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 Collections: Mathematics