i.e. The conclusion to the following argument says If you had an A, then you d have a ~(B v Z).

Transcription

1 7.5 Conditional Proof (CP): Conditional Proof is a different way to do proofs. Using CP will always get you a horseshoe statement, so the best time to use it is when your conclusion is either a horseshoe or triplebar. A Conditional statement is a horseshoe statement (p q). It says If you had a p, then you d have a q. i.e. The conclusion to the following argument says If you had an A, then you d have a ~(B v Z). We are going to test this conclusion by assuming we do have the A (assuming the p of the p q ), and seeing if we can get ~(B v Z) (which is the q ). Any time you make an assumption, to show that you re doing something different than the regular proof, you need to indent (tab in). 1. A ~B 2. A ~Z / A ~(B v Z) 3. A (Assumption for CP) 4. ~B 1,3 MP 5. ~Z 2,3 MP 6. ~B ~Z 4,5 conj. 7. ~(B v Z) 6 DM 8. A ~(B v Z) 3-7 CP (the whole indented area) Line 7 shows that by assuming the A (saying if we had A ), that we could then get the ~(B v Z). So leaving the indented area (tabbing out) automatically gives us the first line of the indented area (our assumption) horseshoed to the last line of the indented area (wherever we decide to step back out). If we had left the indented area after line 5, CP would give us A ~Z (because we would have proven that if we had an A, we could get a ~Z). CP always automatically gives you your assumption horseshoed to where you leave. In the next proof, notice that the main connective of the conclusion is a horseshoe, but also that the q of that statement is another horseshoe statement. So it is saying If you had a ~Z then if you had an M, then you d have ~O. We are going to make 2 assumptions for this proof, one right after the other, because we have more horseshoe statements after our first assumption. 1. (M K) Z 2. ~O v (K v Z) / ~Z (M ~O) 3. ~Z 4. M 5. ~(M K) 1,3 MT 6. ~M v ~K 5 DM 7. ~K 4,6 DS 8. ~K ~Z 3,7 conj.

2 9. ~(K v Z) 8 DM 10. ~O 2,9 DS 11. M ~O 4-10 CP 12. ~Z (M ~O) 3-11 CP Your first assumption is always the p or left side of the main connective horseshoe in the conclusion. Cover that up and look at what s left. If it s another horseshoe statement then assume the p or left side of that one. Keep doing that until you run out of horseshoe statements. Never assume the last statement since it s the one you re trying to find. i.e. If this were your conclusion, this is how you d make your assumptions: / (O J) {(A F) ((~K U) [~O (M Z)])} O J A F ~K U ~O M 1. ~Z R 2. Z ~(O ~T) / (R Z) (~T ~O) 3. R Z 4. ~T 5. ~Z Z 1,3 HS 6. Z v Z 5 Impl. 7. Z 6 taut. 8. ~(O ~T) 2,7 MP 9. ~O v T 8 DM 10. ~O 4,9 DS 11. ~T ~O 4-10 CP 12. (R Z) (~T ~O) 3-11 CP Normally I draw a line from line 4 down to line 10, and then another line from line 3 down to line 11 showing the two indented areas. (I couldn t figure out how to do that in the Word doc). I draw the line for the indented area for 2 reasons. One is to line things up neatly, but the second and most important reason is to remind yourself that once you LEAVE an indented area, every line in that indented area becomes off limits. Doing CP with the triplebar: When you have a triplebar ( ) statement as your conclusion, you ll want to do 2 separate CP s (not one within another). Equivalence your conclusion on scrap paper using the horseshoe/dots one. You are going to do a CP to each half of that statement.

3 If you have a triplebar statement as one of your premises, equiv it first before you make your assumption. If you don t, then you ll have to redo it again for the second half of your CP. 1. ~M O 2. (M K) v Y 3. (~M v B) Z / (~M Z) (O Y) 4. (~M O) (O ~M 1 equiv. 5. ~M O 4 simp. 6. O ~M 4 simp. 7. ~M Z 8. ~M 7 simp. 9. O 5,8 MP 10. ~M v ~K 8 Add 11. ~(M K) 10 DM 12. Y 2,11 DS 13. O Y 9,12 conj. 14. (~M Z) (O Y) 7-13 CP 15. O Y 16. O 15 simp. 17. ~M 6,16 MP 18. ~M v B 17 Add 19. Z 3,18 MP 20. ~M Z 17,19 conj. 21. (O Y) (~M Z) CP 22. [(~M Z) (O Y)] [(O Y) (~M Z)] 14,21 conj 23. (~M Z) (O Y) 22 equiv. 7.6 Indirect Proof (IP): This is another different way to do proofs. IP can be used on any proof (although it s trickier than CP for triplebar conclusions). With IP the assumption you re going to make is always the OPPOSITE of the conclusion (by doing this, it s like you re rendering the proof invalid). Then you will break the statements down until you find a pair of contradictory statements (something and its opposite). The last line within the indented area will be the conjunction of the two opposites. This proves that your assumption lead to a logical impossibility (which is bad), therefore instead of your assumption, you will get the opposite of it (which if you assumed the opposite of your conclusion, will then be your conclusion.)

4 A note on opposites: Remember that the opposite of something is always ONE tilde on the outside of it. Here are some examples: B ~B B H ~(B H) ~Y ~W ~(~Y ~W) ~H v M ~(~H v M) ~[~U v (N A)] ~U v (N A) In each case, the ONLY difference between the 2 statements is that one of them has a (red) tilde on the outside of it. Otherwise they are exactly the same. 1. ~B (B v Y) 2. (Z v O) ~(Y v M) / B v ~Z 3. ~(B v ~Z) AIP (assumption for IP) 4. ~B Z 3 DM 5. ~B 4 simp. 6. Z 4 simp. 7. B v Y 1,5 MP 8. Y 5,7 DS 9. Z v O 6 add 10. ~(Y v M) 2,9 MP 11. ~Y ~M 10 DM 12. ~Y 11 simp. 13. Y ~Y 8,12 conj (the contradiction) 14. B v ~Z 3-13 IP (it gives the opposite of your assumption) IP will always give you the OPPOSITE of your assumption once you find a pair of contradictory statements. Unlike CP, you cannot leave the indented area whenever you want. The last line of the indented area must be the conjunction (with a dot) of the two opposites. 1. (T B) (Y O) 2. (~Z M) (O ~Y) / (M ~T) v (B Z) 3. ~[(M ~T) v (B Z)] AIP 4. ~(M ~T) ~(B Z) 3 DM 5. ~(M ~T) 4 simp. 6. ~(B Z) 4 simp. 7. ~(~M v ~T) 5 Impl. 8. M T 7 DM 9. M 8 simp. 10. T 8 simp. 11. ~(~B v Z) 6 Impl. 12. B ~Z 11 DM

5 13. B 12 simp. 14. ~Z 12 simp. 15. T B 10,13 conj. 16. Y O 1,15 MP 17. Y 16 simp. 18. O 16 simp. 19. ~Z M 9,14 conj. 20. O ~Y 2,19 MP 21. ~Y 18,20 MP 22. Y ~Y 17,21 conj. 23. (M ~T) v (B Z) 3-22 IP If you want, you can use both CP and IP in the same proof: Ex. 1. (M E) v (Z v ~A) 2. (T T) (K A) / ~M (K Z) 3. ~M 4. K 5. ~Z AIP (assume the opposite of what you want) 6. ~M v ~E 3 Add 7. ~(M E) 6 DM 8. Z v ~A 1,7 DS 9. ~A 5,8 DS 10. K ~A 4,9 conj. 11. ~(~K v A) 10 DM 12. ~(K A) 11 Impl. 13. ~(T T) 2,12 MT 14. ~(~T v T) 13 Impl. 15. T ~T 14 DM (the contradiction) 16. Z 5-15 IP (get opposite of your assumption) 17. K Z 4-16 CP 18. ~M (K Z) 3-17 CP