Predicate Calculus CS 270 Math Foundations of Computer Science Jeremy Johnson Presentation uses material from Huth and Ryan, Logic in Computer Science: Modelling and Reasoning about Systems, 2nd Edition
Outline 1. Motivation (program specification & derivation) 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
Hamiltonian Path G = (V,E) is an undirected graph with nodes V = {0,1,,n-1} and edges E V V 1 0 2 V = {0,1,2} E = {(0,1),(1,2)} edge(0,1), edge(1,0), edge(1,2), edge(2,1)
Hamiltonian Path A Hamiltonian path is a sequence of n nodes, where n = V where each node is visited exactly once (i.e. a permutation of the nodes) following edges 1 0 1 2 or 2 1 0 0 2
Hamiltonian Path path(i,j) is true when the jth node in the path is in the ith location Hamiltonian paths path(0,0), path(1,1), path(2,2) path(0,2), path(1,1), path(2,0) 1 0 2 Not Hamiltonian paths path(0,0), path(1,2), path(2,1) path(0,0), path(0,1), path(1,1), path(2,2) path(0,0), path(1,1), path(1,0), path(2,2)
Hamiltonian Path Constraints Every node occurs exactly once in the path j i path(i,j) j i k path(i,j) path(k,j) i = k Every location has exactly one node i j path(i,j) i j k path(i,j) path(i,k) j = k Adjacent nodes must be connected by an edge i j k (i<n-1) (path(i,j) path(i+1,k) edge(j,k))
Simplification Every node occurs exactly once in the path j i k path(i,j) path(k,j) i = k j i k (i k (path(i,j) path(k,j)) j i k (i k) path(i,j) path(k,j) j i k (i < k) path(i,j) path(k,j)
Simplification Adjacent nodes must be connected by an edge i j k (i<n-1) ((path(i,j) path(i+1,k)) edge(j,k)) i j k (i<n-1) ( edge(j,k) (path(i,j) path(i+1,k))) i j k (i<n-1) ( edge(j,k) ( path(i,j) path(i+1,k))) i j k ((i<n-1) & edge(j,k)) ( path(i,j) path(i+1,k)))
Conversion to SAT j i path(i,j) j=0..n-1 i=0..n-1 P ij j i k (i < k) path(i,j) path(k,j) j=0..n-1 i=0..n-1 k=i+1..n-1 P ij P kj i j k ((i<n-1) & edge(j,k)) path(i,j) path(i+1,k)) i=0..n-2 (j,k) E P ij P i+1,k
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
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
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
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
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 )
Example 6 Some people have more than one brother x y1 y2 ( B(y1,x) B(y2,x) (y1 = y2) )
Comparison to Propositional Calculus Proof calculus for predicate calculus φ 1,,φ n ψ [ extend natural deduction ] Provide semantics for predicate calculus φ 1,,φ n ψ [ models needed to evaluate functions and predicates may not be finite ] Soundness and Completeness φ 1,,φ n ψ holds iff φ 1,,φ n ψ is valid Undecidable No program exists that can always determine if φ holds
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
Formulas Formula ::= P is a predicate and t 1,,t n are terms then P(t 1,,t n ) is a formula If ϕ is a formula ϕ is a formula If ϕ 1 and ϕ 2 are formulas, ϕ 1 ϕ 2, ϕ 1 ϕ 2, ϕ 1 ϕ 2 are formulas If ϕ is a formula and x is a variable x ϕ and x ϕ are formulas
Parse Trees x ( ( P(x) Q(x) ) S(x,y) ) x S P Q x y x x
Free and Bound Variables An occurrence of x in ϕ is free if it is a leaf node in the parse tree for ϕ with no quantifier as an ancestor x x S Q P Q x y P Q P y x x x x x
Substitution Given a variable x, a term t and a formula ϕ, ϕ[t/x] is the formula obtained by replacing each free occurrence of x by t x ϕ[f(x,y)/x] x Q Q P Q P y P Q P y x x x x x f x y
Variable Capture t is free for x in ϕ if no free x occurs in ϕ in the scope of any quantifier for any variable y occurring in t. S x P y Q x y
Variable Capture t is free for x in ϕ if no free x occurs in ϕ in the scope of any quantifier for any variable y occurring in t. S x P y Q f y y y
Equality Rules Introduction Rule t = t = i Elimination Rule t 1 = t 2 φ[t 1 /x] φ[t 2 /x] =e
Equivalence Relation Symmetry: tt 1 = tt 2 tt 2 = tt 1 1 tt 1 = tt 2 premise 2 tt 1 = tt 1 =i 3 tt 2 = tt 1 =e 1,2 Transitivity: tt 1 = tt 2, tt 2 = tt 3 tt 1 = tt 3 1 tt 1 = tt 2 premise 2 tt 2 = tt 3 premise 3 tt 1 = tt 3 =e 2,1
Conjunction Rules Introduction Rule φ ψ φ ψ i Elimination Rule φ ψ φ e1 φ ψ ψ e2
Universal Quantification Rules Introduction Rule x 0 φ[x 0 /x] x φ x i Elimination Rule x φ x e φ[t/x]
Illegal Substitution Leads to False Reasoning x ϕ ϕ = y (x < y) ϕ[y/x] = y (y < y) y is not free for x in ϕ
Example Proof xx P xx Q xx, xxp(xx) xxq(xx) 1 xx P xx Q xx, premise 2 xxxx(xx) premise 3 x 0 P(x 0 ) Q(x 0 ) xx e1 4 P(x 0 ) xx e2 5 Q(x 0 ) e3,4 6 xx Q(xx) xx iii 5
Disjunction Rules Introduction Rule φ φ ψ i1 ψ φ ψ i2 Elimination Rule (proof by case analysis) φ ψ φ χ χ ψ χ e
Existential Quantification Rules Introduction Rule φ[t/x] x φ x i Elimination Rule (proof by case analysis) x φ x 0 χ φ[x 0 /x] χ e
Example Proof xx Q xx R xx, xx(p xx Q(xx)) xx(p xx R(xx)) 1 xx Q xx R xx, premise 2 xx(p xx Q(xx)) premise 3 x 0 P(x 0 ) Q(x 0 ) assumption 4 Q xx 0 R xx 0 xx e1 5 Q(x 0 ) e 2 3 6 R xx 0 ee 4,5 7 P(x 0 ) e 1 3 8 P(x 0 ) R(x 0 ) i7,6 9 xx(p xx R(xx)) xxi 8 10 xx(p xx R(xx)) xxe 2,3-9
Quantifier Equivalences 1. x ϕ x ϕ 2. x ϕ x ϕ 3. x ϕ ψ x (ϕ ψ) [x not free in ψ] 4. x ϕ ψ x (ϕ ψ) 5. x ϕ ψ x (ϕ ψ) 6. x ϕ ψ x (ϕ ψ) 7. x (ψ ϕ) ψ x ϕ 8. x (ϕ ψ) x ϕ ψ 9. x (ϕ ψ) x ϕ ψ 10. x (ψ ϕ) ψ x ϕ
Quantifier Equivalences 1. x ϕ x ψ x (ϕ ψ) 2. x ϕ x ψ x (ϕ ψ) 3. x y ϕ y x ϕ 4. x y ϕ y x ϕ
De Morgan s Law (p 1 pp 2 ) ( p 1 pp 2 ) 1 (p 1 pp 2 ) premise 2 ( p 1 pp 2 ) assumption 3 p 1 assumption 4 p 1 pp 2 i 1 3 5 e4,2 6 p 1 e 3-5 7 p 2 assumption 8 p 1 pp 2 i 2 7 9 i 4,2 10 p 2 e 7-9 11 (p 1 pp 2 ) i 6,10 12 i 11,1 13 ( p 1 pp 2 ) e 2-12
Generalized De Morgan s Law x P(x) x P(x) 1 x P(x) premise 2 x P(x) assumption 3 x 0 4 P(x 0 ) assumption 5 x P(x) x i 4 6 i 5,2 7 P(x 0 ) e 4-6 8 x P(x) x i 3-7 9 i 8,1 10 x P(x) e 2-9
Generalized De Morgan s Law x ϕ x ϕ 1 xϕ premise 2 x
Exercise Prove the reverse x ϕ x ϕ
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
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}
Example 1 y R(i,y) F(i) x y z (R(x,y) R(x,z) y = z ) x y R(x,y)
Example 2 F = {e, } and P = { } e a constant function, a binary function, a binary predicate Model string prefix A = {binary strings} e M = ε, M concatenation, M prefix ordering [011 is a prefix of 011001
Example 2 x ((x x e) x e x)) x y (y x) y x (y x) x y z ((x y) (y z) (x z)) x y z ((x y) (x z y z))
Satisfaction Given a model M for (F,P) and given an environment l : var A the satisfaction relation M l ϕ P(t 1,,t n ) (a 1,,a n ) and M l ϕ iff (a 1,,a n ) R M M l x ψ iff M l [x a]ψ holds for all a A M l x ψ iff M l [x a]ψ holds for some a A
Satisfaction Given a model M for (F,P) and given an environment l : var A the satisfaction relation M l ϕ M l ψ iff M l ψ does not hold M l ψ 1 ψ 2 iff M l ψ 1 and M l ψ 2 holds M l ψ 1 ψ 2 iff M l ψ 1 or M l ψ 2 holds M l ψ 1 ψ 2 iff M l ψ 2 holds whenever M l ψ 1 holds
Semantic Entailment For propositional calculus: If for all valuations (assignments of variables to truth values) for which all φ 1,,φ n evaluate to true, ψ also evaluates to true then the semantic entailment relation φ 1,,φ n ψ holds Decidable using truth tables
Semantic Entailment Let Γ be a set of formulas (possibly infinite) and ψ be a formula from predicate calculus Γ ψ holds iff for all models M and lookup tables l, whenever M l ϕ holds for all ϕ Γ then M l ψ holds as well ψ is satisfiable iff there is some model M and lookup table l such that M l ψ holds ψ is valid iff M l ψ holds for all models M and lookup tables l
Soundness and Completeness φ 1,,φ n ψ holds iff φ 1,,φ n ψ is valid In particular, ψ, a tautology, ψ is valid. I.E. ψ is a tautology iff ψ is provable Soundness you can not prove things that are not true in the truth table sense Completeness you can prove anything that is true in the truth table sense
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 in = t i1 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 101110011
Undecidability Theorem. The decision problem of validity for predicate calculus is undecidable: no program exists which, given any ϕ, can determine in a finite amount of time if ϕ Proof reduce to Post Correspondence problem. I.E. show that if the decision problem is solvable, we could solve the Post Correspondence problem. This is a contradiction.
Consequences of Undecidability From the soundness and completeness of predicate logic, which states that ϕ iff ϕ, we infer that we can not decide provability either Since ϕ is satisfiable iff ϕ is valid, we infer that satisfiability for predicate calculus is also undecidable.
Proof Construct a formula ϕ such that ϕ holds iff the corresponding Post correspondence problem has a solution. ϕ = ϕ 1 ϕ 2 ϕ 3 ϕ 1 = i=1..k P(f si (e),f ti (e)) ϕ 2 = v w (P(v,w) i=1..k P(f si (v),f ti (w))) ϕ 3 = z P(z,z)
Proof Find model which tells us the Post Correspondence problem has a solution e M = empty string f 0 (s) = s0, f 1 (s) = s1 P M = {(s,t) there is a sequence of indices i 1,i 2,,i n such that s i 1 s in = t i1 t in } Since ϕ holds M ϕ holds Both M ϕ 1 and M ϕ 2 holds so it follows that M ϕ 3 holds. This tells us there is a solution to the given PC problem
Proof Conversely assume that the PCP C has a solution i 1,i 2,,i n Show that if M is any model having a constant e M, two unary functions f0 M and f1 M, and a binary predicate P M, then the model has to satisfy ϕ Need to show if M ϕ 1 ϕ 2 then M ϕ 3
Proof Encode strings in A interpret(ε) = e M interpret(s0) = f 0M (interpret(s)) interpret(s1) = f 1M (interpret(s)) Interpret(011) = f 1 M (f 1 M (f 0M (e M ))) Since M ϕ 1, (interpret(s i ),interpret(t i )) P M Since M ϕ 2, If (s,t) P M then (interpret(ss i ),interpret(tt i )) P M
Proof Therefore, (interpret(s i1 s in ),interpret(t i1 t in )) P M Since i 1,i 2,,i n is a solution to the PCP interpret(s i1 s in ) = interpret(t i1 t in ) and z P(z,z) in M and thus M ϕ 3
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 s3 A = {s0,s1,s2,s3} R M = {(s0,s1), (s1,s0), (s1,s1),(s1,s2), (s2,s0),(s3,s0),(s3,s2)} s1 s2
Compactness Theorem Let Γ be a set of sentences of predicate calculus. If all finite subsets of Γ are satisfiable, then so is Γ. Proof uses soundness and completeness and finite length of proofs.
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.
Reachability is Not Expressible Theorem. There is no predicate-logic formula ϕ with u and v as its only free variables and R its only predicate such that ϕ holds in directed graphs iff there is a path from u to v. Proof. By contradiction. Suppose there is such a formula. Let ϕ n be the formula expressing that there is a path from c to c ϕ n = x 1 x n-1 (R(c,x 1 ) R(x n-1,c)).
Reachability is Not Expressible Proof. By contradiction. Suppose there is such a formula ϕ. Let ϕ n be the formula expressing that there is a path from c to c ϕ n = x 1 x n-1 (R(c,x 1 ) R(x n-1,c)). = { ϕ i I 0} ϕ[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 ϕ.
Reachability via HOL Encode Transitive closure P x y z (C 1 C 2 C 3 C 4 ) C 1 = P(x,x) C 2 = P(x,y) P(y,z) P(x,y) C 3 = P(u,v) C 4 = R(x,y) P(x,y) Note quantifier applied to predicate P (Existential second order logic)
Reachability via HOL 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.