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Definitions and Proofs Structural Induction Three approaches to semantics compositional definitions Prove that the property holds for all the basis elements of the syntactic cat- natural semantics egory. structural operational

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Definitions and Proofs Three approaches to semantics

compositional definitions
natural semantics
structural operational semantics

Three proof principles

structural induction
induction on the shape of derivation

trees

induction on the length of derivation

sequences

IV.1

Structural Induction

Prove that the property holds for all

the basis elements of the syntactic cat- egory.

Prove that the property holds for all

the composite elements by assuming that the property holds for the imme- diate constituents of the element (this is called the induction hypothesis) and proving that it also holds for the ele- ment itself.

IV.2

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Ind. on Shape of Derivation Trees
Prove that the property holds for all

the simple derivation trees by showing that it holds for all the axioms of the transition system

Prove that the property holds for all

the composite derivation trees: For each rule assume that the property holds for its premises (this is called the induction hypothesis) and prove that it also holds for the conclusion of the rule provided that the conditions

f the rule are satisfied

IV.3

Ind. on Length of Derivation Seq.
Prove that the property holds for all

derivation sequences of length 0

Prove that the property holds for all
ther derivation sequences:

Assume that the property holds for derivation sequences of length at most k (this is called the induction hypothesis) and prove that it holds for derivation se- quences of length k + 1

IV.4

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Proof by Structural Induction Intuitively: The value of an arithmetic ex- pression only depends on the values of the variables that occur in it. Free variables in arithmetic expressions 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) Lemma 1.11: Let s and s′ be two states satisfying s x = s′ x for all x ∈ FV(a). Then A[a]s = A[a]s′

IV.5

A : Aexp → State → N A[n]s = N[n] A[x]s = s x A[a1 + a2]s = A[a1]s + A[a2]s A[a1 ∗ a2]s = A[a1]s ∗ A[a2]s A[a1 − a2]s = A[a1]s − A[a2]s

IV.6

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Proof by Ind. on Shape of Derivation Trees Theorem 2.9: The natural semantics of While is determin- istic, that is for all statements S of While and all states s, s′ and s′′ if (S, s) → s′ and (S, s) → s′′ then s′ = s′′. Proof: We assume (S, s) → s′. We prove that if (S, s) → s′′ then s′ = s′′. We proceed by induction on the inference

f (S, s) → s′.

IV.7

Natural Semantics (x := a, s) → s[x → A[a]s] (skip, s) → s (S1, s) → s′, (S2, s′) → s′′ (S1; S2, s) → s′′ (S1, s) → s′ (if b then S1 else S2, s) → s′ if B[b]s = tt (S2, s) → s′ (if b then S1 else S2, s) → s′ if B[b]s = ff (S, s) → s′, (while b do S, s′) → s′′ (while b do S, s) → s′′ if B[b]s = tt (while b do S, s) → s if B[b]s = ff

IV.8

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Proof by Ind. on Length of Derivation Seq. Lemma 2.19: If (S1; S2, s) ⇒k s′′ then there exists s′, k1 and k2 such that (S1, s) ⇒k1 s′, (S2, s′) ⇒k2 s′′ and k = k1 + k2 Proof: We proceed by induction on the number k.

IV.9

Structural Operational Semantics (x := a, s) ⇒ s[x → A[a]s] (skip, s) ⇒ s (S1, s) ⇒ (S′

1, s′)

(S1; S2, s) ⇒ (S′

1; S2, s′)

(S1, s) ⇒ s′ (S1; S2, s) ⇒ (S2, s′) (if b then S1 else S2, s) ⇒ (S1, s) if B[b]s = tt (if b then S1 else S2, s) ⇒ (S2, s) if B[b]s = ff (while b do S, s) ⇒ (if b then (S; while b do S) else skip, s)