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? {x})) for ? ? {?, ?} 6. (? 1 ?? 2 , first) ? ? (? i , first) for ?, p.2 ,
, ? ? (Win, (a, S)) if x ? S 8. (¬a(x), (b, S)) ? ? (Win, (b, S)) if x ? S and b = a
, 1, r) such that w, ? |= ?x.? , then by construction of f ? we have v k+1 = (c( ? , w, ? ), i + 1, r) and v k+2 = (c(? , w, ? ), 1, r + 1) for some 1 ? i ? n and ? = ? {x ? i}
, 1, r) such that w, ? |= ?x.? , then for some 1 ? i ? n (defined by f ) we have ? = ? {x ? i} such that v k+1 = (c( ? , w, ? ), i + 1, r) and v k+2 = (c(? , w, ? ), 1, r + 1). By definition since w, ? |= ?x
, 1, 2), we deduce that v n+2·|Var?| = (c(? , w, ?), 1, B) where ? has no quantifiers and w, ? |= ?
, 1, B) and w, ? |= ? 1 ? ? 2 then by definition of f ? , v k+1 = (c(? i , w, ?), 1, B) with i ? {1
, 1, B) and w, ? |= ? 1 ? ? 2 then for some i ? {1, 2} defined by f , v k+1 = (c(? i , w, ?), 1, B). By definition of satisfiability
, Using those two properties, we deduce that there exists m ? n + 2 · |Var ? |
, t = a(x): as w, ? |= t we know that a ?(x) = a, and v m = (c(t, w, ?), 1, B) so v m +1 = (c(Win, w, ?), ?(x) + 1, B)
, similarly, we have v m +1 = (c(Win, w, ?), ?(x) + 1, B)
, B) where ? = ? ?{{x ? ?(x)} | x ? Var ? }. Since ?(x) = ?(y), ? (y) = ?(y) > ?(x), we know that ?(x) < ?(y) so v m +1 = (c(Win-if-y, w, ? ), ?(x
, this case ?(y) ? ?(x) and we have v m +1 = (c(Win-if-x, w, ?), ?(y), B) and v m +2 = (c(Win, w, ?), ?(x), B)
, Every case ends in an accepting node, therefore ? is winning