User:Jon Awbrey/SANDBOX

Grammar Stuff

1.  The parse of the concatenation Conc^k of the k sentences S_j,
    for j = 1 to k, is defined recursively as follows:

    a.  Parse(Conc^0)        =  Node^0.

    b.  For k > 0,

        Parse(Conc^k_j S_j)  =  Node^k_j Parse(S_j).

2.  The parse of the surcatenation Surc^k of the k sentences S_j,
    for j = 1 to k, is defined recursively as follows:

    a.  Parse(Surc^0)        =  Lobe^0.

    b.  For k > 0,

        Parse(Surc^k_j S_j)  =  Lobe^k_j Parse(S_j).

1. The parse of the concatenation \(\operatorname{Conc}_{j=1}^k\) of the \(k\!\) sentences \((s_j)_{j=1}^k\) is defined recursively as follows:
1. \(\operatorname{Parse} (\operatorname{Conc}^0) ~=~ \operatorname{Node}^0.\)

2. For \(k > 0,\!\)

   \(\operatorname{Parse} (\operatorname{Conc}_{j=1}^k s_j) ~=~ \operatorname{Node}_{j=1}^k \operatorname{Parse} (s_j).\)

2. The parse of the surcatenation \(\operatorname{Surc}_{j=1}^k\) of the \(k\!\) sentences \((s_j)_{j=1}^k\) is defined recursively as follows:
1. \(\operatorname{Parse} (\operatorname{Surc}^0) ~=~ \operatorname{Lobe}^0.\)

2. For \(k > 0,\!\)

   \(\operatorname{Parse} (\operatorname{Surc}_{j=1}^k s_j) ~=~ \operatorname{Lobe}_{j=1}^k \operatorname{Parse} (s_j).\)

Table Stuff

fi‹x, y›	u = v =	1 1 0 0 1 0 1 0	= u = v	fj‹u, v›
	x = y =	1 1 1 0 1 0 0 1	= f‹u, v› = g‹u, v›

A	u = v =	1 1 0 0 1 0 1 0	= u = v	B
	x = y =	1 1 1 0 1 0 0 1	= f‹u, v› = g‹u, v›

u = v =	1 1 0 0 1 0 1 0	= u = v
x = y =	1 1 1 0 1 0 0 1	= f‹u, v› = g‹u, v›

u = v = x = y =

1 1 0 0 1 0 1 0 1 1 1 0 1 0 0 1

= u = v = f‹u, v› = g‹u, v›