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By P. B Andrews

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The proof, using 153, 147 and 144, is similar to the proof of 154. 156 tVaVx,A, = Vx,VaAo, provided this is a wff, and a does not occur in p. (Note that if this is a wff, either a does not occur in or xa does not occur free in Ao. 3 t- Vx,VaAo 3 VaVxpAo 152, 144: 147. 3. 157 t-VaAo = Vb[StAol,provided that (1) VaA, is a wff; (2) 6 does not occur free in A,,; (3) all free occurrences of a in A. are free for 6. 4 by Rule P. 3. %Ao = VaBo,provided this is a wff. 2 151 151. 6 by Rule P. , xn are distinct primitive variables; (2) for each i (n 2 i 2 l), no xi-variable other than xi, occurs free in A, or B,.

Are free fur all type variables in y, ; (2) u' is not free in any member of 32'; (3) for each wff-variable xp such that xp occurs free in A . and u' oc' curs in p, no x-variable occurs free in a member of 2 and no xvariable other than x, occurs free in A o . , yn, or in members of 2. % t- A . 2. This is the desired theorem. One must of course check that the conditions attached to rule 139 are satisfied each time the rule is applied in the above proof. A little reflection will convince the reader that this is so.

This we see as follows: We define certain type symbols to be critical. The definition is by induction on the length of the type symbol: (1) The type symbol 1 is critical, but no other type symbol of length 1 is critical. (2) The type symbol (ap) is critical if and only if a is critical and critical. p is not Note that l(O1) is critical. We easily see by induction on the length of a that if a > for some type symbol p, then a is not critical. For 2 is not critical, and if a has the where 8 > E. But form (8y) and a > p, then must have the form (q), by inductive hypothesis 6 is not critical, so a is not critical.

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