prop(el ? (q)) ? q ? ?u, El(?), (prop(u) ? prop(v)) ? |u = El(?) v| ,
?) v|: prop(u) if and only if u(?) if and only if v(?), if and only if prop(v) ,
(u)|. We also have prop(f (u)) ? ? and prop(el ? )(?) ? ?. Hence prop(f (u)) ? prop((T rue ? * )(u)). By Theorem 18 we have |f (u) = El(?) (T rue ? * )(u)|. Then we can conclude, Properties of {u ? A | F (u)}) Assume A is an effective set and F ? SP (El(A)). Let B := {u ? A|F (u)}. Then: ?u, v ? |El(A)|, |u = El(B) v| ? (|u = El(A) v| ? F (u)) ,
Therefore u(a ? ) ,
If u(a) and u(a ? ), then |a = B a ? | by uncicity for u in B ,
A) v|. So we have |u = El(A) u|. Then there exists a ? |A| such that |u = El(A) el A (a)|. Hence u(a) and |el A (a) = El(A) u| ,
u(a ? ) then |a = A a ? | by unicity in A. We also have |u = El(A) u|. Then |u = El(A) el A (a)| by Lemma 5.3. Hence F (el A (a)) ,
Theorem 21 (C has pullbacks of T rue) Assume A is an effective set and f : El(A) ? El(?) ,
B) u| then by Theorem 20 we have prop(f (u)). So, prop(f (i(u))). Hence ?u ? El(B), prop(f (i(u))). Therefore, by Theorem 19 ,
X) w|, then |m(p 1 (w)) = El(B) m(p 2 (w))| by Theorem 20, Hence |m(f (w)) = El(B) m(g(w))|. Therefore, m ? f ? m ? g ,
El(A) such that ? m ? ? ? T rue ? * . Then ?u ? El(A), prop(? m (?(u))). For every w ? |El(X)| we construct F w ? |El(A)| ? P rop defined by: F w (u) := |u = El(A) u| ? |m, El(B) ?(w)|. We construct ? ? |El(X)| ? |El(A)| defined by ?(w) := d(F w ) ,
So prop(? m (?(w))). Hence, there exists u ? |El(A)| such that |u = El(A) u| and |m(u) = El(B) ?(w)|. Therefore ,
El(A) such that m ? ? ? ? ?. If |w = El(X) w| then we can prove that F w (?(w)) and the unicity of F w . We also have |m(? ? (w)) = El(B) ?(w)|. So, F w (? ? (w)) By unicity of F w we have |? ,
El(N at) S(el N at )(y)|: By definition of S, |S(el N at (x)) = El(N at) el N at (s(x))| and |S(el N at (y)) = El(N at) el N at (s(y))|. Hence |el N at (s(x)) = El(N at) el N at (s(y))|. Therefore, E(s(x), s(y)). By Lemma 26, Hence |el N at (x) = El(N at) el N at (y)| ,
El(N at) Z|: By definition of S, |S(el N at (x)) = El(N at) el N at (s(x))|. Hence |el N at (s(x)) = El(N at) el N at (0)|. Therefore E(s(x), 0) By Lemma 26 ,