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Proof Tactics, Strategies and Derived Rules CS 270 Math Foundations of CS Jeremy Johnson.

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Presentation on theme: "Proof Tactics, Strategies and Derived Rules CS 270 Math Foundations of CS Jeremy Johnson."— Presentation transcript:

1 Proof Tactics, Strategies and Derived Rules CS 270 Math Foundations of CS Jeremy Johnson

2 Outline 1.Review Rules 2.Positive subformulas and extraction 3.Proof tactics Extraction, Conversion, Inversion, Division, and Refutation Finding contradictions 4.Proof strategy Search tree and an algorithm to find a proof 5.Derived rules

3 Conjunction Rules Introduction Rule Elimination Rule    I     ER      EL 

4 Implication Rules Introduction Rule Assume  and show  Elimination Rule (Modus Ponens)      E   …   I   

5 Disjunction Rules Introduction Rule Elimination Rule (proof by case analysis)   IR     E    IL    …… ……

6 Negation Rules

7 Introduction and elimination rules Double negation     E 

8 Law of the Excluded Middle … 1 p  p

9 Law of the Excluded Middle 1  (p  p) assumption … 2 3 p  p

10 Law of the Excluded Middle 1  (p  p) assumption … 2 p  p 3 4

11 Law of the Excluded Middle 1  (p  p) assumption … 2pGoal 3 p  p 4 5

12 Law of the Excluded Middle 1  (p  p) assumption 2 … 3Goal 4p 5 p  p 6 7

13 Law of the Excluded Middle 1  (p  p) assumption 2 3 (p  p) 4 5p 6 p  p 7 8

14 Search Tree ? P   P  IR  IL EE ?  P ? P

15 Search Tree ? P   P  IR  IL EE ?  P ? P EE

16 Search Tree  IL ? P EE

17 Search Tree ? P   P  IR  IL EE ?  P ? P

18 Proof Tactics Systematically search for a proof Apply ( , ,  ) elimination rules forward Apply introduction rules backwards No extraneous steps Backtrack when dead-end reached 1.Extraction 2.Conversion 3.Inversion 4.Division 5.Refutation

19 Positive Subformulas PS(  ) If  is an atom return  If  =  return  If  =    then return   PS(  )  PS(  ) If  =    then return   PS(  )  PS(  ) If  =    then return   PS(  )

20 Extraction Apply elimination rules forward in order to extract goals that occur as positive subformulae of the formulae on available lines.

21 Conversion Use disjunction elimination in order to obtain goal disjunctions.

22 Inversion Invert non-atomic goals by applying introduction rules backward to them.

23 Division Use disjunction elimination on any goals for which the previous three tactics have either not applied, or not been successful.

24 Refutation Apply negation elimination backward to goals that cannot be obtained by any other means.

25 Possible Contradictions Form a list of all negations that appear as a positive subformulas of all premises and available assumptions. Pair each negation  with its immediate subformula . These pairs are the only possible contradictions that must be considered.

26 Exercise Prove the definition of conditional (    )         (    )

27 Deadend

28 Solution

29

30 Algorithm

31 Using Derived Rules Once you have proven a rule from the basic rules you may use it in your proofs Derive M from (M  O)  M 1 (M  O)  M premise 2 MM assumption 3  M  O 4 M  ODf  I 5M  E1,4 6 7M

32 Derived Rules Commutative rules               Associative rules (    )    (    )   (    )    (    )   Idempotence rules      and           and     

33 Derived Rules Distributive rules   (    )  (    )  (    ) (    )  (    )    (    )   (    )  (    )  (    ) (    )  (    )    (    ) Disjunctive syllogism (    ),    Cut (resolution) (    ), (    )  (    )

34 Derived Rules DeMorgan’s rules  (    )          (    )  (    )          (    )

35 Derived Rules Modus Tollens (   ,  )   Transposition (    )  (    ) Hypothetical Syllogism (   ,    )  (    ) Exportation and Importation ((    )   )  (   (    )) (   (    ))  ((    )   )

36 Derived Rules Definition of conditional (    )         (    ) Negated conditional  (    )          (    )

37 Exercise Prove the definition of the conditional using Disjunctive Syllogism and LEM

38 Solution 1  P  Q premise 2Passumption 3QDSL 1,2 4 P  Q  I3 1 P  Q premise 2 P  PP  P LEM 3Passumption 4Q  E1,3 5  P  Q  IL4 6 PP assumption 7  P  Q  IR6 8  P  Q  E2,5,7


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