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Proving Invalidity in Predicate Logic Kareem Khalifa Department of Philosophy Middlebury College Overview • Refresher on invalidity • How this bears on predicate logic • How to prove invalidity – Play God! – Rules for playing God • Sample exercises Refresher on invalidity • An argument is invalid if there is some way that all of its premises are true when its conclusion is false. Example of an invalid argument in predicate logic • Some philosophers have receding hairlines. • Therefore all philosophers have receding hairlines. + = Invalidity in predicate logic • An argument in predicate logic is invalid if it there is some way all of its premises can be true when its conclusion is false. • However, we can’t quantify over everything! • So we get to… – MAKE OUR OWN WORLDS! – Though they’re kind of lame worlds… Back to our example • Imagine a world with two people, a and b, both of whom are philosophers. However a has a receding hairline and b does not. • Then in this world, it would be true that some philosophers have receding hairlines (a does!) but false that all philosophers have receding hairlines (b doesn’t!). • So the argument is invalid. Playing God • To invalidate an argument, you get to create very simple worlds, often consisting of no more than two objects (let’s say a and b). • You then use the following rules: • Replace every statement of the form (x)(Фx), with a disjunction Фa v Фb • Replace every statement of the form (x)(Фx), with a conjunction of the form Фa & Фb • Now invalidate the argument as you would in propositional logic with the disjunctions and conjunctions. Motivation for the “Existential God Rule” Let this be the whole wide world. At least one thing is an F So EITHER a is an F... F a F b a b …OR b is an F. Motivation for the “Universal God Rule” Let this be the whole wide world. Everything is an F So both a AND b are F’s. a F b a b Formal treatment of our example • (x)(Px & Rb)├ (x)(PxRx) Rb) (Pa & Ra) v (Pb &Rx)(Pa Ra) & (Pb • Create your world: – Let there be exactly two objects {a, b} • Replace the (x) statements with v statements. • Replace your (x) statements with & statements • Now treat this just like a formal proof of invalidity in propositional logic… Continued • (Pa & Ra) v (Pb & Rb)├ (Pa Ra) & (Pb Rb) 1. Let Pa, Pb, and Rb be true; Ra be false. 2. Since Pa and Ra are false, Pa Ra is false. 3. Since Pa Ra is false, the conclusion, (Pa Ra) & (Pb Rb) is also false. 4. Since Pb and Rb are true, Pb & Rb is true. 5. Since Pb & Rb is true, the premise, (Pa & Ra) v (Pb & Rb), is also true. 6. By 3 and 5, all of the premises are true and the conclusion is false. 7. So the argument is invalid. Pa Ra Pb Rb (Pa & Ra) v (Pb & Rb) (Pa Ra) & (Pb Rb) T F T T T F How many objects should I put in my world? • Quick answer: As few as possible to prove it invalid. • So start with a one-object world. If you can invalidate the argument, stop. • If you can’t, create a two-object world. If you can invalidate the argument, stop. • If you can’t, create a three-object world… Potential problem • Suppose you don’t know whether the argument is valid or invalid. • Then this procedure could go on ad infinitum for a valid argument! Rules of thumb to avoid this problem • Try to sketch a proof of validity. If you “see the proof,” then avoid playing God. • In general, start by assuming the bare minimum required to make either one premise true or a conclusion false. Then fill in the rest as you need it. • If all the premises and the conclusion are universally quantified, then the argument can be invalidated using only one object (467, A2). • For each existentially quantified premise, posit a unique object that makes exactly one premise true. (467, A5) • When there are both universally and existentially quantified statements in the argument, you generally need 2+ objects. (467, A10) (x)(Dx → ~Ex), (x)(Ex → Fx) ├ (x)(Fx → ~Dx) • Da → ~Ea, Ea → Fa ├ Fa → ~Da 1. Let there be a model with domain {a}, such that Da and Fa are true, and Ea is false. 2. Since Da is true, ~Da is false. 3. Since Fa is true and ~Da is false, the conclusion Fa → ~Da is false. 4. Since Ea is false, the premise Ea → Fa, is true. 5. Since Ea is false, ~Ea is true. 6. Since ~Ea is true, the premise Da → Ea is true. 7. Thus by 3, 4, and 6, it is possible for all of the premises to be true and the conclusion to be false. 8. So the argument is invalid. 5. (x)(Mx & Nx), (x)(Mx & Ox) ├ (x)(Ox → Nx) • (Ma & Na) v (Mb & Nb), (Ma & Oa) v (Mb & Ob) ├ (Oa → Na) & (Ob → Nb) 1. Let there be a model {a, b} such that Ma, Na, Mb, and Ob are true, and Nb is false. 2. Since Ma and Na are true, Ma & Na is true. 3. Since Ma & Na is true, the premise, (Ma & Na) v (Mb & Nb), is true. 4. Since Mb and Ob are true, Mb & Ob is true. 5. Since Mb & Ob is true, the premise, (Ma & Oa) v (Mb & Ob), is true. 6. Since Ob is true and Nb is false, Ob → Nb is false. 7. Since Ob → Nb is false, the conclusion (Oa → Na) & (Ob → Nb) is false. 8. By 3, 5, and 7, it is possible for all of the premises to be true and the conclusion to be false. 9. So the argument is invalid. 10. (x)(Bx & ~Cx), (x)(Dx → ~Cx) ├ (x)(Dx → Bx) • (Ba & ~Ca) v (Bb & ~Cb), (Da → ~Ca) & (Db → ~Cb) ├ (Da → Ba) & (Db → Bb) 1. Let there be a model with domain {a, b} such that Ba and Db are true, and Ca, Bb, and Cb are false. 2. Since Ca is false, ~Ca is true. 3. Since Ba and ~Ca are true, Ba & ~Ca is true. 4. Since Ba & ~Ca is true, the premise (Ba & ~Ca) v (Bb & ~Cb) is true. 5. Since Db is true and Bb is false, Db → Bb is false. 6. Since Db → Bb is false, the conclusion (Da → Ba) & (Db → Bb)is false. 7. By 2, ~Ca is true, so Da → ~Ca is true. 8. Since Cb is false, ~Cb is true. 9. Since ~Cb is true, Db → ~Cb is true. 10. By 7 and 9, the premise (Da → ~Ca) & (Db → ~Cb) is true. 11. Thus, by 4, 6, and 10, it is possible for all of the premises to be true and the conclusion to be false. 12. So the argument is invalid.