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Comment on Assignment 1-B

Comment on Assignment 1-B. To say that P1, P2, -----, Pn, and C is a theorem is to say the following: ( P1 Λ P2 Λ P3 Λ - - - - Λ Pn) -> C is a tautology And we can express a theorem as follows: P1, P2, -----, Pn => C

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Comment on Assignment 1-B

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  1. Comment on Assignment 1-B • To say that P1, P2, -----, Pn, and C is a theorem is to say the following: ( P1 Λ P2 Λ P3 Λ - - - - Λ Pn) -> C is a tautology And we can express a theorem as follows: P1, P2, -----, Pn => C • This only mean that whenever P1, P2, - - - , Pn are all True, then C must be True. Remember ---- it is possible that C is True and have one of the P’s to be False. • This also means that if C is False, then at least one of the P’s is false. • Therefore, just because P1, P2, -----,Pn => C, it does NOT mean that • ( P1 Λ P2 Λ P3 Λ - - - - Λ Pn) is the same or equal to C

  2. Comment on Assignment 1-B (cont.) • We know that: a -> b, a => b (the modus ponens theorem) • Thus [ (a->b) Λ a ] -> b must be a tautology • But [ (a->b) Λ a ] may Not Equal b a b a-> b (a->b) Λ a [(a->b)Λa] ->b b-> [(a->b) Λ a] T T T T T T T F F F T T F T T F T F F F T F T T tautology not tautology thus [(a->b) Λ a] ≠ b Caution: Do not use the Replacement Rule to substitute the Conclusion for the Premises of a theorem because they are not equal !! (see Replacement Rule 2)

  3. Forward/Backward Chaining approach(using only 2 symbols: R and O) P1 : open /\ receiving -> ~closed : (O/\R) -> O P2 : ~ open : ~O P3 : receiving -> closed : R -> ~O Concl : ~(receiving /\ open) : ~(R /\ O) Note: that Open is used for ~Closed and ~Open is used for Closed Legend: O - open R – receiving C – closed which same as ~O O R P1: O /\ R -> O P2: ~O P3: R -> ~O Concl: ~(R /\ O) (1) (2) (3) (4) T T T F F F F T F T T T T T F T T T F F T T T T Forward chaining : rows 3 and 4 where P1,P2,and P3 are all True and so is Conclusion Backward chaining: Concl is False in row 1 and at least one of the premises is also False

  4. Forward/Backward Chaining approach(using 3 symbols: R, O, C) P1 : open /\ receiving -> ~closed : O/\R -> ~C P2 : ~ open : ~O P3 : receiving -> closed : R -> C Concl : ~(receiving /\ open) : ~(R /\ O) O - open R – receiving C - closed O R C P1: O /\ R -> ~C P2: ~O P3: R -> C Conc: ~(R /\ O) (1) (2) (3) (4) (5) (6) (7) (8) T T T F F T F T T F T F F F T F T T F T T F T T F T F T F T T T T T T F T F T T F T T T T T F F T F F F T T T T Forward chaining : P1,P2,and P3 are all True in rows 5,7 and 8; so is Concl Backward chaining: Concl is False in rows 1 and 2; at least one of the premises is also False

  5. Manipulating Syntax (a la pages: 53-59 of text) • A Fairly “Simple Approach”: • (O/\R) -> ~C, ~O, R -> C ├ ~(R/\O) given problem • [ ((O/\R)-> ~C) /\ (~O) /\ (R ->C) ] -> ~(R/\O) inference, p55 • ~ [ ((O/\R)-> ~C) /\ (~O) /\ (R ->C) ] \/ ~(R/\O) implies law, p48 • ~ [ “ ] \/ ~R \/ ~O DeM law, p47 • [~ ((O/\R) -> ~C) \/ ~(~O) \/ ~(R->C) ] \/ ~R \/ ~O DeM law, p47 • ~ ((O/\R) -> ~C) \/ ~(~O) \/ ~(R->C) \/ ~R \/ ~O assoc law, p47 • ~((O/\R) -> ~C) \/ ~(R->C) \/ ~(~O) \/ ~R \/ ~O comm. law, p46 • ~ ((O/\R) -> ~C) \/ ~(R->C) \/ ~R \/ ~(~O) \/ ~O comm. law, p46 • [ ~((O/\R) -> ~C) \/ ~(R->C) \/ ~R ] \/ [~(~O) \/ ~O] assoc. law, p 47 • [ “ ] \/ [ O \/ ~O ] neg rule, p 47 • [ “ ] \/ TRUE exmid rule, p47 • TRUE or2 rule , p46

  6. Manipulating Syntax (a la pages: 53-59 of text) • A “ Lengthier” Approach: • (O/\R) -> ~C, ~O, R -> C ├ ~(R/\O) given problem • [ ((O/\R)-> ~C) /\ (~O) /\ (R ->C) ] -> ~(R/\O) inference, p55 • (~(O/\R) \/ O) /\ (~O) /\ (R->C) -> “ note : replace ~C with O • ( (~O\/~R) \/ O) /\ “ -> “ DeM rule • (~O \/ ~R \/ O) /\ “ -> “ assoc. rule • (~R \/ ~O \/ O) /\ “ -> “ comm rule • (~R \/ True ) /\ “ -> “ exmid • True /\ “ -> “ or 2 rule • True /\ (~O) /\ (R -> ~O) -> “ note : replace C with ~O • True /\ ~O /\ (R-> ~O) -> “ • ~O /\ (R-> ~O) -> “ and 2 rule • ~O /\ (~R \/ ~O) -> “ impl rule • (~O /\ ~R) \/ (~O /\ ~O) -> “ dist. Rule • (~O /\ ~R ) \/ (~O) -> “ and 1 rule • ~(O \/ R) \/ (~O) -> “ DeM rule • ~ [ (O\/R) /\ O ] -> “ DeM rule • ~[ (O/\O) \/ (R/\O) ] -> “ distr. Rule • ~ [ O \/ (R/\O) ] -> “ and 1 rule • ~ { ~ [O \/ (R/\O) ] } \/ ~(R /\ O) impl rule • [ O \/ (R/\O) ] \/ ~(R/\O) neg rule • O \/ (R/\O) \/ ~(R/\O) assoc rule • O \/ True exmid rule • True or 2 ~O /\ ~ (R /\ O) DeM rule ~ [ O v (R /\ O) ] DeM rule

  7. Solution borrowed from student, Carolyn (spring ’08) Let O= open ; R=receiving ; C= closed Problem: Show that: (OΛR -> ~C), ~O, (R->C)├ ~(RΛ O) Premise: (OΛR -> ~C), ~O, (R->C) are all True • Solution: • Show that { (OΛR -> ~C) Λ ~O Λ(R -> C) } -> ~(RΛ O) is a tautology • {(OΛR -> O) Λ ~O Λ (R -> ~O)}-> ~(RΛ O):substituting O = ~C • {[(OΛR -> O) Λ ~O] Λ (R -> ~O)}-> ~(RΛ O) : assoc. law • { TRUE Λ ~O Λ (R -> ~O)} -> ~(R ΛO) : a Λ b -> a is a tautology • finish as previous page (from step 8 of previous page and go down)

  8. Solution borrowed from student (Rick - Spring2012) Let O= open ; R=receiving ; C= closed Problem: Show that: (OΛR -> ~C), ~O, (R->C)├ ~(RΛ O) Premise: (OΛR -> ~C), ~O, (R->C) are all True • Start with the conclusion : ~(R Λ O) • via DeMorgan, convert to : ~ R V ~ O • We know ~ O is True given above: ~R V TRUE • (Anything V TRUE) : TRUE

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