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ln2-page6

# ln2-page6 - CD the new SHOW line will itself be an if-then...

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We have now shown R in the normal way. We can cross off that “SHOW”: 1. 2. 3. 4. 5. 6. 7. P ! Q Q ! R SHOW: P ! R P SHOW: R Q R Pr Pr CD Ass DD 1,4 ! O 2,6 ! O Where are we now? We have assumed P , and have successfully shown R on the basis of that assumption. We are now entitled to regard ourselves as having shown P ! R and can cross off the “SHOW” at line 3. That’s the way CD works. The final problem looks like this: 1. 2. 3. 4. 5. 6. 7. P ! Q Q ! R SHOW: P ! R P SHOW: R Q R Pr Pr CD Ass DD 1,4 ! O 2,6 ! O One thing that is important to remember is that, once a new SHOW line is introduced, your task has changed. You should focus on proving what it says to prove at the new SHOW line, and forget for the moment about trying to get the final conclusion. Here are some additional examples. 1. 2. 3. 4. 5. 6. 7. 8. P ! ⇠ Q SHOW: ( Q & R ) ! ( S _ ⇠ P ) Q & R SHOW: S _ ⇠ P Q ⇠⇠ Q P S _ ⇠ P Pr CD Ass DD 3 &O 5 DN 1,6 ! O 7 _ I 1. 2. 3. 4. 5. 6. 7. 8. 9. E \$ G H _ E SHOW: H ! ( G & E ) H SHOW: G & E E E ! G G G & E Pr Pr CD Ass DD 2,4 _ O 1 \$ O 6,7 ! O 6,8 &I Sometimes when we introduce a new SHOW line as part of a
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Unformatted text preview: CD, the new SHOW line will itself be an if-then statement. If so, we may do a conditional derivation within a conditional derivation. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. A ! D ( D & ⇠ B ) ! C ⇠ ⇠ ⇠ SHOW: A ! ( ⇠ B ! C ) A ⇠ ⇠ ⇠ SHOW: ⇠ B ! C ⇠ B ⇠ ⇠ ⇠ SHOW: C D D & ⇠ B C Pr Pr CD Ass CD Ass DD 1,4 ! O 6,8 &I 2,9 ! O 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. B _ ( C ! E ) ⇠ D !⇠ ( E & A ) ⇠ ⇠ ⇠ SHOW: ( A & ⇠ B ) ! ( C ! D ) A & ⇠ B ⇠ ⇠ ⇠ SHOW: C ! D C ⇠ ⇠ ⇠ SHOW: D ⇠ B C ! E E A E & A ⇠⇠ ( E & A ) ⇠⇠ D D Pr Pr CD Ass CD Ass DD 4 &O 1,8 _ O 6,10 ! O 4 &O 10,11 &I 12 DN 2,13 ! O 14 DN We can use CD whenever we wish to prove a conditional (if-then) statement. It does not need to be the Fnal conclusion of our argument; 6...
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