Midterm II Sample Questions

Transcription

1 Midterm II Sample Questions Constructive Logic Frank Penning November 7, 2017 Name: Sample Solution Andrew ID: p Instructions This exam is closed book, closed notes. You have 80 minutes to complete the exam. There are only 4 problems! Admissibility Logic Deriving o Cut Prolog Programming Rules Prob 1 Prob 2 Prob 3 Prob 4 Prob 5 Total Score Max

2 1 Admissibility o Cut (30 pts) In unctional languages, we have parametric polymorphism and abstract types, which corresponds to universal and existential quantiication over types, respectively. The corresponding constructs in logic are universal and existential quantiication over propositions. In this problem we consider the existential quantiier, given by the ollowing rules in the sequent calculus. Γ = [B/x]A Γ = x:prop. A R Γ, x:prop. A, [p/x]a = C Γ, x:prop. A = C L p In the R rule, B must be a proposition. In the L rule, p must be a resh propositional variable. The substitution principle or propositions says that we can always substitute a proposition or a propositional variable throughout a proo; you will need that below. We state the admissibility o cut as D Γ = A E Γ, A = C Γ = C cut A Task 1 (20 pts). Show what would be the critical case in the proo o the admissibility o cut, when the cut ormula is inerred by R in D and by L in E. You should only be concerned about eliminating the cut at proposition x:prop. A; do not be concerned about whether you are actually justiied in applying the induction hypothesis (see Task 3). Case: D = D E Γ = [B/x]A Γ = x:prop. A R Γ, x:prop. A, [p/x]a = C and E = Γ, x:prop. A = C F Γ, [p/x]a = C by cut on x:prop. A, D, and E [B/p]F Γ, [B/x]A = C by substitution property or propositions and [B/p][p/x]A = [B/x]A F Γ = C by cut on [B/x]A, D, and [B/p]F L p Task 2 (5 pts). The admissibility o cut is shown by a nested induction, which can also be ormulated as a well-ounded induction over a lexicographic order. Complete the ollowing paragraph, using Γ, A, C, D, and E rom the statement o the admissibility o cut. The proo is by nested induction, irst over A, then over D and E. This means in an appeal to the induction hypothesis, either A becomes smaller and D and E are arbitrary, or A remains the same and one o D and E becomes smaller while the other remains the same. 2

3 Task 3 (5 pts). When adding quantiiers over propositions, this particular induction argument ails. Explain why by completing the ollowing paragraph. The case or R against L would require us to appeal to the induction hypothesis on [B/x]A, D, and [B/p]F. However, this is not smaller than x:prop. A, D, and E because [B/x]A may be bigger than x:prop. A. 3

4 2 Prolog (30 pts) One issue with lists in unctional programs is that we can access them only at one end. In logic programming, so-called dierence lists allows us to access both ends! Recall that Prolog notation or a list is [] or the empty list, and [X Xs] or the list with head X and tail Xs. We abbreviate [X1 [X2 [X3... Xs]]] as [X1,X2,X3,... Xs]. A dierence list with elements x 1,..., x n has the orm di([x1,...,xn L],L) where L is a logic variable, subject to uniication. Note that it must be the same logic variable L at the tail end o the list, and as the second argument to the constructor di, but that dierent dierence lists may have dierent variables (ha!). Complete the ollowing programs, always assuming that the given dierence lists are in the orm shown above. Task 1 (5 pts). I to_list(di(k,l), Xs) and di(k,l) represents [x1,...,xn] then the ordinary list Xs is [x1,...,xn]. to_list(di(k,l), K) :- L = []. % or % to_list(di(k,[]), K). Task 2 (5 pts). I prepend(y, di(k,l), D) and di(k,l) represents [x1,...,xn], then the dierence list D represents [Y,x1,...,xn]. prepend(y, di(k,l), di([y K], L)). Task 3 (10 pts). I postpend(di(k,l), Y, D) and di(k,l) represents [x1,...,xn], then the dierence list D represents [x1,...,xn,y]. postpend(di(k,l), Y, di(k,m)) :- L = [Y M]. % or % postpend(di(k,[y M]), Y, di(k,m)). 4

5 Task 4 (10 pts). I concat(di(k,l), di(m,n), D) and di(k,l) represents [x1,...,xi] and di(m,n) represents [y1,...,yj] then dierence list D represents [x1,...,xi,y1,...,yj]. concat(di(k,l), di(l,m), di(k,m)). 5

6 3 Logic Programming (30 pts) We explore some programming over heap-ordered binary trees. A binary tree is heap-ordered i the key at every node is less than or equal to the key at its children. We assume we have predicate N <= K which succeeds i N K. Both N and K must be ground, that is, ree o variables subject to uniication. We represent binary trees with lea and node constructors. tree(lea). tree(node(let, N, Right)) :- tree(let), nat(n), tree(right). Task 1 (15 pts). Deine a backward chaining predicate ordered(tree) in Prolog that tests i a given tree is heap-ordered. It should have mode ordered(+tree) and always terminate in this mode. You may deine auxiliary predicates as needed and use N <= K. ordered(lea). ordered(node(let, N, Right)) :- below(n, Let), below(n, Right). below(n, lea). below(n, node(let, K, Right)) :- N <= K, ordered(node(let, K, Right)). 6

7 Task 2 (15 pts). Now deine a orward chaining predicate ord(t) that tests i a given tree is heapordered. The assertion ord(t) or a given (ground) tree T should derive no i the assumption that T is heap-ordered is inconsistent. You may generate additional assertions leq(n,k), assuming that this will prove no i N is not less than or equal to K. Your program must saturate. You may give your answer in the orm o inerence rules (read rom the premises to the conclusion), propositions with all positive atoms, or in Prolog syntax (but keeping in mind the clauses will be applied with orward chaining). ord(node(let, N, Right)) below(n, Let) below(n, node(let, K, Right)) leq(n, K) ord(node(let, N, Right)) below(n, Right) below(n, node(let, K, Right)) ord(node(let, K, Right)) 7

8 4 Deriving Inerence Rules (30 pts) We can use orward chaining in order to derive big-step inerence rules rom propositions. Recall a ew characteristic rules or orward chaining, where all atoms are positive. Γ, [D(X)] Γ, [ x.d(x) ] L Γ, P + Γ, [ P + ] L Γ, P + [P + ] id + Γ, [D 1 ] Γ, [D 1 D 2 ] L 1 Γ, [D 2 ] Γ, [D 1 D 2 ] L 2 Γ [G + 1 ] Γ, [D 2 ] Γ, [G + 1 D 2 ] L where X is a reshly chosen variable subject to uniication in L. Task 1 (5 pts). When is an inerence rule with judgment K in the premise and judgment J in the conclusion derivable? Circle all that apply. Correct: (i), (iii) (i) There is a hypothetical proo o the conclusion J rom hypothesis K. (ii) I we add this rule to the system we do not change the provable judgments. (iii) I we substitute a proo or K in the premise we obtain a proo or J. (iv) There is a dierentiable unction rom K to J. (v) I we add the rule the system remains in harmony. Now consider the negative propositions numbits(e, z), n. k. numbits(n, k) ( numbits(b0(n), s(k)) numbits(b1(n), s(k))) where all atomic propositions numbits(, ) are positive. Using these two propositions as antecedents we can deduce numbits(n, k) i the number o bits b0 and b1 in the binary number n is the unary number k. Task 2 (5 pts). Derive all inerence rules that arise rom let ocusing on numbits(e, z). Γ, numbits(e, z) Γ, [ numbits(e, z)] L Γ, numbits(e, z) G+ Γ R 0 8

9 Task 3 (15 pts). Derive all inerence rules that arise rom let ocusing on n. k. numbits(n, k) ( numbits(b0(n), s(k)) numbits(b1(n), s(k))) Γ, numbits(b0(n), s(k)) L numbits(n, K) Γ Γ, [ numbits(b0(n), s(k))] id + L 1 Γ [numbits(n, K)] Γ, [( numbits(b0(n), s(k)) numbits(b1(n), s(k)))] L Γ, [numbits(n, K) ( numbits(b0(n), s(k)) numbits(b1(n), s(k)))] L 2 Γ, [ n. k. numbits(n, k) ( numbits(b0(n), s(k)) numbits(b1(n), s(k)))] and, symmetrically, using L 2 instead o L 1 : Γ, numbits(n, K), numbits(b0(n), s(k)) Γ, numbits(n, K) R 1 Γ, numbits(n, K), numbits(b1(n), s(k)) Γ, numbits(n, K) R 2 Task 4 (5 pts). Use your rules to prove that numbits(b0(b1(e)), s(s(z))), that is, the number o bits in the binary number 2 is 2. Deine G 0 = numbits(b0(b1(e)), s(s(z))) numbits(e, z), numbits(b1(e), s(z)), numbits(b0(b1(e)), s(s(z))) G 0 id numbits(e, z), numbits(b1(e), s(z)) G 0 R 1 numbits(e, z) G 0 R 2 G 0 R 0 To just use derived rules, we should also derive a rule by ocusing on G 0 on the right. This new rule would just replace the use o id in the last step. Γ, numbits(b0(b1(e)), s(s(z))) G 0 R 3 9