Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Tet(a) ᴧ Large(a)Med(b) ᴧ Dodec(b)(Large(a) ᴧ Dodec(b)) ᴠ Cube(c) Always first ask: ‘Is this argument valid?’Look at what you know from the premises: Both premises are conjunctions, so we know every one of the conjuncts is true. And it’s easy to break apart conjunctions to gain access to the individual conjuncts (just use ᴧ Elim).Look at what you are trying to prove (the goal): It’s a combination of one of the conjuncts from each premise, disjoined to an atomic sentence that’s not in either of the premises.1 Tet(a) ᴧ Large(a)2 Med(b) ᴧ Dodec(b)3 Large(a) ᴧ Elim: 1(Large(a) ᴧ Dodec(b)) ᴠ Cube(c) We break open the premises one at a time and take just the conjuncts that we need in order to reach the goal.1 Tet(a) ᴧ Large(a)2 Med(b) ᴧ Dodec(b)3 Large(a) ᴧ Elim: 14 Dodec(b) ᴧ Elim: 2(Large(a) ᴧ Dodec(b)) ᴠ Cube(c) Now that we have the two conjuncts we need, we can combine them at the next step to build the new conjunction we need for the goal …1 Tet(a) ᴧ Large(a)2 Med(b) ᴧ Dodec(b)3 Large(a) ᴧ Elim: 14 Dodec(b) ᴧ Elim: 25 Large(a) ᴧ Dodec(b) ᴧ Intro: 3,4(Large(a) ᴧ Dodec(b)) ᴠ Cube(c) This gives us the first disjunct of the goal, but how can we get the other disjunct (Cube(c)) in order to reach the goal? We need to create a disjunction … but where do we get Cube(c) from, since it’s not mentioned in the premises?1 Tet(a) ᴧ Large(a)2 Med(b) ᴧ Dodec(b)3 Large(a) ᴧ Elim: 14 Dodec(b) ᴧ Elim: 25 Large(a) ᴧ Dodec(b)ᴧ Intro: 3,46 (Large(a) ᴧ Dodec(b)) ᴠ Cube(c) ᴠ Intro: 5We can pull Cube(c) ‘out of thin air’, as it were, because we are creating a disjunction, and only one disjunct of a disjunction need be true for the entire disjunction to be true. We already know that the first disjunct we created is true (because we drew that information from the premises), so that is enough to ensure the truth of the whole disjunction. We’re done!(A ᴧ B) ᴠ (C ᴧ D)Notice that both your premise and your goal are disjunctions. Obviously, you’ll likely need to use one or both disjunction rules, and ᴠ Elim would be a promising overall strategy. The most direct thing to try would be to use the overall disjunction in the premise as the basis for two subproofs and try to prove the goal B ᴠ D as the conclusion of each subproof …B ᴠ D(A ᴧ B) ᴠ (C ᴧ D)A ᴧ B Here’s the structure that the overall proof-by-cases (i.e., disjunction elim) strategy requiresB ᴠ DC ᴧ DB ᴠ DB ᴠ D ᴠ ElimLet’s begin by fleshing out the first ‘case’, the first subproof. What can we do with the premise A ᴧ B that will help us get to the conclusion of the subproof, B ᴠ D?1 (A ᴧ B) ᴠ (C ᴧ D)2 A ᴧ B3 B ᴧ Elim: 2B ᴠ DC ᴧ DB ᴠ DB ᴠ D ᴠ ElimWe can break the conjunction A ᴧ B apart using conjunction elim to isolate just B, citing step 2 (i.e., the conjunction we broke apart) as input. Now, which of our rules can we use next to get from just B to the disjunction B ᴠ D?1 (A ᴧ B) ᴠ (C ᴧ D)2 A ᴧ B3 B ᴧ Elim: 24 B ᴠ D ᴠ Intro: 3C ᴧ DB ᴠ DB ᴠ D ᴠ ElimCreating the disjunction B ᴠ D is easy using the rule disjunction intro, which allows us to disjoin anything to an element that we already know to be true. We‘ve proven B is true (supposing A ᴧ B is true), so we can assert that B ᴠ D is true, and we’ve reached our desired subproof conclusion.Next, we turn to the second subproof ‘case’. We can flesh out this subproof in a way parallel to how we completed the first subproof above …1 (A ᴧ B) ᴠ (C ᴧ D)2 A ᴧ B3 B ᴧ Elim: 24 B ᴠ D ᴠ Intro: 35 C ᴧ D6 D ᴧ Elim: 5B ᴠ DB ᴠ D ᴠ Elim1 (A ᴧ B) ᴠ (C ᴧ D)2 A ᴧ B3 B ᴧ Elim: 24 B ᴠ D ᴠ Intro: 35 C ᴧ D6 D ᴧ Elim: 57 B ᴠ D ᴠ Intro: 6B ᴠ D ᴠ Elim1 (A ᴧ B) ᴠ (C ᴧ D)2 A ᴧ B3 B ᴧ Elim: 24 B ᴠ D ᴠ Intro: 35 C ᴧ D6 D ᴧ Elim: 57 B ᴠ D ᴠ Intro: 68 B ᴠ D ᴠ Elim: 1, 2-4, 5-7Last, we just fill in the step citations for the disjunction elim rule that justifies the final step, based on the proof-by-cases we’ve now completed. Cite the original disjunction on which the strategy was based, and cite in their entirety each of the subproofs that served as the
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