# Proofs using all 18 rules, from Hurley 9th ed. pp. 376-378 1. 1. Q > (F > A) 2. R > (A > F) 3. Q ∙ R / F ≡ A 2. 1. (J ∙ R) > H 2. (R > H) > M 3. ~(P v.

## Presentation on theme: "Proofs using all 18 rules, from Hurley 9th ed. pp. 376-378 1. 1. Q > (F > A) 2. R > (A > F) 3. Q ∙ R / F ≡ A 2. 1. (J ∙ R) > H 2. (R > H) > M 3. ~(P v."— Presentation transcript:

Proofs using all 18 rules, from Hurley 9th ed. pp. 376-378 1. 1. Q > (F > A) 2. R > (A > F) 3. Q ∙ R / F ≡ A 2. 1. (J ∙ R) > H 2. (R > H) > M 3. ~(P v ~J) / M ∙ ~P 3. 1. F > (A ∙ K) 2. G > (~A ∙ ~K) 3. F v G / A ≡ K 4. 1. T > G 2. S > G / (T v S) > G Let’s walk through how to do these, one at a time.

1. Q > (F > A) 2. R > (A > F) 3. Q ∙ R / F ≡ A Start by thinking about the conclusion. It’s a bi-conditional, so there are two Rules of Equivalence that pertain. The last step will surely be EQ, rewriting one of them into a triple bar. So you should be able to see that the consequent of line 1, and the consequent of line 2, if joined together by a dot, would meet one of the versions of the EQ rule. That means that the task is to get those two consequents onto lines by themselves, so they can be joined together by the rule CN. What rule will deliver the consequent of a line on to a new line? MP What is the missing piece for an MP to 1?Q And Q can be got from line 3 by SM.CM and SM to 3 will get R, And that allows MP to line 2.

1. (J ∙ R) > H 2. (R > H) > M 3. ~(P v ~J) / M ∙ ~P Look at the conclusion and think about a likely candidate for the final step. Like CN? You’d shoot for getting M on one line, ~P on another, and then conjoining them. Hint: Whenever you spot the sort of thing we have here in Line 3, do DM to the line. DM will turn this from a negative wedge statement into a dot statement, which is much easier to do something to! 4. ~P. ~~J 3, DM 5. ~P. J 4, DN 6. ~P 5, SM Half the job is done. We have ~P. How to get M? GO for an MP to line 2, since M is the consequent of 2. The missing piece is R > H. Look at the rule called exportation, and look at line 1. 7. J > (R > H) 1 EXP 8. J 5, CM, SM 9. R > H MP 7,8 10.M MP 9, 2 11.M. ~P CN 6,10

1. F > (A ∙ K) 2. G > (~A ∙ ~K) 3. F v G / A ≡ K Like an earlier example, since the conclusion is a triple bar statement, the last move is probably going to be EQ. Do you see the pieces of the other version of EQ in the premises? The triple bar means: “Either both or neither” as well as meaning “is necessary and sufficient for.” Line 1 says “If F, then both A and K.” Line 2 says “If G, then neither A nor K.” Line 3 is the missing piece for a CD. 4. [F> (A. K)]. [G>(~A. ~K)] CN, 1,2 5. (A. K) v (~A. ~K) CD 3,4 6. A ≡ K EQ 5

1. T > G 2. S > G / (T v S) > G Here’s a chance to use IMP. IMP is extremely useful: it lets us rewrite wedges as horseshoes and vice versa. Whenever you do this, just change the left hand side by a tilde. This change is very intuitive: “We’ll either have soup or egg roll” means “We’ll have soup if we don’t have egg roll.” But you can’t CM horsehoes the way you can wedges. 3. ~T v G IMP 1 4. ~S v G IMP 2 Now we’ll use DIST, after putting 3 and 4 together 5. (~T v G). (~S v G) CN 3,4 6. (G v ~T). (G v ~S) CM 7. G v (~T. ~S) DIST 6 Now use IMP again and DM 8. ~G > ~(T v S) IMP, DM 7 And TRANS flips this over and reverses the tildes 9. (T v S) > G TRANS 8 This is much easier to do by Conditional Proof!

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