Assignment 1 Solutions Structural Induction and First-Order Logic Due: 11am on Monday 26th August 2013

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1 Deartment of Comuter Science, Australian National University COMP2600 Formal Methods in Software Engineering Semester 2, 2013 Assignment 1 Solutions Structural Induction and First-Order Logic Due: 11am on Monday 26th August 2013 The submission of your assignment must be done via the assignment boxes in the student foyer Failure to submit by the due date will result in late enalties of ten ercent er weekday Other arrangements that are reuired because of truly excetional circumstances need to be negotiated before your deadline Your submission must be well-resented on clean A4 aer with a fully comleted standard cover age, including your tutor s name and your tutorial grou Failure to follow this instruction will result in a enalty The COMP2600 Assignments age has a link to an aroriate cover age 1 Structural Induction [8 marks] Exercise 11 counttrue [] = 0 -- (C1) counttrue (x:xs) ( x == True ) = 1 + counttrue xs -- ( C2) otherwise = counttrue xs -- ( C3) easycounttrue xs = length ( filter (== True ) xs) -- ( EC1 ) length [] = 0 -- (L1) length (x:xs) = 1 + length xs -- (L2) filter [] = [] -- (F1) filter ( x : xs) x = x : filter xs -- ( F2) otherwise = filter xs -- ( F3) Prove by structural induction the following roerty about the functions easycounttrue and counttrue: easycounttrue xs = counttrue xs State clearly what roerty P is being roved by induction, including any uantifiers needed in the statement of P and in the inductive hyothesis COMP2600 Semester , Assignment 1 Solutions 1

2 We will rove P (xs) : easycounttrue xs = counttrue xs by structural induction on xs Base case: xs = [] Show that easycounttrue [] = counttrue [] Proof: easycounttrue [] = length (filter (== True) []) -- by (EC1) = length [] -- by (F1) = 0 -- by (L1) = counttrue [] -- by (C1) Ste case: Show that if xs = a:as easycounttrue as = counttrue as -- (IH1) then easycounttrue (a:as) = counttrue (a:as) We rove by cases on a Proof (case when a == True): easycounttrue (a:as) = length (filter (== True) a:as) -- by (EC1) = length (a:(filter (== True) as)) -- by (F2) = 1 + length (filter (== True) as) -- by (L2) = 1 + easycounttrue as -- by (EC1) = 1 + counttrue as -- by (IH1) = counttrue (a:as) -- by (C2) Proof (case when a == False): easycounttrue (a:as) = length (filter (== True) a:as) -- by (EC1) = length (filter (== True) as) -- by (F3) = easycounttrue as -- by (EC1) = counttrue as -- by (IH1) = counttrue (a:as) -- by (C3) Having shown P (as) P (a : as), we can generalise to x xs(p (xs) P (x : xs)) Having shown P ([]) and x xs(p (xs) P (x : xs)), we have that xsp (xs) COMP2600 Semester , Assignment 1 Solutions 2

3 Exercise 12 count Nul = 0 -- (C1) count ( Node a t1 t2) = 1 + count t1 + count t2 -- ( C2) counta t = countb t 0 -- ( CA) countb Nul acc = acc -- ( CB1 ) countb ( Node a t1 t2) acc = countb t1 (1 + ( countb t2 acc )) -- ( CB2 ) Here is the usual Haskell definition of a binary tree: data Tree a = Nul Node a ( Tree a) ( Tree a) Prove by structural induction the following roerty about the functions count and counta: count t = counta t State clearly what roerty P is being roved by induction, including any uantifiers needed in the statement of P and in the inductive hyothesis We will show a stronger roerty P first P is t acc (count t) + acc = countb t acc Do this by induction on t Base case: Show acc t = Nul (count Nul) + acc = countb Nul acc Proof: (count Nul) + acc = 0 + acc -- by (C1) = acc -- by (Arith) = countb Nul acc -- by (CB1) Ste case: t = (Node a t1 t2) Assume acc (count t1) + acc = countb t1 acc -- (IH1) acc (count t2) + acc = countb t2 acc -- (IH2) Prove acc (count (Node a t1 t2)) + acc = countb (Node a t1 t2) acc COMP2600 Semester , Assignment 1 Solutions 3

4 Proof: countb (Node a t1 t2) acc = countb t1 (1 + (countb t2 acc)) -- by (CB2) = (count t1) (countb t2 acc) -- by (IH1) * = (count t1) (count t2) + acc -- by (IH2) = 1 + count t1 + count t2 + acc -- by Associativity and Commutativity of + = (count (Node a t1 t2)) + acc -- by C2 (*) Note, acc in IH1 is instantiated to 1 + (countb t2 acc) when it is used in the roof Having shown that P (N ul) and P (t1) P (t2) P(Node a t1 t2), we have t acc(count t) + acc = countb t acc Secifically, for acc = 0, we have (count t) + 0 = countb t 0 -- P1 Then count t = (count t) by Arithmetic = countb t 0 -- by (P1) = counta t -- by (CA) 2 FOL Secification [4 marks] Exercise 21 Use the redicates S(x) - x is succesful, F (x) - x fails, C - the transaction is committed, to translate the following sentences into first-order logic: i The transaction is committed unless some oeration fails ii The transaction is committed if every oeration succeeds i ( of (o)) C ii ( os(o)) C Exercise 22 Use the redicates S(x) - x is succesful, F (x) - x fails, C(x) - x is committed, O(x, y) - y belongs to x, to translate the following sentences into first-order logic: i Any transaction is committed unless some oeration belonging to it fails ii Any transaction is committed if all oerations belonging to it succeed Comare the statements given here to those in 21 That is, comare (21i) to (22i) and (21ii) to (22ii) Which version is more likely to be useful when describing a realistic system, and why? COMP2600 Semester , Assignment 1 Solutions 4

5 i t( oo(t, o) F (o)) C(t) ii t( oo(t, o) S(o)) C(t) The statements given in this exercise uantify over all transactions in the system, not just one transaction, and relate their success to the sucess of the individual oerations belonging to the articular transaction Thus, these statements are more exressive and allow to describe a more realistic system than those in 21 3 Truth Tables [1 mark] Exercise 31 Use truth-tables to rove or disrove whether the roosition ((a b) (b c)) (a c) is a tautology The roosition is a tautology a b c a b b c (a b) (b c) a c result T T T T T T T T T T F T F F F T T F T F T F T T T F F F T F F T F T T T T T T T F T F T F F T T F F T T T T T T F F F T T T T T 4 Natural Deduction [7 marks] Exercise 41 Prove the following derived rule using natural deduction: ( ) You may only use the rules given in the Aendix Do not use algebraic laws, or any of the derived rules obtained in lectures 1 ( ) 2 3 -I, 2 4 ( ) ( ) -I, 1, 3 5 -E, 2 4 COMP2600 Semester , Assignment 1 Solutions 5

6 Exercise 42 Prove the following derived rule using natural deduction: x (P (x) Q(x)) x P (x) y Q(y) In addition to the rules given in the Aendix, you may use the following rule: (contradiction) 1 x (P (x) Q(x)) 2 y Q(y) 3 a P (a) Q(a) 4 Q(a) -E, 2 5 P (a) 6 x P (x) -I, 5 7 Q(a) 8 Q(a) Q(a) -I, 4, 7 9 x P (x) contradiction, 8 10 x P (x) -E, 3, 5 6, x P (x) -E, 1, 3 10 Aendix 1 Truth Table Values T T T T T F T T F T F F F F F T T F T T F F F F F T T T COMP2600 Semester , Assignment 1 Solutions 6

7 Aendix 2 Natural Deduction Rules ( I) ( E) [] [] ( I) ( E) r r r [] ( I) ( E) [] [ ] ( I) ( E) ( I) P (a) (a arbitrary) x P (x) ( E) x P (x) P (a) ( I) P (a) xp (x) [P (a)] ( E) xp (x) (a is not free in ) (a arbitrary) COMP2600 Semester , Assignment 1 Solutions 7