Semantics with Applications

1. Introduction

Hanne Riis Nielson, Flemming Nielson

(thanks to Henrik Pilegaard)

[SwA] Hanne Riis Nielson, Flemming NielsonSemantics with Applications: An AppetizerSpringer, 2007

1 / 31

Syntax and SemanticsReading material: Chapter 1 of SwA

2 / 31

Language While: Syntactic Categories

I Numerals n ∈ Num

n ::= 0 | 1 | n0 | n1

I Variables x ∈ Var – e.g. x, y, z, . . .

I Arithmetic expressions a ∈ Aexp

a ::= n | x | a1 + a2 | a1 ? a2 | a1 − a2

I Boolean expressions b ∈ Bexp

b ::= true | false | a1 = a2 | a1 ≤ a2 | ¬b | b1 ∧ b2

I Statements S ∈ Stm

S ::= x := a | skip | S1;S2 | if b then S1 else S2 | while b do S

3 / 31

Example Programs

I The factorial programI If x = n > 0 initially, then y = n! after program termination

y := 1; while ¬(x = 1) do (y := y ? x ; x := x − 1)

4 / 31

Compositional Definitions

I The syntactic category is specified by an abstract syntax giving theI basis elements

I composite elements – decomposition into their immediateconstituents

a ::= n | x | a1 + a2 | a1 ? a2 | a1 − a2

5 / 31

Abstract vs. Concrete Syntax

I Abstract syntax focuses on the structure of expressions andstatements

I ignores scanning and parsing aspects

I formalises a set of allowed parse trees

I Concrete syntax deals with the scanning and parsing aspectsI provides sufficient information to construct unique parse trees

6 / 31

Abstract vs. Concrete Syntax

I In this part of the course, we are not concerned with concrete syntax

I We use brackets to disambiguate the syntax where necessary

while x < y do (x := x + 1; y := 0)or

(while x < y do x := x + 1); y := 0

I Assume familiar precedence rules for arithmetic and boolean operators

write 1 + x ? 2 for 1 + (x ? 2) and not for (1 + x) ? 2

7 / 31

Semantics of Arithmetic and Boolean Expressions

8 / 31

States

I We have used states before in the vending machine example

I Now we can be more formal: a state is an element of the set

State = Var→ Z

where Z is the set of integers, i.e. . . . ,−1, 0, 1, 2, . . .

I Other representations used: “table” as in the vending machineexample, or “list” [x 7→ 3, y 7→ 0]

I Lookup: Each state s specifies a value s x for every x ∈ VarWe use the variant notation s(x) as well

I Update:

(s[y 7→ n])(x) =

{n if x = ys(x) if x 6= y

9 / 31

Semantic Functions

I Crucial distinction between the syntax (grammar) and the semantics(meaning) of a program

I A semantic function is a total function that maps syntactic

elements to a mathematical object representing their meaning

10 / 31

Semantics of Numerals

Numerals

I Assume that numerals n ∈ Num (in binary form)

I Function N : Num→ Z maps numerals to integers Z

I N [[0]] = 0,

I N [[1]] = 1,

I N [[n0]] =?,

I N [[n1]] =?,

I Notation: “syntactic brackets” [[ and ]] to enclose syntactic elements

11 / 31

Semantics of Numerals

Numerals

I Assume that numerals n ∈ Num (in binary form)

I Function N : Num→ Z maps numerals to integers Z

I N [[0]] = 0,

I N [[1]] = 1,

I N [[n0]] = 2 · N [[n]],

I N [[n1]] = 2 · N [[n]] + 1,

I Notation: “syntactic brackets” [[ and ]] to enclose syntactic elements

12 / 31

Semantics of Arithmetic Expressions

I The value of an expression such as 2 ? x + y depends on the values ofthe variables x and z – these are determined by the current state

I The semantics of an arithmetic expression a in current state s is givenby the semantic function

A : Aexp→ (State→ Z)

I The function A takes its parameters one at a time:

I A[[x + 1]] has functionality State→ Z

I Given a state s with s x = 2 we have A[[x + 1]]s = 3

13 / 31

Semantics of Arithmetic Expressions

The semantics of arithmetic expressions is given by:

14 / 31

Semantics of Arithmetic Expressions

The semantics of arithmetic expressions is given by:

15 / 31

Semantics of Arithmetic Expressions

I Operators +, ?,− on the right hand side are the usual arithmeticoperations, on the left hand side they are just pieces of syntax

I Note that we have one clause for each of the different forms ofarithmetic expressions

Given a state s with s x = 3 and s y = 1 we have:

A[[2 ? x + y]]s =

A[[2 ? x]]s +A[[y]]s

=

A[[2]]s · A[[x]]s + 1

=

N [[2]] · 3 + 1 = 2 · 3 + 1 = 7

16 / 31

Semantics of Arithmetic Expressions

I Operators +, ?,− on the right hand side are the usual arithmeticoperations, on the left hand side they are just pieces of syntax

I Note that we have one clause for each of the different forms ofarithmetic expressions

Given a state s with s x = 3 and s y = 1 we have:

A[[2 ? x + y]]s = A[[2 ? x]]s +A[[y]]s=

A[[2]]s · A[[x]]s + 1

=

N [[2]] · 3 + 1 = 2 · 3 + 1 = 7

17 / 31

Semantics of Arithmetic Expressions

I Operators +, ?,− on the right hand side are the usual arithmeticoperations, on the left hand side they are just pieces of syntax

I Note that we have one clause for each of the different forms ofarithmetic expressions

Given a state s with s x = 3 and s y = 1 we have:

A[[2 ? x + y]]s = A[[2 ? x]]s +A[[y]]s= A[[2]]s · A[[x]]s + 1=

N [[2]] · 3 + 1 = 2 · 3 + 1 = 7

18 / 31

Semantics of Arithmetic Expressions

I Operators +, ?,− on the right hand side are the usual arithmeticoperations, on the left hand side they are just pieces of syntax

I Note that we have one clause for each of the different forms ofarithmetic expressions

Given a state s with s x = 3 and s y = 1 we have:

A[[2 ? x + y]]s = A[[2 ? x]]s +A[[y]]s= A[[2]]s · A[[x]]s + 1= N [[2]] · 3 + 1 = 2 · 3 + 1 = 7

19 / 31

Compositional Definitions

I The semantics is defined by compositional definitions of a function:I A semantic clause for each of the basis elements

I A semantic clause for each of the methods for constructingcomposite elements – defined in terms of the semantics of their

immediate constituents

20 / 31

Semantics of Boolean Expressions

I Analogous to the development for arithmetic expressions

I Semantic function for boolean expressions

B : Bexp→ (State→ T)

I Set of (semantic) truth values T consists of tt (for true) and ff (forfalse)

21 / 31

Semantics of Boolean ExpressionsThe semantics of boolean expressions is given by:

22 / 31

Semantics of Boolean ExpressionsThe semantics of boolean expressions is given by:

23 / 31

Properties of the Semantics

24 / 31

Mathematical Induction

Mathematical induction is used to prove properties of things that can begiven a size as a natural number (being a non-negative integer).

I Prove that 2n ≤ (n + 1)! ≤ (n + 1)n.

I Basis: for n = 0 we must prove 1 ≤ 1 ≤ 1 which is immediate.

I Induction step: for n = m + 1 > 0 we must prove2m+1 ≤ (m + 1 + 1)! ≤ (m + 1 + 1)m+1

and may use2m ≤ (m + 1)! ≤ (m + 1)m

25 / 31

Structural Induction

Structural induction is used to prove properties of the semantics. It worksin a similar way as mathematical induction:

I Prove that the property holds for all the basis elements (formathematical induction: 0)

I Prove that the property holds for all the composite elements :Assume that the property holds for all the immediate constituents ofthe element (this is called the induction hypothesis) and prove that italso holds for the element itself (for mathematical induction: n+1)

Structural induction can alternatively be explained as mathematicalinduction on an upper bound on the size of the element (think number ofASCII characters).

26 / 31

Semantics of Numerals

QuizIs the function N [[]] total? I.e., for each n ∈ Num, there exists exactly onei ∈ Z such that N [[n]] = i?

Yes. Proof by structural induction.

I case n = 0: ?

I case n = n′0: ?

I cases n = 1 or n = n′1 are similar

27 / 31

Semantics of Numerals

QuizIs the function N [[]] total? I.e., for each n ∈ Num, there exists exactly onei ∈ Z such that N [[n]] = i?

Yes. Proof by structural induction.

I case n = 0: ?

I case n = n′0: ?

I cases n = 1 or n = n′1 are similar

28 / 31

Free Variables

I We want to formalise the fact that the value of an arithmeticexpression only depends on the values of variables occurring in it

I The set of free variables FV (a) of expression a:

FV (n) = ∅FV (x) = {x}FV (a1 + a2) = FV (a1) ∪ FV (a2)FV (a1 ? a2) = FV (a1) ∪ FV (a2)FV (a1 − a2) = FV (a1) ∪ FV (a2)

LemmaLet s and s ′ be two states satisfying that s x = s ′ x for all x ∈ FV (a).Then A[[a]]s = A[[a]]s ′.

Proof by structural induction.

29 / 31

Substitutions

I We want to formalise the fact that a substitution within an expressioncan be mimicked by a similar change of the state

I Substitution of a variable y with a0 in the expression a:

n[y 7→ a0] = n

x [y 7→ a0] =

{a0 if x = yx if x 6= y

(a1 + a2)[y 7→ a0] = (a1[y 7→ a0]) + (a2[y 7→ a0])(a1 ? a2)[y 7→ a0] = (a1[y 7→ a0]) ? (a2[y 7→ a0])(a1 − a2)[y 7→ a0] = (a1[y 7→ a0])− (a2[y 7→ a0])

LemmaA[[a[y 7→ a0]]]s = A[[a]](s[y 7→ A[[a0]]s]) for all states s.

Proof: Do Exercise 1.14

30 / 31

Summary

I Language While: syntax and semantics

I Proof technique: structural induction

Exercise Class

I Exercises 1.2, 1.14, 1.15 from SwA.

31 / 31