Chapter 3: Semantics PHIL 121: Methods of Reasoning March 13, 2013 Instructor:Karin Howe Binghamton University

Recall our definition of a statement: A statement is a sentence that is either true or false. But what does that mean? When would we say each of the following is true … –conjunctions –disjunctions –conditionals –negations We take the logical operators (&, , , ,  ) to be truth-functional

Conjunctions pqp & q TTT TFF FTF FFF A conjunction is true only when both conjuncts are true.

Disjunctions pq p  q TTT TFT FTT FFF A disjunction is false only when both disjuncts are false.

Negations p pp TF FT The truth value of a negated statement is always the opposite of the truth value of the unnegated statement.

Conditionals pq p  q TTT TFF FTT FFT Conditionals are false only when it is the case that the antecedent is true and the consequent false.

Biconditionals "if and only if" statements Example: –You may GO to the movies if and only if you CLEAN up your room. Expanding this: –You may GO to the movies if you CLEAN up your room and you may go to the movies only if you clean up your room. –(C  G) & (G  C) The "lazy way:" –You may GO to the movies iff you CLEAN up your room. –G  C

Compare Truth Tables pq (q  p) & (p  q) pq p  qp  q TTTTTT TFFTFF FTFFTF FFTFFT Biconditionals are true only if the truth values for both sides of the biconditional match.

Determining the truth-values of statements Given A, B, C are all true; X, Y, Z are all false; G, H, I have unknown truth values –(X  A)   B –(X & A)  (B  Y) –(A  B) & (B  X) –(G  H) &  A –(A & G)  (B & H) –H  (G  H) –  (A & G)

Full Truth Table Method Put all of the atomic statements, premises and the conclusion in a table together. Under each of the atomic statements, fill in all of the possible truth values for these atomic statements. –General method: working from the right, fill in the values as follows: T F T F (etc.) –In the next row, double the number of T's and the number of F's (e.g. T T F F etc,) –Continue doubling the number of T's and the number of F's until all of the columns are filled in This is called the base column General formula for the number of rows in a full truth table: 2 n, where n = number of atomic statements

Validity and Truth Tables An argument is valid iff any truth-value assignment that makes all the premises true also makes the conclusion true. Alternatively, an argument is valid iff there is NO truth- value assignment that makes all the premises true while the conclusion is false. KF K  F K  F TTTTT TFFTF FTTFT FFTFF

Invalidity and Truth Tables An argument is invalid iff there is some truth- value assignment that makes the all premises true, but the conclusion false. TP T  P P  T T TTTTF TFTFF FTTTT FFFFT

Prove the following arguments valid or invalid P  Q,  P   Q   P  Q P   Q,  (P  Q)  P,  (P & Q)  (Q  P)   (P & Q)   P P  (Q & R),  P   (Q & R), Q &  P   R (P  Q)   R, R   Q  Q  P (P  Q)  ( R  S), P   (R & S), Q   (P & R)  (S & P)   (P v Q)

There has to be a quicker way … (P  Q)  (R  S), P   (R & T), S  T  P  T P, Q, R, S, T -- 5 variables! That's 2 5 = 32 rows. Yuck!! All we really care about is the row that proves invalidity, right? What if we just try to find that row??

The Short Method for Proving Invalidity 1.First, try to find truth value assignments that will make the conclusion false.  Important: Look for any other occurrences of those letter(s) in the other formulas in the argument and give them the same truth value assignment. (we need to make sure that we are consistent in our truth value assignments or this won't work!) 2.Next, try to find truth value assignments that will make all the premises true. (making sure to keep all truth value assignments consistent throughout) If you succeed, that shows that the argument is invalid!

What if you fail? Does that mean the argument is valid? Fact A: We prove validity by showing that it is impossible for all the premises to be true while the conclusion is false. However, showing ONE assignment of truth values where it is not the case that the premises are all true while the conclusion is false does not in itself show that it is impossible to find such an assignment. Fact B: When we are doing the short method, basically what we are doing is trying to show that an argument is invalid, and we either fail or succeed. If we succeed, then the argument is invalid. However, we can fail for one of two reasons: 1.We can fail because it really is valid (we can fail because there is no possible way to succeed) 2.We can fail because we just got unlucky – we picked the wrong starting set of truth-value assignments during our guess-and- check portion of the method, and if we choose a different option then we will see that it really is invalid, and we just missed the boat on the first try.

This makes using the short method to prove validity tricky, but not impossible Fact A means we need to do more than just show the truth value assignments that we got using the short method (because this does not prove that another assignment would not produce different results – see Fact B for why this is). Thus, we need some kind of argument that makes it clear to the reader why this assignment of truth values shows that it is impossible for it to work out otherwise – why it cannot be the case that there is a line on the truth table where all the premises are true and the conclusion false. Thus, the short method is tricky for proving validity because it requires a bit more work than the short method when proving invalidity. Fact B means that we could be wrong in thinking the argument is valid – we could have just missed the counter-example. That's another reason for the argument that shows why it couldn't be otherwise is important and useful – it can help you double-check your work.

P  Q, Q  P P  ~Q, Q   P   (P   Q) (P  Q)   R, R   Q  Q  P (P  Q)  (R  S), P   (R & S), Q   (P & R)  (S & P)   (P  Q) P  (Q   P), P  Q   P &  Q (P  Q)  (R   S),  (T  S)  (W  X), S  X, X  (  W   Z),  T  (R & Z)  P  (X & R)

Truth Trees (p & q) p q (p  q) p q (p  q)  p q (p  q) p  p q  q  (p & q)  p  q (p  q)pq(p  q)pq (p  q)pq(p  q)pq  (p  q) p  p  q q  p p

Ch 3 Lab, Problem 5 (E  P), [(E & P)  T], T  E [(B  D) & (  B  D)]  D (F  D), (S  D), (F  S)  D P   Q, Q v P   Q v  P P  Q, P v  Q  P  Q (P & Q) &  R, (P v Q)   R,  (P -> Q)   (Q  P)   (P  R)