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**Additional info for A Completeness Theorem in Modal Logic (paper)**

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5 Cod(A;5 ) E;(s); +(a> by E Pr otherwise. Assume A D . 1. Proof. 5 + 5' A , if A x(A) = order type E 5 5' < 0 ( ~ 2 ) . t+h : R -+ct;xe)2. 2. Assume A D . I f there is 5' < 0 ( ~ 2 ) . a surjection 4: R+,5 which i s ordinal definable j r o m a real (in L [ R ] ) then there is a surjection $: R + { 2 which is also ordinal dejinable f r o m a real (in L[R])sothat each subset of 5 is ordinal definable from a reaI (in L [ R ] ) . L [ R ] is t+h Proof L[R]) I 0(~2) nA 0(~2). t 1967, 1969. 2 o(R2), no 6 < x x 2 3.

2 supremum {q(fi): /? R } = g, is f, ( x , ~ ) ~ C o d (5 g ,)~* ; ~P[(~*[P],x,Y)EG], Zt(6) Cod(g,;x) is Case 2. 11 has a winning strategy G 3 x $9 by (x, y ) E o w I w & 3z[z S w & i ( w 5 z ) & { E } ( z ) i s defined (z), x, Y ) GI 7 82 {E}(z)N X R. Ae,w = w & ~ ( c 1 ) ~= z & x(a) R E { E } ~ , ~ ( Z ) , on Ei(S) N 3ct[n(~),, on z E Field(5). w w, 5 w , g:R x X R c> E,W, A&,W = G g k w ) = { ( x ,Y ) : (g(E, w), x, Y ) E by 4 z) i(w x 13 , o E E R ,W E X , T. GI * by G Zi(x) &, v, ~CX, on w EX, E* {&*I (w) _N on R x 3 , [g(E*, w ) ] * T 21 ME*, 4 1*7 5 g as = SUBLEMMA.

Proof. R by Xi(x) G(x) X's by (3-3) X:(x) ~ ( 6 N) x, , Proof of the theorem. us on YIANNIS 44 MOSCHOVAKIS by by F on 5 >w, 5 5 F 5 <5; L, w. F {aA} L, K 5

### A Completeness Theorem in Modal Logic (paper) by Kripke Saul

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