1. Proof of Invalidity Kareem Khalifa Department of Philosophy Middlebury College

2. Overview Why this matters Proving invalidity –Discovery –Proof Exercises

3. Why this matters Not every argument you encounter will be valid. But, just as with valid arguments, you want to avoid the cumbersome nature of truth tables and the unreliable nature of constructing counterexamples. Just as with formal proofs of validity, formal proofs of invalidity strike a balance between these two poles.

4. Proving invalidity Take a well known example of an invalid argument: denying the antecedent. –P  Q, ~P ├ ~Q How do we prove this is invalid? –Recall: An argument is invalid if it is possible for all of the premises to be true and the conclusion to be false. –So we set the truth-values of the simple propositions in the argument (in this case, P and Q) such that all of the premises are true and the conclusion is false.

5. The two aspects of a formal proof of invalidity 1.Your process of discovery. This is how you came to the answer you arrived at. This belongs on your SCRAP paper. 2.The proof. This is how you show that the a certain combination of truth values yields a scenario in which the all of the premises are true and the conclusion is false. I ONLY care about the proof. This is your final answer. These two aspects often require thinking in opposite directions.

6. Discovery in 3 steps 1.Set up a one-row truth table, with a column for each simple proposition, each premise, and the conclusion. 2.Fill in T’s under the premise-columns and F under the conclusion column. 3.Using your knowledge of &, ~, v, etc., make the columns under the simple propositions fit with the T’s and F you wrote in Step 2.

7. Steps 1-2 P  Q, ~P ├ ~Q 1.Set up a truth table with only 1 row. The number of columns depends on the number of simple propositions and premises. 2.Fill in all P-columns as true, and the C- column as false. P1P1 P2P2 C PQ P  Q ~P~Q TTF

8. Step 3 P1P1 P2P2 C PQ PQPQ ~P~Q TTF P  Q, ~P ├ ~Q 3.Using your knowledge of the logical operators, fill in the remaining blanks in the truth table such that all of the premises remain true and the conclusion remains false. If the conclusion, ~Q, is false, then Q is true. T If the premise, ~P, is true, then P is false. If Q is true, then the premise, P  Q, is also true. √F NOTE: You don’t need to write all of this out.

9. Tips for discovery Use DM, MI, etc. to find simpler formulations of negated &, v, and  propositions. o Ex. If you have ~(PvQ), switch it to ~P&~Q. Try to minimize wiggle room, i.e. find simple propositions that can only have one truth-value. When the main operator of the conclusion is ‘~,’ ‘v,’ ‘ ,’ or a simple proposition, the conclusion has minimal wiggle room. When a premise is ‘~,’ ‘&,’ or a simple proposition, the premise has minimal wiggle room.

10. The Proof 4.Write down your completed truth-table from Step 3. 5.Begin by stating the truth-values of all the simple propositions. 6.Show that according to these simple truth-values and the rules for &, ~, v, etc., all of the premises are true and the conclusion is false.

11. Step 4-6 5.Begin by stating the truth-values of all the simple propositions. Let P be false and Q be true. 6.Show that according to these simple truth-values, all of the premises are true and the conclusion is false. P1P1 P2P2 C PQ PQPQ ~P~Q FTTTF Since P is false and Q is true, the premise P  Q is also true. Since P is false, the premise ~P is true. Since Q is true, the conclusion ~Q is false. So there is at least one way in which the premises can be true and the conclusion can be false. So the argument is invalid. This is your proof

12. Contrast Discovery (scrap) Start by assuming the conclusion is false and all premises are true. Then show how the simple propositions must assume certain truth values. Proof (final answer) Start by assigning certain truth values to the simple propositions. Then show how it follows from this assignment of truth values that the conclusion is false and all premises are true.

13. For the test… A single line of a truth-table, in which truth- values are assigned to all of the simple propositions, is sufficient. (No need for the numbered proof.) P1P1 P2P2 C PQ PQPQ ~P~Q FTTTF

14. 2) ~(E&F), (~E&~F)  (G&H), H  G ├ G CP1P1 P2P2 P3P3 EFGH~(E&F) (~E&~F)  (G&H)HGHG TFFFTTT FTFFTTT 1.Let E be true and F, G, and H be false. 2.Since F is false, E&F is false. 3.By 2, the premise ~(E&F) is true. 4.Since E is true, ~E is false. 5.Since ~E is false, ~E&~F is false. 6.By 5, the premise (~E&~F)  (G&H) is true. 7.Since H is false, the premise H  G is true. 8.The conclusion G is false. 9.So, by 3,6,7, and 8, the premises can be true when the conclusion is false. 10.So the argument is invalid.

15. 7) D  (EvF), G  (HvI), ~E  (IvJ), (I  G)&(~H  ~G), ~J├ D  (GvI) DEFGHIJ D  (EvF)~E  (IvJ)G  (HvI)(I  G)& (~H  ~G) ~J D  (GvI) TTFFFFTTTF 1.Let D and E be true; F, G, I, and J be false. (You can let H be either T or F) 2.Since G and I are false, G v I is false. 3.Since D is true and G v I is false, the conclusion, D  (GvI) is false. 4.Since J is false, the premise ~J, is true. 5.Since I is false, I  G is true. 6.Since G is false, ~G is true. 7.By 6, ~H  ~G is true. 8.By 5 and 7, the premise (I  G)&(~H  ~G) is true. 9.Since G is false, the premise G  (HvI) is true. 10.Since E is true, E v F is true. 11.Since D and E v F are both true, the premise D  (EvF) is true. 12.Since E is true, ~E is false. 13.By 12, the premise ~E -> (I v J) is true. 14.By 3, 4, 8, 9, 11, and 13 it’s possible for all the premises to be true when the conclusion is false. 15.So the argument is invalid. T

16. 9.(S  T)&(T  S), (U&T)v(~T&~U), (UvV)v(SvT), ~U  (W&X), (V  ~S)&(~V  ~Y), X  (~Y  ~X), (UvS)&(VvZ) ├ X&Z 1.Let S, T, U, W, and Z be true; V,X,Y be false. 2.Since X is false, the conclusion X&Z is false. 3.Since T is true, S  T is true. 4.Since S is true, T  S is true. 5.By 3 and 4, the premise (S  T)&(T  S) is true. 6.Since U and T are true U&T is true. 7.By 6, the premise (U&T) v(~T&~U) is true. 8.Since U is true U v V and UvS are both true. 9.By 8, the premise (UvV)v(SvT) is true. 10.Since U is true, ~U is false. 11.By 10, the premise ~U  (W&X) is true. 12.Since V is false V  ~S is true. STUVWXYZ (S  T)& (T  S) (U&T)v (~T&~U) (UvV)v (SvT) ~U  (W&X) (V  ~S)& (~V  ~Y) X  (~Y  ~X) (UvS) & (VvZ) X&ZX&Z TTTFTFFTTTTTTTTF TTTFFFFTTTTTTTTF 13.Since Y is false, ~Y is true. 14.Since ~Y is true, ~V  ~Y is true. 15.By 12 and 14 are true, the premise (V  ~S)&(~V  ~Y) is true. 16.Since X is false, the premise X  (~Y  ~X) 17.Since Z is true, VvZ is true. 18.By 8 and 17, the premise (UvS) & (VvZ) is true. 19.By 2, 5, 7, 9, 11, 16, and 18 it is possible for all the premises to be true when the conclusion is false. 20.So the argument is invalid.