M khong vuot qua so tat cacac hinh thai khac nhau Ct = UtqiVt
Big nX ngay sau luong tii chinh la dSi tU0ng cua lugng tii aỵ
2341 Do phurc tap khong gian Thi du, khi xem xet c^c so ttr nhien, menh de \/x[x + 1 > x] c6 nghia la phan t i i ke tigp x + l cua so tU nhien x Idn hon chinh so x. Dg tliay rang, menh de nay dung. Tuy vay, menh de 3y[2y = 5] ro rang la saị K h i khao sat tinh dung d^n cua nhflng menh de bao ham cac luong tut, ta can can nh3,c den mien gid tri cua cac bien. Trong hai menh dg viia neu, mien gia t r i bao gom cac so tU nhien, nhung neu nhu ta thay the bdi cac so thuc t h i menh de thiJ hai vdi luong tijf ton tai se trd nen dilng.
Mot menh dg c6 t h i chiJa nhieu lirong t i i khac nhau, t h i du Vx3y[y > x]. Doi vdi mien xac dinh bao gom cac so tU nhien, menh de nay bigu t h i rang moi so tu nhien diu c6 mOt so tu nhien khdc
Idn han nọ T h i i t U cua cac luong t i i trong menh de 1^ rat quan trong. Viec dao nguoc thii t u ay, chang han nhu 3yVx[t/ > x ] , cho ta menh dg vdi mot ngil nghIa hoan toan khac, cu the la ton tai mot
so tu nhien ndo do Idn han moi so tU nhien khdc. Ro rang, menh
de tru6c la diing va menh de sau la saị
Luong tut CO thg xuat hien d bat cut vi t r i n^o trong menh dẹ M6i luqng tut deu c6 mOt pham vi {scope) anh hu5ng cua no, do la
phan menh de ngay sau bien cua Itrong tut ay, khong kg nhflng cap cac dau ngoac. Thong thudng, that la thuan tien khi ma tat ca cac luong tut deu xuat hien 6 phan dau menh de va khi do pham v i cua tutng luong tut chinh la nhung gi ngay sau laong tut aỵ Nhiing menh de nhu vay dildc goi la dang chudn tien liCdng {prenex normal
form). Moi menh de deu c6 thg duoc bien d6i ve dang chuin tign luong mOt each dg dang. B6i vay, sau day ta chi xem xet cac menh de d dang chudn tien ludng.
Cong thiic Boole chiia cac luong tut duqc goi la cong thiHc Boole dinh liidng {quantified Boolean formula). K h i moi bien cua cong thutc Boole deu duoc dinh luong, ta noi rkng do la cong
thvCc Boole dinh IvCdng hodn todn {totally quantified Boolean
formula). D6i vdi cac cong thiic Boole, mien gia t r i ciia cac bign la ^
tap nhung gia t r i chan ly va thudng duoc bigu t h i b5i tap { 1 , 0 } . ^
3.3 Do phUc tap khong gian da thiic
239 Thi du,
0 = Vx3y[(x V y ) A (x V y )
la mot cong thiic Boole dinh luong hoan toan, duoc thg hien dudi dang chuin tien dinh luong. De thay rang cong thiic 0 la diing, nhung khi hoan doi vi t r i hai luong tut trong cong thiic ay, ta thu duoc mot cong thiic sai:
0' = 3yVx[(x V y ) A (x V y ) .
Noi dung cua bai toan T Q B F la xac dinh tinh dung dan ciia bat ky mot cong thiic Boole dinh luong hoan toan 6 dang chuin tien dinh, tiic cong thiic:
trong do B ( x i , . . . , x ^ ) la cong thiic Boole chiia cdc bien x i , . . . , x ^ va moi Qi la luong tut 3 ho&c luong tut V tren mien nhiing gia t r i chan ly { 1 , 0 } .
Bay gid ta hay xac dinh ngon ngU tuong utng vdi bai toan nay
va chiing minh tinh PS-day dii cua ngon ngfl ay trong Idp phiic tap PS. Ngon ngu: tuong ling vdi bai toan T Q B F duoc xac dinh nhu sau:
TQBF = {{(p) \ Ik cong thiic Boole dinh luong hoan toan dung}.
D i n h ly 3.3.4 TQBF Id PS-day dụ
Chitng minh Dg chiing to r^ng TQBF thuoc PS, ta xay dUng
may Turing tat dinh khong gian da thiic sau daỵ
T = "Trgn dau vao (0), trong đ (p la cong thiic Boole dinh luqng hoan toan:
1. Neu 0 khong chiia luong tut nao t h i no la bigu thiic gom chi nhiing gia t r i chan ly va khi đ neu (p diing t h i chap