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


fixy
u =
v =
1 1 0 0
1 0 1 0
= u
= v
fjuv
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›