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1. Equivalence is preserved under substitution. We fix a first order logic F and we let S be its set of statements.
 

Summary: 1. Equivalence is preserved under substitution.
We fix a first order logic F and we let S be its set of statements.
Theorem 1.1. Suppose A, B and Ci, Di, i = 1, . . . , N for some positive integer N
satisfy the following conditions.
(i) A S;
(ii) for each i = 1, . . . , N, Ci, Di S and (Ci Di);
(iii) for each i = 1, . . . , N, Ci is a substring of A;
(iv) if 1 i < j N then the substrings Ci and Cj of A are disjoint;
(v) B is the string resulting from replacing each occurrence of Ci in A by Di,
i = 1, . . . , N.
Then
B S and (A B).
Proof. Well, it's a sketch. Let T be a parse tree for A.
Suppose 1 i N. Consider the leaf node in T corresponding to the leftmost
letter in Ci. Note that it is the leftmost member of the children of its parent. Thus
the subtree of T corresponding to this parent must parse Ci. It follows that B S.
To prove that (A B) we induct on the depth of T . Owing to previous
theory we may assume without loss of generality that none of , , , occur in
A. In case A = E for some E S or in case A = (E F) for some E, F S
the assertion follows by earlier work (and is pretty clear regardless). So suppose

  

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

 

Collections: Mathematics