2. Truth Tables and Validity Kareem Khalifa Department of Philosophy Middlebury College pqrSp v qr & s~(p v q)~p~q~p & ~q~(p v q) -> (r & s) (~p & ~q) -> (r & s) TTTTTFFFTT TTTFTFFFTT TTFTTFFFTT TTFFTFFFTT TFTTTFFFTT TFTFTFFFTT TFFTTFFFTT TFFFTFFFTT FTTTTFTFFTT FTTFTFTFFTT FTFTTFTFFTT FTFTTFTFFTT FFTTFTTTTTTT FFTFFFTTF FFFTFFTTF FFFFFFTTF P1C

3. Overview Why this matters How truth tables work How to work truth tables –The basic method –The shortcut Sample Exercises

4. Reminder: The Official Definition of Deductive Validity  A deductive argument is valid when, if all of its premises are true, its conclusion must be true.  This is the SINGLE MOST IMPORTANT CONCEPT IN THE CLASS!!!!!!!!!  Failure to define validity properly is an automatic 5 point penalty on anything you do! You’ll also be very confused if you don’t get this concept.

5. Why this matters Thus far, testing for validity has involved devising counterexamples. –This method has limitations We may not be creative enough to devise a counterexample for an invalid argument. So we will not always be reliable in testing for validity. Truth tables provide an algorithm for testing for validity, in which no creativity is required, however… Truth tables can also enhance our creativity in devising counterexamples.

6. How truth tables work Recall: An argument is valid if, when all of its premises are true, its conclusion must also be true. Equivalently, there is no way in which all of the premises are true and the conclusion is false. Truth tables specify all the different ways in which premises and conclusions can be true and false. –“Where there’s a row, there’s a way.” Thus, an argument is valid if there is no row on a truth table in which all of its premises are true and its conclusion is false.

7. Where there’s a row, there’s a way Recall that the truth/falsity of compound statements is a function of the truth/falsity of simple statements Truth tables specify all of the different combinations of truth and falsity of simple statements So truth tables tell us all the different ways compound statements can be true or false.

8. Where there’s a row, there’s a way: Illustration with ‘&’ pqp & q TT T F F T F F T F F F

9. More on how truth tables work Note that validity says nothing about cases in which: –Some of the premises are false –All of the premises are false Lesson: Focus on the rows in which all of the premises are true.

10. How to work truth tables in 5 easy steps 1.Set up guide columns. 2.Set up an additional column for every premise and the conclusion. Clearly mark these columns with “P’s” for each premise, and “C” for the conclusion. 3.Add more columns to help interpret complex propositions. 4.Using your knowledge of the truth-tables for conjunction ‘&’, disjunction ‘v’, negation ‘~’, material implication ‘→’, and material equivalence ‘↔’, fill out the table. 5.Look at all of the columns with P’s and C’s above them. If there are one or more rows in which all of the P-columns have T’s and the C-column has an F, then you have an invalid argument; otherwise the argument is valid.

11. Step 1: Setting up guide columns First, count the number of simple propositions in the argument or proposition under consideration. Call this number N. Ex A. Suppose you are asked to discern the validity of the following argument: ~q |- [(p → q) → p] → p. There are two simple propositions—p and q—thus N=2.

12. More on guide columns Next, set up a table with 2 N + 1 rows. Place the simple propositions in the top row in the leftmost columns. These columns are called guide columns. Continuing with ~q |- [(p → q) → p] → p, there would 2 N +1 rows= 2 2 +1 = 4+1 = 5 rows, this leads to the following: p q

13. Filling out guide columns For the leftmost guide column, the top 2 N-1 rows will be True, the bottom 2 N-1 rows will be false. In other words, the top half of the leftmost column will be true, the bottom half will be false. In our example, the top 2 N-1 rows will be 2 2-1 = 2 1 = the top 2 rows. Thus we fill in the top half of the leftmost column with T’s, and the bottom with F’s. pq T T F F

14. More on filling out guide columns For the next column, the top 2 N-2 rows will be T, the next 2 N-2 rows will be F, the next 2 N-2 rows will be T, etc. In this case, the top quarter will be T, the next quarter will be F, the third quarter will be T, the last quarter will be F. Continue raising 2 to a lower power until the rows alternate between T and F, i.e., when the (2 0th or) 1st row is T, the next (2 0 = 1) row is F, and so on. In our example, this means just one more step—2 2-2 = 2 0 = 1 so alternate every 1 row between T and F. Thus, you’re done constructing guide columns for ~q |- [(p → q) → p] → p pq T T F F T F T F

15. What you’ve done so far… You’ve presented all of the possible ways in which to interpret the argument you’re out to analyze.

16. Step 2: Columns for premises and conclusions Set up an additional column for every premise and the conclusion. Mark “P” above the columns corresponding to premises and “C” above the column corresponding to the conclusion. In our example, we have one premise, ~q, and one conclusion, [(p → q) → p] → p. Thus, we add two columns to our truth-table. pq~q [(p → q) → p] → p TT TF FT FF PCPC

17. Step 3: Add more columns to interpret complex propositions First, for each complex premise/conclusion, count the number of logical operators. Call this number M. Add M-1 columns to the truth table. There are 3 logical operators in this conclusion, so add 3-1 = 2 additional columns pq~q [(p → q) → p] → p TT TF FT FF →→ PCPC

18. More on additional columns Next, analyze the compound statement, such that each of the additional columns corresponds to the scope of the operator in question. pq~q [(p → q) → p] → p TT TF FT FF → → (p → q)→pp → q PCPC

19. Step 4: Filling out the table Using your knowledge of the truth-tables for conjunction ‘&’, disjunction ‘v’, negation ‘~’, material implication ‘→’, and material equivalence ‘↔’, fill out the table. pq p → q(p → q) → p ~q [(p → q) → p] → p TT TF FT FF TFTTTFTT FTFTFTFT T T F F T T T T PCPC

20. Step 5: Reading the truth table Look at all of the columns with P’s and C’s above them. If there are one or more rows in which all of the P-columns have T’s and the C-column has an F, then you have an invalid argument; otherwise the argument is valid. IMPORTANT: This does NOT mean that if you have ONE row with F’s in the P- columns or T’s in the C-column that you have a valid argument—you need to prove that there are NO rows in the ENTIRE truth table that have all T’s in the P- columns and an F in the C-column in order to have a valid argument. In this example, there is NO row in which P = T and C = F. Therefore the argument is valid. pq p → q(p → q) → p ~q [(p → q) → p] → p TTTTFT TFFTTT FTTFFT FFTFTT PCPC

21. Shortcut in Step 4 First, determine which of the premises and conclusion is the easiest in terms of figuring out the truth values. Fill this column first. pq p → q(p → q) → p ~q [(p → q) → p] → p TT TF FT FF PCPC FTFTFTFT

22. Shortcut, Continued Since an invalid argument can only occur when the premises are true and the conclusion is false, any rows in which either a premise is false or the conclusion is true can be ignored, since it will be irrelevant to determining the invalidity of the argument. pq p → q(p → q) → p ~q [(p → q) → p] → p TTF TFT FTF FFT PCPC IGNORE! F T T FT T

23. Exercise B2 (C v D)  (C & D), C & D├ C v D CDC v DC & D (C v D)  (C & D) TT TF FT FF CP2P2 P1P1 TTTFTTTF IGNORE! F VALID!

24. Exercise B10 U  (V v W), (V&W)  ~U ├ ~U UVW~UV v WV & W U  (V v W) (V&W)  ~U TTTTFFFFTTTTFFFF TTFFTTFFTTFFTTFF TFTFTFTFTFTFTFTF C P1P2P1P2 FFFFTTTTFFFFTTTT IGNORE! TTTFTTTF TTTFTTTF TFFTFF FTTFTT INVALID!

25. Exercise C7 If terrorists’ demands are met, then lawlessness will be rewarded. If terrorists’ demands are not met, then innocent hostages will be murdered. So either lawlessness will be rewarded or innocent hostages will be murdered. T  L, ~T  I |- L v I TLI~T T  L~T  I L v I TTT TTF TFT TFF FTT FTF FFT FFF IGNORE TTTFTTTFTTTFTTTF VALID! IGNORE P1P2CP1P2C F TTF

26. Exercise B8 (O v P)  Q, Q  (O & P)├ (O v P)  (O & P) OPQO v PO & P (OvP)  Q Q  (O&P) (OvP)  (O&P) TTT TTF TFT TFF FTT FTF FFT FFF TTTTTTFFTTTTTTFF TTFFFFFFTTFFFFFF TTFFFFTTTTFFFFTT IGNORE! FTFTFTFT F F VALID! P1P2CP1P2C