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Using propositional logic, prove that the argument AA (B → C)A [(A A B) → (D V C')] A B→D is valid, We must produce a proof sequence that begins with the hypotheses and ends with the conclusion. There are four hypotheses, so this gives us lots of “ammuni- tion" to use in the proof. The beginning of the proof is easy enough because it just involves listing the hypotheses: 1. A hyp 2. B→C hyp 3. (АЛ В) — (DЛC') hyp 4. B hyp Our final goal is to arrive at D, the conclusion. But without even looking ahead, there are a couple of fairly obvious steps we can take that may or may not be helpful. 5. C 2, 4, mp 6. AAB 1, 4, con 7. DV C' 3, 6, mp At least at this point we have introduced D, but it's not by itself. Note that from step 5 we have C, which we haven't made use of. If only we had C→D, we'd be home free. Ah, look at the form of step 7; it's a disjunction, and the implication rule says that we can transform a disjunction ofa certain form into an implication. The disjunction must have a negated wff on the left We can do that: 8. C' VD 7, COIII 9. C-D 8, imp so 10. D 5,9, mp

Using propositional logic, prove that the argument AA (B → C)A [(A A B) → (D V C')] A B→D is valid, We must produce a proof sequence that begins with the hypotheses and ends with the conclusion. There are four hypotheses, so this gives us lots of “ammuni- tion" to use in the proof. The beginning of the proof is easy enough because it just involves listing the hypotheses: 1. A hyp 2. B→C hyp 3. (АЛ В) — (DЛC') hyp 4. B hyp Our final goal is to arrive at D, the conclusion. But without even looking ahead, there are a couple of fairly obvious steps we can take that may or may not be helpful. 5. C 2, 4, mp 6. AAB 1, 4, con 7. DV C' 3, 6, mp At least at this point we have introduced D, but it's not by itself. Note that from step 5 we have C, which we haven't made use of. If only we had C→D, we'd be home free. Ah, look at the form of step 7; it's a disjunction, and the implication rule says that we can transform a disjunction ofa certain form into an implication. The disjunction must have a negated wff on the left We can do that: 8. C' VD 7, COIII 9. C-D 8, imp so 10. D 5,9, mp

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Using propositional logic, prove that the argument AA (B → C)A [(A A B) → (D V C')] A B→D is valid, We must produce a proof sequence that begins with the hypotheses and ends with the conclusion. There are four hypotheses, so this gives us lots of “ammuni- tion" to use in the proof. The beginning of the proof is easy enough because it just involves listing the hypotheses: 1. A hyp 2. B→C hyp 3. (АЛ В) — (DЛC') hyp 4. B hyp Our final goal is to arrive at D, the conclusion. But without even looking ahead, there are a couple of fairly obvious steps we can take that may or may not be helpful. 5. C 2, 4, mp 6. AAB 1, 4, con 7. DV C' 3, 6, mp At least at this point we have introduced D, but it's not by itself. Note that from step 5 we have C, which we haven't made use of. If only we had C→D, we'd be home free. Ah, look at the form of step 7; it's a disjunction, and the implication rule says that we can transform a disjunction ofa certain form into an implication. The disjunction must have a negated wff on the left We can do that: 8. C' VD 7, COIII 9. C-D 8, imp so 10. D 5,9, mp
Transcribed Image Text:Using propositional logic, prove that the argument AA (B → C)A [(A A B) → (D V C')] A B→D is valid, We must produce a proof sequence that begins with the hypotheses and ends with the conclusion. There are four hypotheses, so this gives us lots of “ammuni- tion" to use in the proof. The beginning of the proof is easy enough because it just involves listing the hypotheses: 1. A hyp 2. B→C hyp 3. (АЛ В) — (DЛC') hyp 4. B hyp Our final goal is to arrive at D, the conclusion. But without even looking ahead, there are a couple of fairly obvious steps we can take that may or may not be helpful. 5. C 2, 4, mp 6. AAB 1, 4, con 7. DV C' 3, 6, mp At least at this point we have introduced D, but it's not by itself. Note that from step 5 we have C, which we haven't made use of. If only we had C→D, we'd be home free. Ah, look at the form of step 7; it's a disjunction, and the implication rule says that we can transform a disjunction ofa certain form into an implication. The disjunction must have a negated wff on the left We can do that: 8. C' VD 7, COIII 9. C-D 8, imp so 10. D 5,9, mp
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