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Definition 1.1. Suppose A is an alphabet, a A, s (A) and n = |s|. We let
 

Summary: 1. Scope.
Definition 1.1. Suppose A is an alphabet, a A, s (A)
and n = |s|. We let
o(a, s) = {i I(n) : si = a}.
A member of o(a, s) is called an occurrence of a in s. We say a occurs in s if
o(a, s) = .
Definition 1.2. Suppose S is a statement. B is substatement of S if B is a
statement which is a substring of S.
Definition 1.3. Suppose S is a statement and n = |S|. We define
scopeS : I(n) 2I(n)
as follows. Let Q = (N, , p, <, f) be a parse tree for S and let
0 < 1 < < n-1
be the leaf nodes. Suppose i I(n) and let be the parent of i. Let j I(n) and
m N be such that l, j l < j + m are the leaf nodes of the subtree with root
node . Then scopeS(i) = {l I(n) : j l < j + m}.
Note that
i scopeS(i) for i I(n).
Proposition 1.1. Suppose S is a statement, n = |S|, i I(n), j, m N are such
that
scopeS(i) = {l N : j l < j + m}

  

Source: Allard, William K. - Department of Mathematics, Duke University

 

Collections: Mathematics