### Story Post: Challenge Post (5)

If the axioms in the post “Pairs and lists” are true, then there exists a list X that has all of the following properties:

* Every element of X is a list.

* For all elements a of X, if b and c are lists such that b ++ c = a, then there does not exist d such that d is an element of both b and c.

* For all pairs (a,b) of elements of X, there does not exist c such that c is an element of both a and b.

* X has at least two elements.

* Every element of X has at least two elements.

* Every element of X has the same number of elements.

* The number of x such that x is an element of an element of X is exactly 35 844 088 534 668 175 608 533 550 469 325 416 140 711 880 836 358 433 520 741 183 341 559 409 752 958 417 564 900 205 152 763 659 819 317 338 304 041 228 758 958 269 744 936 094 178 001 500 112 469 790 418 773 135 543 301 899 194 065 881 801 585 777 676 220 908 841.

Did I do that right,

## Clucky: he/him

(unfortunately, it doesn’t look like I get any points for proving you wrong but I do not believe this to be true.

1: Property 1: If b and c are lists such that b ++ c = a, then there does not exist d such that d is an element of both b and c.

2: Claim 1: For all a, [if a is a list and a contains a finite number of elements, then the empty list is an element of a].

3: Subclaim A: If a list has a finite number of elements, one of them must be the empty list.

4: Suppose c is not the empty list. Then c = a:b where b is a list and a is an element of c. (axiom 5)

5: Subclaim B: If c is a list, then b:{} ++ c = b:c

6: b:{} ++ c = b:{{} ++ c} (axiom 10)

7: {} ++ c = c (axiom 11)

8: so b:{} ++ c = b:c

9: Subclaim C: If a is a list s.t. Property 1 holds, and a = b:c then if c has x elements, b has x + 1 elements.

10: For all d, if d = b then

11: b:{} ++ c = b:c = a (Subclaim b)

12: b is an element of b:{} (Axiom 8)

13: b is not an element of c (Property 1)

14: If d is an element of c, then d is an element of a (Axiom 8)

15: And if d != b, and d is not an element of c, then d is not an element of a. (Axiom 8)

16: So if c contains x elements, a contains every element of c plus b, or x + 1 elements.

17: Subclaim D: There does not exist a finite n > 0 (such that there exists a list a (such that a has exactly n elements and the empty list is not an element of n))

18: If there does exist such an n, there must be an m + 1 (m > 0) such that m + 1 is the smallest such n.

19: Let c be a list which contains m + 1 elements.

20: c is non empty as m > 0 (Subclaim A) so c = a:b where b is a list.

21: b contains n elements for some n

22: n + 1 = m + 1 (Subclaim C)

23: So n = m

24: m < m + 1 so the empty list is an element of b

25: Thus the empty list is an element of a (Axiom 8) And we have a contradiction

26: Claim 2: Every element of X cannot contain a finite number of elements:

27: Let c be an element of X.

28: c is a list, so b:{} ++ c = b:c (Subclaim 2)

29: {} is an element of b:{} (Axiom 8)

30: So {} cannot be an element of c (Property 1)

31: So c does not contain a finite number of elements (Claim 1)

32: But in order for the number of x such that x is an element of an element of X to be finite, then each element of X must have a finite number of elements. I have shown this to be not true, so the whole claim is false.