There are ___ satisfying truth assignments to (¬w∧x∧¬y)∨(¬w∧¬x∧y∧¬z)(\neg w \wedge x \wedge \neg y) \vee (\neg w \wedge \neg x \wedge y \wedge \neg z)(¬w∧x∧¬y)∨(¬w∧¬x∧y∧¬z).
參考答案與解析
共有 3 組滿足指派。窮舉16組(w,x,y,z)(w,x,y,z)(w,x,y,z)驗證:(¬w∧x∧¬y)∨(¬w∧¬x∧y∧¬z)(\neg w\wedge x\wedge\neg y)\vee(\neg w\wedge\neg x\wedge y\wedge\neg z)(¬w∧x∧¬y)∨(¬w∧¬x∧y∧¬z)在w=0w=0w=0時成立的組合為(x,y,z)=(1,0,0),(1,0,1),(0,1,0)(x,y,z)=(1,0,0),(1,0,1),(0,1,0)(x,y,z)=(1,0,0),(1,0,1),(0,1,0)共3組。