greater than or equal to a/(1-a)).

Now, we will work with the inequality k-1 < a/(1-a) to show that a <= k/(k+1).

3. Multiply both sides of the inequality by (1-a) to get rid of the denominator: (k-1)(1-a) < a.
4. Expand the inequality: k - k*a - a + a^2 < a.
5. Rearrange the inequality to isolate a: a^2 - k*a + a < 0.
6. Factor a out of the left side: a(a - k + 1) < 0.

Now, let's analyze the inequality a(a - k + 1) < 0.
- Since 0 < a < 1, we know that a > 0.
- Also, since k = ceiling(a/(1-a)), k must be at least 1. Thus, a - k + 1 must be less than or equal to 0.

Considering these two points, we have a positive term (a) and a non-positive term (a - k + 1) on the left side of the inequality. Therefore, their product must be non-positive:

a(a - k + 1) <= 0.

Now, let's rewrite the inequality in terms of a and k:

7. Divide both sides by a (since a > 0, we don't need to worry about changing the inequality sign): a - k + 1 <= 0.
8. Add k and subtract 1 from both sides: a <= k - 1 + 1.
9. Simplify: a <= k.

Finally, we can show that a <= k/(k+1):

10. Since k >= 1, we know that k+1 > 1.
11. Dividing k by (k+1) must yield a value smaller than or equal to k: k/(k+1) <= k.
12. Combining the inequalities from steps 9 and 11, we get: a <= k/(k+1).

So, for 0 < a < 1 and k = ceiling(a/(1-a)), we have proven that a <= k/(k+1).

--------------------------------------
Me: Please formalize this proof using the ACL2 theorem prover.  You can do this by filling in the code where the ... are in the following snippet:

(defthm step-1-kens-proof
        (implies (and (pos-rationalp a)
                      (< a 1)
                      (equal k (ceiling (/ a (- 1 a)) 1)))
                 (<= a (/ k (1+ k))))
        :instructions ...)
	
GPT4:

-------
While I cannot directly interact with the ACL2 theorem prover, I can provide you with a complete definition for the theorem in ACL2's Lisp-like syntax. You can then use this definition to work with ACL2:

```
(defun pos-rationalp (x)
  (and (rationalp x) (< 0 x)))

(defun step-1-kens-proof-helper (a k)
  (<= a (/ k (1+ k))))

(defthm step-1-kens-proof
  (implies (and (pos-rationalp a)
                (< a 1)
                (equal k (ceiling (/ a (- 1 a)))))
           (step-1-kens-proof-helper a k)))
```

I have defined two utility functions, `pos-rationalp` and `step-1-kens-proof-helper`, to make the theorem definition cleaner. `pos-rationalp` checks if a number is a positive rational number, while `step-1-kens-proof-helper` checks if a is less than or equal to k divided by (k+1).

The main theorem, `step-1-kens-proof`, states that if `a` is a positive rational number less than 1 and `k` is equal to the ceiling of a/(1-a), then the helper function `step-1-kens-proof-helper` returns true for `a` and `k`.

Please note that this theorem definition only states the problem and does not provide a proof. You'll need to interact with the ACL2 theorem prover to actually prove the theorem.
------------

My actual proof in ACL2s:

(property divide-ineq (x :pos-rational y :pos-rational z :pos-rational)
  (implies (and (< 0 x) (<= y z))
	   (<= (/ y x) (/ z x))))

(property multiply-ineq (x :pos-rational y :pos-rational z :pos-rational)
  (implies (and (< 0 x) (<= y z))
	   (<= (* y x) (* z x))))

(defthm step-1-kens-proof
        (implies (and (pos-rationalp a)
                      (< a 1)
                      (equal k (ceiling (/ a (- 1 a)) 1)))
                 (<= a (/ k (1+ k))))
        :instructions (:pro (:claim (<= (/ a (- 1 a)) k))
                            (:use (:instance multiply-ineq (x (- 1 a))
                                             (y (* a (/ (+ 1 (- a)))))
                                             (z k)))
                            :pro
                            (:claim (<= (* (* a (/ (+ 1 (- a)))) (+ 1 (- a)))
                                        (* k (+ 1 (- a)))))
                            (:drop 1)
                            (:claim (<= a (* k (- 1 a))))
                            (:drop 5)
                            (:claim (equal (* k (+ 1 (- a))) (- k (* k a))))
                            (:claim (<= a (+ k (- (* k a)))))
                            (:drop 5 6)
                            (:claim (<= (+ a (* k a)) k))
                            (:claim (equal (+ a (* k a)) (* a (1+ k))))
                            (:claim (<= (* a (1+ k)) k))
                            (:use (:instance divide-ineq (x (1+ k))
                                             (y (* a (1+ k)))
                                             (z k)))
                            :pro
                            (:claim (<= (* (* a (+ 1 k)) (/ (+ 1 k)))
                                        (* k (/ (+ 1 k)))))
                            (:drop 1)
                            :prove))