By P. B Andrews

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**Example text**

A, em-, m r 2 0. If s = n then either b is not free in A, or B,, or else b is already included among a', ---, a', so that any free occurrences of b in A, or B, are not free in Eo or Eb. If s = n - 1 then a" is b. -Vxir[Aa= B,]. 4 I- Val ... Eo = EA by inductive hypothesis ... Van-'VanVxp', ... 6 I- Val ... 7I-Val VaWx;, ... 9 I- Val 3 ... PbE,] = [\dbEb] 0 . 8. , is free in [VbE,] or [VbEb] but Note that each of that the occurrence of [VbE,] in C, is not free in C, for any of these xp", (where k 2 0) be a complete k variables.

Do]],so this is the desired theorem. ), provided that for each i (n 2 i 2 I), no x'-variable other than xLi occurs free in A,. ) The proof is by induction on n. If n = 1 we have 104 t- [Ax&& t- [Ax,AJx, 2 A , by 103, since our conditions assure that the conditions on 103 are satisfied. If n > 1 by inductive hypothesis we have 33 BASIC LOGIC IN Q 105 If 2 I- Bp e C, then &' I- Si;[BP= Cp],provided that B a (I) y 2 a and all free occurrences of xy in [BP= CP]are free for all free variables of A,; P (2) xy is the only x-variable free in [B, = Cp] ; (3) no x-variable occurs free in a member of 2.

Note that no wff-variables occur free in these wffs. Now consider the system Q' which is obtained when Axiom Schema 5 is deleted from the list of axioms of Q. It is easy to see by induction on the length of the proof of A o , that if A . is a theorem of Q', then Bo is also a theorem of Q', where Bo is the result of simultaneously replacing all occurrences of the wffs in A . by the wffs Ja(oa)for each type symbol a. (Note that no trouble arises when A . Qoaay,] y, were derivable in Q . I. Qoaaya] ya would also be derivable, so by Rule R and the axiom Ja(oa)[Qoaay,]2 C, (an instance of 4,) we would obtain C, & y, as a theorem of Q'.