实用标准文案
(2)T(Q)∨T(S), (3)T(S)∨T(L) (4)┑T(Z)∨┑T(S) (5)┑T(Q)∨┑T(L) 添加求解字句
(6): ┑T(X) ∨ANSWER(X) (7): (1)+(4)= T(Q)∨┑T(S) (8):(1)+(5)= T(Z)∨┑T(L) (9):(2)+(4)= ┑T(Z)∨T(Q) (10):(2)+(5)= T(S)∨┑T(L) (11):(3)+(4)= T(L)∨┑T(S) (12):(3)+(5)= ┑T(Q)∨T(S) (13):(2)+(7)= T(Q)
(14):(6)+(13)= ANSWER(Q)/(Q/x) (15):(3)+(10)=T(S)
(16):(6)+(13)= ANSWER(S)/(S/x) 3.由 r4 得到：
CF( E1 ) = 0.7*max { 0, CF [ E4 and (E5 or E6 ) } = 0.7 *max { 0, min { CF(E4) ,CF (E5 or E6 ) } } = 0.7*max { 0, min { CF(E4) , max {CF ( E5 ) , CF( E6 ) } } } = 0.7*max { 0, min { 0.5 , max { 0.6 , 0.7 } } } = 0.7 *0.5
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实用标准文案
= 0.35 由 r5 得到：
CF( E3 ) = 0.9 *max { 0, CF ( E7 and E8 ) } = 0.9*0.6= 0.54 由 r1 得到：
CF1( H ) = 0.8*max { 0, CF ( E1 ) } = 0.8*0.35
= 0.28 由 r2 得到：
CF2( H ) = 0.6*max { 0, CF ( E2 ) } = 0.6*0.8 = 0.48 由 r3 得到：
CF3( H ) = - 0.5*max { 0, CF ( E3 ) } = - 0.5*0.54 = - 0.27 结论不确定性的合成算法
CF1，2( H ) = CF1 ( H ) + CF2 ( H )–CF1 ( H )*CF2 ( H ) = 0.28 + 0.48–0.28*0.48 = 0.63
CF1,2,3 ( H ) = (CF1,2 ( H ) + CF3( H ))/(1 – min { | CF1,2 ( H ) | , | CF3( H ) |)
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实用标准文案
= 0.49
即：CF( H ) = 0.49
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