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2 We demonstrate (show) that an argument is valid deriving by deriving (deducing) its conclusion from its premises using a few fundamental modes of reasoning.
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3 Modus (Tollendo) Tollens (denying by denying) –––––– Modus (Ponendo) Ponens (affirming by affirming) –––––– Modus Tollendo Ponens (1) (affirming by denying) –––––– Modus Tollendo Ponens (2) (affirming by denying) –––––– aka disjunctive syllogism
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4 (10) (9) (8) (7) (6) (5) (4) (3) (2) (1) 5,9, Q 4,8, P 3,7, T 1,2, R DD : Q Pr P Q Pr P T Pr R T Pr R S Pr SS S ; R S ; R T ; P T ; P Q / Q MP MTP2 MTP1 MT 1.write all premises 3.apply rules to available lines 4.box and cancel (high-five!) 2.write ‘show’ conclusion
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5 Tek gives Manny a high-fiveTek gives A-Rod a high-five
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6 available on course web page (textbook) provided on exams keep this in front of you when doing homework don’t make up your own rules
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8 Every connective has (1)a take-out rule also called an: elimination-rule OUT-rule (2)a put-in rule also called an: introduction-rule IN-rule
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9 conjunction & if you have a conjunction & then you are entitled to infer–––––– first conjunct its first conjunct conjunction & if you have a conjunction & then you are entitled to infer–––––– second conjunct its second conjunct ‘have’ means ‘have as a whole line’ rules apply only to whole lines not pieces of lines
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10 formula if you have a formula formula and you have a formula then you are entitled to infer–––––– conjunction & their (first) conjunction & formula if you have a formula formula and you have a formula conjunction & then you are entitled to infer–––––– their (second) conjunction &
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11 disjunction if you have a disjunction negation1 st disjunct and you have the negation of its 1 st disjunct then you are entitled to infer –––––– second disjunct its second disjunct disjunction if you have a disjunction negation2 nd disjunct and you have the negation of its 2 nd disjunct then you are entitled to infer –––––– first disjunct its first disjunct what we earlier called ‘modus tollendo ponens’
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12 formula if you have a formula then you are entitled to infer–––––– disjunctionany its disjunction with any formula to its right formula if you have a formula then you are entitled to infer–––––– disjunctionany its disjunction with any formula to its left
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13 conditional if you have a conditional antecedent and you have its antecedent then you are entitled to infer –––––– consequent its consequent conditional if you have a conditional negationconsequent and you have the negation of its consequent then you are entitled to infer –––––– negationantecedent the negation of its antecedent What we earlier called ‘modus ponens’ and ‘modus tollens’
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14 THERE IS NO SUCH THING AS ARROW-IN ( I) what we have instead is CONDITIONAL DERIVATION (CD) which we examine later
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15 if you have a formula then you are entitled to infer––––– its double-negative if you have a double-negative then you are entitled to infer––––– the formula rules apply only to whole lines not pieces of lines
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16 : ° ° ° In Direct Derivation (DD), one directly arrives at DD The Original and Fundamental -Rule the very formula one is trying to show.
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17 (10) (9) (8) (7) (6) (5) (4) (3) (2) (1) 5,9, Q 4,8, P 3,7, T 1,2, R DD : Q Pr P Q Pr P T Pr R T Pr R S Pr SS S ; R S ; R T ; P T ; P Q / Q OO OO OO OO
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18 If you have a line of the form , arrow-out O then try to apply arrow-out ( O), which requires a second formula as input, in particular, either or . deduce find have or
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19 (10) (9) (8) (7) (6) (5) (4) (3) (2) (1) 5,9, Q 4,8, P 3,7, T 1,2, R DD : Q Pr P Q Pr P T Pr R T Pr R S Pr SS S ; R S ; R T ; P T ; P Q / Q OO OO OO OO
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20 If you have a line of the form , wedge-out O then try to apply wedge-out ( O), which requires a second formula as input, in particular, either or . deduce find have or
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21 (8) (7) (6) (5) (4) (3) (2) (1) 5,7, R 3,6, Q R 1,5, P Q 1,2, R DD : Q Pr (P Q) (Q R) Pr [(P Q) R] R Pr (P Q) R (P Q) R ; [(P Q) R] R ; (P Q) (Q R) / Q OO OO OO OO
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22 (8) (7) (6) (5) (4) (3) (2) (1) 2,7, T 6, R S 1,5, S 3, P Q DD : T PrP (R S) T Pr (P Q) S (P Q) S ; (R S) T) ; P / T OO II OO II
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23 If you need then look for either disjunct: find OR find then v I apply v I to get
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24 (8) (9) (10) (11) (12) (7) (6) (5) (4) (3) (2) (1) 1,7, Q 3,6, R 2,9, Q S 8,10, S 8,11, Q & S P 4,&O T DD Q & S Pr T & P Pr R T Pr R (Q S) Pr P Q P Q ; R (Q S) ; R T ; T & P / Q & S OO OO OO OO &I&I
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25 If you have a line of the form &&,&&, ampersand-out then apply ampersand-out (&O), which can be applied immediately to produce and
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26 If you are trying to find or show &&&& then look for both conjuncts: find AND find then & I apply & I & to get &
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