Proposal: Bracketeering
Times Out and Passes 4-0-1. -RaichuKFM
Adminned at 27 Jun 2013 10:44:11 UTC
In the list of Axioms, replace
For all a and b, if b is a list, then a:b is a non-empty list.
with
For all a and b, if b is a list, then {a}:b is a non-empty list.
Replace
For all a, if a is a non-empty list, then there exist b and c such that a = b:c and c is a list.
with
For all a, if a is a non-empty list, then there exist b and c such that a = {b}:c and c is a list.
Replace
For all a, b, c, and d, if b and d are lists, then [a:b = c:d if and only if [a = c and b = d]].
with
For all a, b, c, and d, if b and d are lists, then [{a}:b = {c}:d if and only if [a = c and b = d]].
Replace
For all a and b, if b is a list, then [the head of a:b is a, and the tail of a:b is b].
with
For all a and b, if b is a list, then [the head of {a}:b is a, and the tail of {a}:b is b].
Replace
For all a, b and c, if c is a list, then [a is an element of b:c if and only if [a = b or a is an element of c]]
with
For all a, b and c, if c is a list, then [a is an element of {b}:c if and only if [a = b or a is an element of c]]
And replace
For all a, b and c, if b and c are lists, then (a:b) ++ c = a:(b ++ c).
with
For all a, b and c, if b and c are lists, then ({a}:b) ++ c = {a}:(b ++ c).
Travos and I were having this discussion on his Challenge Post. I think the list containing a_1:a_2, and the list containing a_1 *and* a_2 are both legally written a_1:a_2:{} despite not being the same list. This would force us to write them as {a_1:a_2}:{} and {a_1}:{a_2}:{}, making them clearly different lists.
nqeron: