Predicate Logic. Xinyu Feng 11/20/2013. University of Science and Technology of China (USTC)
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1 University of Science and Technology of China (USTC) 11/20/2013
2 Overview Predicate logic over integer expressions: a language of logical assertions, for example x. x + 0 = x Why discuss predicate logic? It is an example of a simple language It has simple denotational semantics We will use it later in program specifications
3 Abstract Syntax (intexp) e ::= x e e+e e-e... (assert) p, q ::= true false e=e e < e e < e... p p p p p p p p p x. p x. p
4 Resolving Notational Ambiguity Using parentheses: ( x. (((x + 0)+0) =x)) Using precedence and associativity: x. x = x arithmetic operators ( /...) with the usual precedence relational operators(= <...) The body of a quantified term extends to a delimiter.
5 Carriers and Constructors Carriers (or Syntactic Categories): sets of abstract phrases (e.g. intexp, assert) Constructors: specify abstract grammar productions (intexp)e ::= 0 c 0 intexp (intexp)e ::= x c var var intexp (intexp)e ::= e+e c + intexp intexp intexp Note: Independent of the concrete pattern of the production: (intexp)e ::= plus (e, e ) c + intexp intexp intexp Constructors must be injective and have disjoint ranges Carriers must be either predefined or their elements must be constructable in finitely many constructor applications
6 Inductive Structure of Carrier Sets With these properties of constructors and carriers, carriers can be defined inductively: intexp (0) intexp (j+1) assert (0) assert (j+1) def = {} def = {c 0 } {c + (e 0, e 1 ) e 0, e 1 intexp j } {...} def = {} def = {c true, c false } def = {c = (e 0, e 1 ) e 0, e 1 intexp} def = {c p p assert (j) } def = {c (p, q) p, q assert (j) } intexp def = assert j=0 def = j=0 intexp (j) assert (j)
7 Denotational Semantics The meaning of a term e intexp is e intexp, i.e. the function intexp maps intexp objects to their meanings. What is the set of meanings? The meaning of intexp of the term (i.e. c + (c 5, c 37 )istheinteger 42. However, the term x + 37 contains the free variable x, sothe meaning of an intexp e in general cannot be an integer...
8 Environments or States To give meanings to x + 37, we need an environment (also called variable assignment or state): σ Σ def = var Z which gives meanings to free variables. The meaning of a term (intexp or assert) is a function from the states to Z or B. intexp intexp Σ Z assert assert Σ B If σ = {(x, 3), (y, 4)}, then x + 5 intexp σ = 8, z. x < z z < y assert σ = false
9 Direct Semantics Equations for 0 intexp σ def = 0 x σ def = σ x e 0 +e 1 intexp σ def =( e 0 intexp σ)+( e 1 intexp σ) true assert σ def = true e 0 =e 1 assert σ def =( e 0 intexp σ)=( e 1 intexp σ) p assert σ def = ( p assert σ) p q assert σ def =( p assert σ) ( q assert σ) x. p assert σ def = n Z. p assert (σ{x n})
10 Example: The Meaning of a Term x. x+0=x assert σ
11 Properties of the Semantics Equations They are Syntax Directed (Homomorphic): exactly one equation for each constructor results expressed using semantics (meanings) of subterms (arguments of constructors) only. they have exactly one solutions (proof by induction on the structure of terms) They define compositional semantic functions (depending only on the meanings of subterms) equivalent terms can be substituted.
12 Validity of Assertions p holds/is true in σ p assert σ = true σ satisfies p p is valid p is unsatisfiable p is stronger than q σ Σ. p holds in σ σ Σ. p assert σ = false p is valid σ Σ. q holds in σ if p holds in σ p q is valid p and q are equivalent p is stronger than q and q is stronger than p
13 Inference Rules Axioms and Rules p p 0, p 1, p n 1 p They are all schemas. Judgments p We also use p to mean there exists a proof or derivation of p following the axioms and rules.
14 Inference Rules Examples x + 0 = x (xpluszero) e 0 = e 1 e 1 = e 0 (SymmObjEq) p p q q (ModusPonens) p x. p (Generalization)
15 Formal Proofs A set of inference rules define a Logical Theory. A Formal Proof in a logical theory: a sequence of instantiations of axioms and inference rules, where a premise of each rule occur as conclusions of another rule earlier in the sequence. Example: 1. x + 0 = x (xpluszero) 2. x + 0 = x x = x + 0 (SymmObjEq) 3. x = x + 0 (ModusPonens, 1, 2) 4. x. x = x + 0 (Generalization, 3)
16 Tree Representation (Derivation Trees) x + 0 = x (xpz) x + 0 = x x = x + 0 (Symm) (MP) x = x + 0 x. x = x + 0 (Gen)
17 Sequent Style Rules Γ, p p, Δ (id) Γ p, Δ Γ, p Δ (negl) Γ, p Δ Γ p, Δ (negr) Γ, p, q Δ Γ, p q Δ (conjl) Γ p, Δ Γ q, Δ Γ p q, Δ (conjr) Γ, p Δ Γ, q Δ Γ, p q Δ Γ p, Δ Γ, q Δ Γ, p q Δ (disjl) (impl) Γ p, q, Δ Γ p q, Δ (disjr) Γ, p q, Δ Γ p q, Δ (impr)
18 Sequent Style Rules (cont d) Γ, p(x) Δ Γ, x.p(x) Δ (all) Γ p(z), Δ Γ x.p(x), Δ (alr) Γ, p(z) Δ Γ, x.p(x) Δ (exl) Γ p(x), Δ Γ x.p(x), Δ (exr)
19 Soundness of a Logical Theory An inference rule is sound if in every instance of the rule the conclusion is valid if all the premises are. A logical theory is sound if all the inference rules in it are sound. If is sound and there is a formal proof of p, then p is valid. Object vs. Meta implication: p x. p p x. p unsound sound.
20 Completeness of a Logical Theory A logical theory is complete if for every valid assertion p there is a formal proof of p. A logical theory is axiomatizable if there exist a finite set of inference rules from which formal proofs of assertions in can be constructed. No first-order theory of arithmetic is axiomatizable and complete (Gödel s incompleteness theorem).
21 Variable Binding x. y. x < y x. x > y Binder and scope Bound and free occurrences of variables α-renaming and substitution similar to λ-calculus
22 Only Assignment of Free Variables Matters Coincidence Theorem: If σ x = σ x for all x fv(p), then p θ σ = p θ σ (θ {intexp, assert} and p θ). Proof: By induction over the structure of p. Induction Hypothesis: The statement of the theorem holds over all phrases of depth less than that of p. Base cases: p = 0: p intexp σ = 0 = p intexp σ. p = x: p intexp σ = σ x = σ x = p intexp σ,sincefv(x) ={x}.
23 Proof of Coincidence Theorem (cont d) Coincidence Theorem: If σ x = σ x for all x fv(p), then p θ σ = p θ σ Inductive cases: p = e 0 + e 1 :ByIH, e i intexp σ = e i intexp σ for i {0, 1}. So e intexp σ = e 0 intexp σ + e 1 intexp σ = e 0 intexp σ + e 1 intexp σ = e intexp σ p = x. q: σ z = σ z, for all z fv(p) =fv(q) {x} So σ{x n} z = σ {x n} z, for all n Z and z fv(q). By IH, q assert σ{x n} = q assert σ {x n}, foralln Z. Therefore ( n Z. q assert σ{x n}) =( n Z. q assert σ {x n}) That is p assert σ = p assert σ.
24 Substitution /δ intexp intexp /δ assert assert } where δ var intexp 0/δ = 0 (-e)/δ = -(e/δ) (e 0 +e 1 )/δ =(e 0 /δ)+(e 1 /δ)... x/δ = δ x (p q)/δ =(p/δ) (q/δ) ( x. p)/δ = x. (p/δ{x x }) where x fv(δ z) z fv( x. p) Examples: (x < 0 x. x y)/{(x, y + 1)} = y + 1 < 0 x. x y (x < 0 x. x y)/{(y, x + 1)} = x < 0 z. z x + 1
25 Substitution Theorems Substitution Theorem: If σ = intexp σ δ (i.e. σ = λx var. δ x intexp σ), then p assert σ = p/δ assert σ Finite Substitution Theorem: Let δ = {(x 0, e 0 ),...,(x n 1, e n 1 )}, and σ = σ { x 0 e 0 intexp σ,...,x n 1 e n 1 intexp σ }, then p assert σ = p/δ assert σ.
26 Acknowledgments Slides follow Chapter 1 of Reynolds, and are taken from Zhong Shao s lecture notes on Formal Semantics.
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