Intro to Logic notes9

Intro to Logic notes9 - 9 Monday, July 23, 2007 10:14 AM...

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Unformatted text preview: 9 Monday, July 23, 2007 10:14 AM Review of methods we have covered for testing validity. Most recent method: Derivation. Conditional derivation 1) 2) 3) 4) 5) 6) 7) PQ Pr QR Pr SHOW: PR CD P As SHOW: R DD Q 1,4,O R 2,6,O 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) PR Pr QS Pr SHOW: (P&Q)(R&S) CD P&Q As SHOW: R&S DD P 4,&O R 1,6,O Q 4,&O S 2,8,O R&S 7,9,&I 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) QR Pr R(PS) Pr SHOW: (P&Q)S CD (P&Q) As SHOW: S DD P 4,&O Q 4,&O R 1,7,O PS 2,8,O S 6,9,O 1) 2) 3) 4) 5) 6) 7) 8) (P&Q)R Pr SHOW: P(QR) CD P As SHOW: QR CD Q As SHOW: R DD P&Q 3,5,&I R 1,7,O Intro to Logic Page 1 Indirect Derivations Let's think of a way we might try to prove that the argument is valid Consider what the argument is saying: PQ; P~Q / ~P If we assume P, then Q If we assume P then ~Q So if we assume P then both Q and ~Q This is an obvious contradiction If we assume P then a contradiction follows So we assume ~P by modus tollens We want to show that P implies a contradiction and infer ~P by modus tollens We try to show by conditional derivation that if P, the (Q&~Q) This is allowed by the show line rule. 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) PQ Pr P~Q Pr SHOW: ~P DD SHOW: P(Q&~Q) CD P As SHOW: Q&~Q DD Q 1,5,O ~Q 2,5,O Q&~Q 7,8,&I ~(Q&~Q) ??? ~P 4,10,O We could just introduce a rule that allows us to infer that this is false. We introduce Indirect Derivation. This gives us the contradiction symbol. XI A ~A _____ X XO X ______ A Reductio ad absurdum 1) 2) 3) 4) 5) 6) 7) PQ Pr P~Q Pr SHOW: ~P ID P As SHOW: X DD Q 1,4,O ~Q 2,4,O Intro to Logic Page 2 8) X 6,7,XI Intro to Logic Page 3 ...
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