【正文】 f(1)=2 f(2)=1 I(P)={1} I(Q): 包含 2,1 不包含 1,1 1,2,2,2未設(shè)定 Logical Foundations of Artificial Intelligence Chapter 2: 45 169。 School of CIT, Beijing JiaoTong University Term joint assignment ? Term joint assignment ?IU : a mapping from terms to objects. (a) If ? is an object constant, then ?IU(?) = I(?). (b) If ? is a variable, then ?IU(?) = U(?). (c) If ? is a term of the form ?(?1,…, ?n) and I(?) = g and ?IU(?i)=xi, then ?IU(?)=g(x1, ... , xn). ? Example: – Hat(C)=?: because I(C)=c, Hat(c)=b (c,b) – Hat(z)=?: if U(z)=b a b c d e Logical Foundations of Artificial Intelligence Chapter 2: 47 169。?)[U] if and only if |?I ?[U]. (b) |=I (?1? ... ??n)[U] if and only if |=I (?i)[U] for all i =1, ..., n. (c) |=I (?1? ... ??n)[U] if and only if |=I (?i)[U] for some i, 1 ? i ? n. (d) |=I (???)[U] if and only if |?I ?[U] or |=I ? [U]. (e) |=I (???)[U] if and only if |=I ?[U] or |?I ? [U]. (f) |=I (???)[U] if and only if |=I (???)[U] and |=I (???)[U]. Logical Foundations of Artificial Intelligence Chapter 2: 50 169。 School of CIT, Beijing JiaoTong University Elementarily equivalent ? Two interpretations I and J are elementarily equivalent (I ? J) if and only if |=I ? ? |=J ? and |=J ? ? |=I ? for any sentence ?. ? Example: |I|= set of real numbers |J|= set of rational numbers (有理數(shù) ) I(R): greater than relation on reals J(R): greater than relation on rationals I and J are elementarily equivalent despite their universes are different. if ?:R(5,3) then |=I ? also |=J ? Logical Foundations of Artificial Intelligence Chapter 2: 54 169。 School of CIT, Beijing JiaoTong University Select a vocabulary Logical Foundations of Artificial Intelligence Chapter 2: 57 169。 School of CIT, Beijing JiaoTong University Expansion of conception ? The above sentences only describe the digital structure and behavior of the circuit. ? To express a gate is malfunctioning – Adding additional relations ? To express a connection is malfunctioning – Describe connections as objects (12 new objects) – Extending the binary connectivity relation into a ternary relation Port1,Port2,Conn. Logical Foundations of Artificial Intelligence Chapter 2: 61 169。 School of CIT, Beijing JiaoTong University 167。 School of CIT, Beijing JiaoTong University 蘊含式的真值表 P Q P?Q T T T T F F F T T F F T ?公式 P?Q ?只有當(dāng) P為真且 Q為假時不滿足 ?即： P為假時Q為任何公式都能從 P推出 Q Logical Foundations of Artificial Intelligence Chapter 2: 66 169。 School of CIT, Beijing JiaoTong University 羅素與教皇 ? （ 1）假定 2＋ 2＝ 5； ? （ 2）由等式兩側(cè)減去 2，得出 2＝ 3； ? （ 3）易位后得出 3＝ 2； ? （ 4）由兩側(cè)減去 1，得出 2＝ 1. ? 請看：教皇與我是二人。 School of CIT, Beijing JiaoTong University 作業(yè) ? P42P43： 10 Logical Foundations of Artificial Intelligence Chapter 2: 69 169。因此我是教皇。”哲學(xué)家問：“你能證明這一點么？”羅素答：“當(dāng)然能。 School of CIT, Beijing JiaoTong University Binary Table ? Excellent for expressing information about binary functions。 Natural Language Example ? The universe of discourse is the set of all plants. Logical Foundations of Artificial Intelligence Chapter 2: 62 169。 School of CIT, Beijing JiaoTong University State representation ? Associate a port with a value ? Example: – Inputs are 1,0,1 – Outputs are 0,1 Logical Foundations of Artificial Intelligence Chapter 2: 59 169。 School of CIT, Beijing JiaoTong University 167。 School of CIT, Beijing JiaoTong University Model ? If for all variable assignment, an interpretation I satisfies a sentence ?, then I is called a model of ?. Denoted as |=I ?. – Example: On(x,y)?Above(x,y) ? A sentence is satisfiable if and only if there is some interpretation and variable assignment that satisfy it. Otherwise, it is unsatisfiable. ? A sentence is valid (正確的 ) if and only if it is satisfied by every interpretation and variable assignment. a b c d e Logical Foundations of Artificial Intelligence Chapter 2: 52 169。 School of CIT, Beijing JiaoTong University Satisfaction of an atomic sentence ? I and U satisfy an atomic sentence : ? Example： |=I On(A,B)[U] ?IU(A)=a ?IU(B)=b a,b?I(On) |=J On(A,B)[U] ?JU (A)=a ?JU(B)=b a,b?J(On) J(On)={b,a, c,b, e,d} a b c d e Logical Foundations of Artificial Intelligence Chapter 2: 49 169。 – Assignment of no