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Predicate Calculus Formal Methods in Verification of Computer Systems Jeremy Johnson

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Outline 1.Motivation 1.Variables, quantifiers and predicates 2.Syntax 1.Terms and formulas 2.Quantifiers, scope and substitution 3.Rules of natural deduction for quantifiers 4.Semantics 1.Models and semantic entailment 5.Undecidability and limitations

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Example 1 Every student is younger than some instructor x ( S(x) y(I(y) Y(x,y) ) S(x) : x is a student I(x) : is an instructor Y(x,y) : x is younger than y

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Example 2 Not all birds can fly x ( B(x) F(x) ) x ( (B(x) F(x) ) B(x) : x is a bird F(x) : x can fly Semantically equivalent formulas

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Example 3 Every child is younger than its mother x y ( C(x) M(y,x) Y(x,y) ) C(x) : x is child M(x,y) : x is y’s mother Y(x,y) : x is younger than y x ( C(x) Y(x,m(x)) m(x) : function for mother of x

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Example 4 Andy and Paul have the same maternal grandmother x y u v ( M(x,y) M(y,a) M(u,v) M(v,p) x = u ) m(m(a)) = m(m(p)) a, b : variables for Andy and Paul = : binary predicate

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Example 5 Everyone has a mother x y ( M(y,x) ) x y ( M(y,x) ) [ not equivalent ] Everyone has exactly one mother x y ( M(y,x) z (M(z,x) z = y )

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Example 6 Some people have more than one brother x y1 y2 ( B(y1,x) B(y2,x) (y1 = y2) )

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Comparison to Propositional Calculus

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Terms Terms are made up of variables, constants, and functions Term ::= Variable If c is a nullary function c is a term If t 1,…,t n are terms and f is an n-ary function then f(t 1,…,t n ) is a term

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Formulas Formula ::= P is a predicate and t 1,…,t n are terms then P(t 1,…,t n ) is a formula If is a formula is a formula If 1 and 2 are formulas, 1 2, 1 2, 1 2 are formulas If is a formula and x is a variable x and x are formulas

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Parse Trees x ( ( P(x) Q(x) ) S(x,y) ) xx S xy P Q x x

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Free and Bound Variables An occurrence of x in is free if it is a leaf node in the parse tree for with no quantifier as an ancestor xx S xy P Q x x xx P Q x x P x Q y

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Substitution Given a variable x, a term t and a formula , [t/x] is the formula obtained by replacing each free occurrence of x by t xx P Q x x P x Q y xx P Q x x P f Q y x y [f(x,y)/x]

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Variable Capture t is free for x in if no free x occurs in in the scope of any quantifier for any variable y occurring in t. yy S x P Q x y

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Variable Capture t is free for x in if no free x occurs in in the scope of any quantifier for any variable y occurring in t. yy S x P Q y f y y

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Equality Rules Introduction Rule Elimination Rule = i t = t t 1 = t 2 [t 1 /x] =e [t 2 /x]

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Equivalence Relation 1premise 2=i 3=e 1,2 1premise 2 3=e 2,1

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Conjunction Rules Introduction Rule Elimination Rule i e1 e2

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Universal Quantification Rules Introduction Rule Elimination Rule x i x x e [t/x] x 0 … [x 0 /x]

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Illegal Substitution Leads to False Reasoning x = y (x < y) [y/x] = y (y < y) y is not free for x in

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Example Proof 1premise 2 3 x 0 P(x 0 ) Q(x 0 ) 4P(x 0 ) 5Q(x 0 ) e3,4 6

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Disjunction Rules Introduction Rule Elimination Rule (proof by case analysis) i1 e i2 …… ……

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Existential Quantification Rules Introduction Rule Elimination Rule (proof by case analysis) [t/x] x i x e x 0 [x 0 /x] …

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Example Proof 1premise 2 3 x 0 P(x 0 ) Q(x 0 ) assumption 4 5Q(x 0 ) e 2 3 6 7 P(x 0 ) e 1 3 8 P(x 0 ) R(x 0 ) i7,6 9 10

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Quantifier Equivalences

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De Morgan’s Law 1premise 2assumption 3 4 i 1 3 5 e4,2 6PBC 3-5 7assumption 8 i 2 7 9 e4,2 10PBC 7-9 11 12 e 11,1 13PBC 2-12

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Generalized De Morgan’s Law 1 x P(x) premise 2assumption 3x0x0 4 5 6 e 5,2 7PBC 4-6 8 9 e 8,1 10 x P(x) PBC 2-9

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Generalized De Morgan’s Law 1 x premise 2assumption 3x0x0 4 5 6 e 5,2 7PBC 4-6 8 9 e 8,1 10 x PBC 2-9

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Exercise

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Models Let F be a set of functions and P a set of predicates. A model M for (F,P) consists of A non-empty set A [universe] of concrete values For each nullary f F an element of A = f M For each n-ary f F a function f M : A n A For each n-ary P P a subset P M A n

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Example 1 F = {i} and P = {R,F} i a constant function, R binary and F unary predicates Model – A set of states, initial state i, state transitions R, final states F A = {a,b,c} i M = a R M = {(a,a),(a,b),(a,c),(b,c), (c,c)} F M = {b,c}

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Example 1 y R(i,y) F(i) x y z (R(x,y) R(x,z) y = z ) x y R(x,y)

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Example 2 F = {e, } and P = { } e a constant function, a binary function, a binary predicate1 Model – A set of states, A = {binary strings} e M = , M concatenation, M prefix ordering [011 is a prefix of 011001

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Example 2 x ((x x e) x e x)) y x (y x) x y (y x) x y z ((x y) (x z y z)) x y ((x y) (y x))

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Satisfaction

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Semantic Entailment

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Soundness and Completeness

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Post Correspondence Given a finite sequence (s 1,t 1 ),…,(s k,t k ) of pairs of binary strings. Is there a sequence of indices i 1,i 2,…,i n such that s i 1 s i n = t i 1 t i n Example s 1 = 1, s 2 = 10, s 3 = 011 t 1 = 101, t 2 = 00, t 3 = 11 Solution (1,3,2,3) 101110011

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Undecidability

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Consequences of Undecidability

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Proof

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Reachabilty When modeling systems via states and state transitions, we want to show that a “bad” state can not be reached from a “good” state. Given nodes n and n’ in a directed graph, is there a finite path of transitions from n to n’. s0 s1 s3 s2 A = {s0,s1,s2,s3} R M = {(s0,s1), (s1,s0), (s1,s1),(s1,s2), (s2,s0),(s3,s0),(s3,s2)}

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Compactness Theorem Let be a set of sentences of predicate calculus. If all finite subsets of are satisfiable, then so is . Proof – uses soundness and completeness and finite length of proofs.

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Reachability is Not Expressible Can reachability be expressed in predicate calculus? u=v x (R(u,x) R(x,v)) x 1 x 2 (R(u,x 1 ) R(x 1,x 2 ) R(x 2,v)) … This is infinite The answer is no! Proof follows from compactness theorem.

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Reachability is Not Expressible Theorem. There is no predicate-logic formula with u and v as its only free variables and R its only predicate such that holds in directed graphs iff there is a path from u to v. Proof. By contradiction. Suppose there is such a formula. Let n be the formula expressing that there is a path from c to c’ n = x 1 … x n-1 (R(c,x 1 ) … R(x n-1,c)).

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Reachability is Not Expressible Proof. By contradiction. Suppose there is such a formula . Let n be the formula expressing that there is a path from c to c’ n = x 1 … x n-1 (R(c,x 1 ) … R(x n-1,c)). = { i | I 0} [c/u][c’/v] is unsatisfiable, but any finite subset is satisfiable. By compactness this leads to a contradiction and hence there is no such .

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Reachability via HOL

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Obtain formula for the existence of a path from u to v by negating previous formula (use DeMorgan’s law) P x y z ( C 1 C 2 C 3 C 4 ) If both and can range over predicates then second order logic.

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