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Wythagoras once considered how to get hydras (or trees) past Buchholz ones. I can come up with a certain ruleset for it.

Definitions

  • # is a remainder of tree and they can be different in one expression as in (#)# unless we state that they are equal.
  • A(#) is the function which reduces tree to smaller one.
  • n is the number of rule-application which we perform.
  • ()# indicates #-leveled node (that is, one tree can express level of another).
  • Rn is the iterating function which was described by Chris Bird.
  • Three dots have obvious meaning and can alternate # sign in the rule 3 and indicate omitted (#)#'s in the rule 2.

Rules

Rule 1.

() is the separated node (it has no root):

A(#()) = #

Rule 2.

() has level 0, but it has a root:

A((#())#) = (#)#(#)#...(#)#(#)#

With n (#)#'s.

Rule 3.

#'s are equal in the string below, ... must not contain (#)# which have (#)#() in its subtree.

A((...()#()...)#) = Rn

Rm+1 = (...Rm...)

R0 = (......)

Rule 4.

#'s is non-empty, rules 1,2 and 3 don't apply.

A((#)) = (A(#))

Rule 5.

Rules 1-4 don't apply.

A(()#) = ()A(#)

Examples

()

(()) => ()

((())) => (()) => ()() => ()

(((()))) = ((())) = (()()()) = (()())(()())(()())(()()) = ...

(()()) = (()) = ()()() = ()() = ()

(()())() = (()()) = ((())) = (()()()) = ...

(()())()() = (()())() = (()()) = (((()))) = ((()()()())) = ...

(()())(()) = (()())()() = (()())() = (()()) = (((((()))))) = ...

(()())(()()) = (()())((())) = (()())(()()()) = ...

(()()()) = (()())(()())

(()()(()())) = (()()((())))

(()()()()) = (()()(()()(()())))

(()()()()()()) = (()()()()(()()()()(()()()())))

((())()) = (()()()())

((()())()) = ((())()(())())

(((()))()) = ((()())())

(((()()))()) = ((((())))())

((()())()) = ((((())()))())

(((()())())()) = ((((((())())()))())())

([{}]) = ([[[]]]) = ([[([[([[]])]])]])

([{}[]]) = ([{}([{}([{}])])])

([{}{}]) = ([{}[{}[{}]]])

([{()}]) = ([{}{}]) = ([{}[{}[{}[{}]]]])

([{()()}]) = ([{()}{()}])

([{[]}]) = ([{([{([{[]}])}])}])

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