Proofs in Predicate Logic A rule of inference applies only if the main operator of the line is the right main operator. So if the line is a simple statement, with a quantifier as the main operator, no rule of inference can be applied (except AD). New Rules: For removing quantifiers or For introducing quantifiers

Free Variables Bound Variables Px > Cx Here the “x” is called “free,” which means it is not bound or controlled by anything that determines its meaning. (x) (Px > Cx) Here the “x” in “Px” and in “Cx” is bound (governed/ regulated) by the quantifier out front. The quantifier tells you how to read the occurrences of “x” inside the parentheses. In a sense, you can’t read this statement form; it doesn’t really say anything. The value of a statement form like this is: that the quantifier is gone, so the main operator is now the “>. ” So now rules like MP and MT might be applied.

Free variables can only result from invoking the rule called Universal Instantiation. (UI). So only A and E statements can be rewritten with free variables. So whenever we see a free variable, we know it came from a universal statement before – this is important because it means that adding a universal quantifier to such a statement form, and making it into a universal statement, will be justified.

Universal statements can be instantiated in two ways: To free variables Or to constants. (x) (Rx > ~Bx) No rabbits are bears Let “a” stand for Adam, “b” for Betty, etc. Rx > ~Bx This is a free variable Ra > ~Ba Rb > ~Bb Rc > ~Bc If Colbert’s a rabbit, he’s not a bear. If Adam’s a rabbit, he’s not a bear If Betty’s a rabbit, she’s not a bear

How do we know which to instantiate to: the free variable or the constant? It depends on where you are trying to go: If your conclusion is a universal statement, chances are you want the free variable instantiation, so you can return to the universal level of predication. If your conclusion is about Adam, or Eve, or Colbert, you’ll want to instantiate quantified statements to their names, so you can spell out inferences about them.

1 (x) (Mx > Ox) 2 Ms / Os 3 Ms > Os UI 1 4 Os MP 3,2 Socrates must be mortal, because all men are, and he’s a man. 1 (x) (Mx > Ox) 2 Ms / Os 3 Ms > Os UI 1 If Socrates is a man, he’s mortal 4 Os MP 3,2 Socrates is mortal Given the conclusion (not a universal statement) there is nothing to be gained by instantiation to a free variable; obviously, the instantiated sentence has to be about Aristotle’s teacher’s teacher.

Universal Instantiation UI (x) Fx Fx or Fa

Since you can’t encounter a statement that looks like this: Cx > Bx unless… Unless UI has been done to a universal that was already given or proven to be true… It will be legitimate to GENERALIZE from such a statement. To Generalize is to predicate at the level of the Universal UG: Fx (x) Fx But this would not be valid: Fa . Ga / (x) (Fx . Gx) HG

Instantiation and Generalization for Existential Statements EI and EG No comedians are viable candidates for president, but some comedians are running, so some people who are running for president are not viable candidates. 1 (x) (Cx > ~ Vx) 2 Эx (Cx . Rx) / (Эx) ~Vx 3 Cc . Rc EI 2 Colbert’s a comedian, and he’s running 4 Cc > ~Vc UI 1 If Colbert’s a comedian, he’s not viable SM 3 Colbert’s a comedian 5 Cc Colbert’s not viable 6 ~Vc MP 4,5 At least one candidate is not viable (some candidates are not viable) 7 (Эx) ~Vx EG 6

EI has an important restriction: You cannot use a constant in an EI line if that constant has already been in play in the proof, even in the premises. So the rule EI is: (Эx) Fx ---------- Fa (where “a” is new) This means that if you have to instantiate twice (see previous slide) you must do the EI before you do the UI. There are no restrictions on UI.

There are no restrictions on EG: you can generalize to a “Some” statement from either a constant, or from a free variable. Fa (Эx) Fx or Fx

UI --Universal Instantiation (x) (Ax > Bx) --------------------- Ax > Bx or Aa > Ba Instantiate to a free variable or to a name. UG --Universal Generalization Ax > Bx not allowed: Aa > Ba -------------- ------------ (x) (Ax > Bx) (x) (Ax > Bx) Generalize to a universal only from a free variable. EI --Existential Instantiation Not allowed: (Зx) (Ax . Bx) (Зx) (Ax . Bx) ------------------- ------------------ Ax . Bx Aa . Ba Only instantiate to a name. Restriction: The name must be a new name EG --Existential Generalization Aa . Ba or Ax . Bx (Зx) (Ax . Bx) Generalize to a particular from a name or a free variable.

Apples and pears grow on trees, so pears grow on trees. 1 (x) [(Ax v Px) > Gx] / (x) (Px > Gx) 2 (Ax v Px) > Gx UI 1 3 Px ACP 4 Ax v Px AD, CM 3 5 Gx MP 2, 4 6 Px > Gx CP 3-5 7 (x) (Px > Gx) UG 6

Only philosophers are logicians, and anyone who's a philosopher is a thinker. Frege and Russell are logicians, so they're also thinkers. (x) (Lx > Px) . (x) (Px > Tx) 2 Lf . Lr / Tf . Tr (x) (Lx > Px) 1 SM 4 (x) (Px > Tx) 1 CM, SM 10 Lr > Pr UI 3 11 Pr > Tr UI 4 5 Lf > Pf UI 3 12 Lr > Tr HS 10, 11 6 Pf > Tf UI 4 13 Lr CM, SM 2 7 Lf SM 2 14 Tr MP 12, 13 8 Pf MP 5, 7 15 Tf . Tr CN 9, 14 9 Tf MP 6, 8

1. (∃x) Kx > (x) (Lx > Mx) 2. Kc . Lc /Mc 3. Kc 2 SM 4. (∃x) Kx 3 EG 5. (x) (Lx > Mx) 1,4 MP 6. Lc > Mc 5, UI 7. Lc 2 CM, SM 8. Mc 6,7 MP

S= is starving H = is healthy A = is alright P = is a panda   1. (x) Ax > (x) (Px > Hx) 2. (x) [(Px . Sx) > ~Hx] 3. (Pg . Pb) . (Sg . Sb) / (Зx) ~Ax If everything is alright, the pandas are healthy. No starving pandas are healthy. George and Bob are pandas that are starving, so something’s not right.

1. (x) Ax > (x) (Px > Hx) 2. (x) [(Px . Sx) > ~Hx] 3. (Pg . Pb) . (Sg . Sb) / (Зx) ~Ax 4. (Pg . Sg) > ~Hg UI 2 5. Pg . Sg AS, CM, AS, SM 3 6. ~Hg MP 4,5 7. Pg SM 5 8. Pg . ~ Hg CN 6,7 9. ~(~Pg v Hg) DM 8 10. ~(Pg > Hg) IMP 9 11. (Зx) ~(Px > Hx) EG 10 12. ~(x) (Px > Hx) CQ 11 13. ~(x) Ax MT 12, 1 14. (Зx) ~Ax CQ 13