80-210: Logic & Proofs July 30, 2009 Karin Howe Review for Midterm 3 80-210: Logic & Proofs July 30, 2009 Karin Howe

Definitions Validity/invalidity Tautology Contradictory statement Contingent statement

Skills you should be comfortable with Translating statements (universal and existential statements, as well as conjunctions, disjunctions, conditionals and negations involving quantified formulas Determining truth or falsity of a statement on a given interpretation Well-formed formulas (WFFs) of predicate logic, and parse trees Truth trees For determining validity/invalidity For determining whether a statement is a tautology/contradictory statement/contingent statement Derivations (using all of the propositional rules) Fill-in-the-blanks Complete derivations

Practice: Translating Statements Dictionary: R(x) = x is red, p = pen Everything is either red or not red Either everything is red or everything is not red Everything is red, although this pen is not red Dictionary: f(x) = x is female Someone is female, but not everyone is. Someone is female, although no one is. If everyone is female, then someone is. Dictionary: M(x) = x is a Muslim, P(x) = x is a plumber, j = Jones If there are Muslim plumbers, then there are Muslims and plumbers. If there are Muslims and plumbers, then there are Muslim plumbers. Jones is a Muslim plumber, although there are no Muslim plumbers.

Practice: Translating Statements Dictionary: L(x,y) = x loves y, f = Fred, a = Alice Fred loves Alice. Fred loves someone. Everyone loves Alice. There is a person who loves everyone. Each person loves someone or other. Everybody loves everybody. Everybody loves somebody. Somebody loves everybody. Dictionary: L(x,y) = x is located in place y Everthing is somewhere. Dictionary: S(x,y) = x scared y, g = Georgie Georgie never scared anybody. Dictionary: P(x,y) = x can please y No one can please everybody.

Practice: Determining Truth/Falsity Under an Interpretation Dictionary: R(x) = x is red Interpretation: Domain: {pen, cat, bull, rabbit, sandwich, kangaroo} I(R) = <bull, sandwich, kangaroo, pen> (x)(R(x)  R(x)) (x)(R(x)  (x)R(x) (x)R(x) & R(p) Dictionary: f(x) = x is female Interpretation: Domain: {our class} I(F) = <Karin, Amelie> (x)F(x) & (x)F(x) (x)F(x) & (x)F(x) (x)F(x)  (x)F(x)

Practice: Determining Truth/Falsity Under an Interpretation Dictionary: L(x,y) = x loves y Interpretation: Domain: {Fred, Alice, George, Samantha, Bruce, Kirsten} L(x,y) = <(f,a), (g,a), (k,a), (a, s), (k, f), (g, f), (f, b), (a, b), (f, k), (s, a)> L(f,a) (x)L(f,x) (x)L(x,a) (x)(y)L(x,y) (x)(y)L(x,y)  (x)(z)L(x,z) (x)(y)L(x,y) (x)(y)L(x,y)

Truth Tree Rules: Propositional (p & q) p q (p  q) p q (p  q) p q p  q p p q p (p & q) p q (p  q) p q (p  q) p q p (p  q) p p q q

Truth Tree Rules: Predicate (x)(P(x) & Q(x)) (P(a) & Q(a))* * Constant introduced must be new to the branch (x)(P(x)  Q(x)) (P(a)  Q(a)) (x)(P(x) & Q(x)) (x)(P(x) & Q(x)) (x)(P(x)  Q(x)) (x)(P(x)  Q(x))

Determining validity/invalidity with truth trees Steps: Place premises at the top of the tree Negate the conclusion and place it below the premises Take apart the statements according to the truth tree rules A branch may be closed …. when? Whenever it contains both p and p on the same branch The argument is valid iff … All the branches in the tree are closed If the argument has at least one open branch: Then it is probably invalid Need to construct a counter-example to show this for sure Constructing a counter-argument from a truth tree: Pick an open branch and "read off" the interpretation for each formula

Practice: Validity/Invalidity with Truth Trees ¬(x)(A(x)  B(x)), (x)(A(x)  C(x))  ¬(x)(¬B(x) & C(x)) ¬(x)[A(x) & (B(x)  C(x)], (x)(A(x) & B(x))  (x)(A(x) & C(x)) (x)(K(x)  F(x), F(j)  K(j) ¬(x)[(A(x) & B(x))  (C(x) & D(x))], ¬(x)(A(x) & C(x))  ¬(x)(B(x) & D(x)) (x)(y)[(N(x) & N(y))  R(x,y)], (x)(y) (z){[((N(x) & N(y)) & N(z)]  [(R(x,y) & R(y,z)  R(x,z)]}  (x)(N(x)  ¬R(x,x)) (x)[A(x)  (C(x)  D(x))], ¬(x)(A(x) & C(x)), (x)(A(x) & B(x))  (x)(A(x)  ¬C(x)) (x)[A(x)  (B(x)  D(x))], (x)(B(x) & D(x))  (x)[A(x)  (C(x)  D(x))] (¬C(b) & ¬C(f)), (x)(¬C(x)  ¬R(x))  (¬R(b) & ¬R(f)) (x)(C(x) & F(x))  (x)(C(x) & ¬F(x)) (x)(E(x)  G(x), E(c)  G(c) (x)(C(x)  G(x)  (x)(G(x)  C(x))

Determining types of statements using truth trees Recall that our truth tree rules operate off the assumption that a statement is true Thus, if we take apart a statement according to the rules, and all of its branch(es) close, what does that tell us about the statement? Contradictory statement! What if we wanted to check if a statement was a tautology? Negate the statement Take it apart according to the rules See if all the branch(es) close What about contingent statements? Have to check both ways! (both the negated and the non-negated form)

Practice: Determining Types of Statements with Truth Trees (x)(R(x)  ¬R(x)) (x)R(x)  (x)¬R(x) (x)R(x) & ¬R(p) (x)F(x) & ¬(x)F(x) (x)F(x) & (x)¬F(x) (x)F(x)  (x)F(x) (x)[A(x)  (¬A(x) & B(x))] (x)[A(x) & (A(x)  B(x))] (x)[(A(x) & B(x))  (B(x) & C(x))] (x)[(A(x) & B(x))  C(x)]  (x)[(A(x) & ¬B(x))  ¬C(x)] (x)(M(x) & P(x))  ((x)(M(x) & (x)(P(x)) ((x)(M(x) & (x)(P(x))  (x)(M(x) & P(x)) (M(j) & M(p)) & ¬(x)P(x)

Practice: Derivations (V(h,l,m) & V(j,l,m)) (V(h,l,m)  V(h,p,m)) (V(j,l,m)  V(j,p,m)) (V(h,p,m)  V(j,p,m))  (V(h,p,m) & V(j,p,m))  (A(p)  A(q))  (A(p) & A(q))  [(H(d) & B(l,r)) & S(s)]  [(H(d)  B(l,r))  (H(d)  S(s))]  (S(o)  S(o))  (R(p,m)  L(p))  (R(p,m)  L(p))

Practice: Derivations  (R(f)  R(h))  [(R(f) & R(f))  (R(f) & R(f))] (x)[(T(x,b)  S(x))  V(b,p,d)], (x)(T(x,b)  S(x))  S(j)  V(b,p,d) [A  (B  C)] [(B & A) & C] (x)(y)V(x,p,y), ((x)(y)V(x,p,y)  (x)(S(x)), (x)(S(x) V(p,b,m) ((x)P(x)  (x)H(x)), (x)H(x), ((x)P(x)  (x)S(x)) (x)S(x)