axiomatic semantics

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Axiomatic Semantics

Predicate Transformers

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Motivation

• Problem Specification• Properties satisfied by the input and expected of the

output (usually described using “assertions”).

• E.g., Sorting problem– Input : Sequence of numbers

– Output: Permutation of input that is ordered.

• Program• Transform input to output.

Input Output

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• Sorting algorithms» Bubble sort; Shell sort;

» Insertion sort; Selection sort;

» Merge sort; Quick sort;

» Heap sort;

• Axiomatic Semantics To facilitate proving that a program satisfies its

specification, it is convenient to have the description of the language constructs in terms of assertions characterizing the input and the corresponding output states.

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Axiomatic Approaches• Hoare’s Proof System (partial correctness)

• Dijkstra’s Predicate Transformer (total correctness)

Assertion: Logic formula involving program variables, arithmetic/boolean operations, etc.

Hoare Triples : {P} S {Q}

pre-condition statements post-condition

(assertion) (program) (assertion)

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Swap Example{ x = n and y = m } t := x; x := y; y := t;{ x = m and y = n }

– program variables vs ghost/logic variables

• States : Variables Values

• Assertions : States Boolean

(= Powerset of States)

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Partial vs Total Correctness{P} S {Q}

• S is partiallypartially correct for P and Q if and only if whenever S is executed in a state satisfying P and the execution terminates, then the resulting state satisfies Q.

• S is totallytotally correct for P and Q if and only if whenever S is executed in a state satisfying P , then the execution terminates, and the resulting state satisfies Q.

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Examples• Totally correct (hence, partially correct)

•{ false } x := 0; { x = 111 }•{ x = 11 } x := 0; { x = 0 }•{ x = 0 } x := x + 1; { x = 1 }•{false} while true do; {x = 0}• {y = 0} if x <> y then x:= y; { x = 0 }

• Not totally correct, but partially correct•{true} while true do; {x = 0}

• Not partially correct• {true} if x < 0 then x:= -x; { x > 0 }

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Axioms and Inference Rules• Assignment axiom

{Q[e]} x := e; {Q[x]}

• Inference Rule for statement composition {P} S1 {R} {R} S2 {Q} {P} S1; S2 {Q}

• Example {x = y} x := x+1; {x = y+1} {x = y+1} y := y+1; {x = y}{x = y} x:=x+1; y:=y+1; {x = y}

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Generating additional valid triples {P} S {Q} from {P’} S {Q’}

P’

States States

P’

P Q’

Q

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Rule of Consequence

{P’} S {Q’} and P=>P’ and Q’=>Q {P} S {Q}

– Strengthening the antecedent – Weakening the consequent

• Example{x=0 and y=0} x:=x+1;y:=y+1; {x = y}{x=y} x:=x+1; y:=y+1; {x<=y or x=5} (+ Facts from elementary mathematics [boolean algebra + arithmetic] )

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Predicate Transformers

• Assignment

wp( x := e , Q ) = Q[x<-e]• Composition

wp( S1 ; S2 , Q) = wp( S1 , wp( S2 , Q )) • Correctness

{P} S {Q} = (P => wp( S , Q))

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Correctness Illustrated

States States

QQPP

wp(S,Q)wp(S,Q)

P => wp( S , Q)P => wp( S , Q)

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Correctness Proof {x=0 and y=0} x:=x+1;y:=y+1; {x = y}

• wp(y:=y+1; , {x = y}) = { x = y+1 }• wp(x:=x+1; , {x = y+1}) = { x+1 = y+1 }• wp(x:=x+1;y:=y+1; , {x = y}) = { x+1 = y+1 } = { x = y }• { x = 0 and y = 0 } => { x = y }

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Conditionals { P and B } S1 {Q}

{P and not B } S2 {Q}{P} if B then S1 else S2; {Q}

wp(if B then S1 else S2; , Q) = (B => wp(S1,Q)) and

(not B => wp(S2,Q))

= (B and wp(S1,Q)) or

(not B and wp(S2,Q))

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“Invariant”: Summation Program

{ s = i * (i + 1) / 2 } i := i + 1; s := s + i; { s = i * (i + 1) / 2 }

• Intermediate Assertion ( s and i different){ s + i = i * (i + 1) / 2 }

• Weakest Precondition{ s+i+1 = (i+1) * (i+1+1) / 2 }

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while-loop : Hoare’s Approach {Inv and B} S {Inv}{Inv} while B do S {Inv and not B}

Proof of Correctness {P} while B do S {Q}= P => Inv andand {Inv} B {Inv} andand {Inv and B} S {Inv} andand {Inv and not B => Q}

+ Loop Termination argument

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{I} while B do S {I and not B}

{I and B} S {I}

0 iterations: {I} {I and not B} ^not B holds

1 iteration: {I} S {I and not B} ^B holds ^ not B holds

2 iterations: {I} S ; S {I and not B} ^B holds ^B holds ^ not B holds

• Infinite loop if B never becomes false.

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Example1 : while-loop correctness { n>0 and x=1 and y=1}while (y < n) [ y++; x := x*y;] {x = n!}

• Choice of Invariant•{I and not B} => Q•{I and (y >= n)} => (x = n!)•I = {(x = y!) and (n >= y)}

• Precondition implies invariant{ n>0 and x=1 and y=1} => { 1=1! and n>=1 }

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• Verify Invariant {I and B} => wp(S,I)wp( y++; x:=x*y; , {x=y! and n>=y})= { x=y! and n>=y+1 }I and B = { x=y! and n>=y } and { y<n }= { x=y! and n>y }

• Termination• VariantVariant : ( n - y ) y : 1 -> 2 -> … -> n(n-y) : (n-1) -> (n-2) -> … -> 0

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while-loop : Dijkstra’s Approach

wp( while B do S , Q)

= P0 or P1 or … or Pn or …

= there exists k >= 0 such that Pk

Pi : Set of states causing i-iterations of while-loop before halting in a state in Q.

P0 = not B and Q P1 = B and wp(S, P0) Pk+1 = B and wp(S, Pk)

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...

P0P0

P1P1

P2P2

States

Q

States

wpwp

P0 => wp(skip, Q)

P0 subsetsubset Q

P1 => wp(S, P0)

P0P0

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Example2 : while-loop correctnessP0 = { y >= n and x = n! }Pk = B and wp(S,Pk-1)P1 = { y<n and y+1>=n and x*(y+1) = n! }

Pk = y=n-k and x=(n-k)!Weakest Precondition Assertion:

Wp = there exists k >= 0 such that P0 or {y = n-k and x = (n-k)!}Verification :

P = n>0 and x=1 and y=1 For i = n-1: P => Wp

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Induction ProofHypothesis : Pk = {y=n-k and x=(n-k)!}

Pk+1 = { B and wp(S,Pk) }

= y<n and (y+1 = n-k) and (x*(y+1)=(n-k)!)

= y<n and (y = n-k-1) and (x = (n-k-1)!) = y<n and (y = n- k+1) and (x = (n- k+1)!) = (y = n - k+1) and (x = (n - k+1)!)

Valid preconditions:– { n = 4 and y = 2 and x = 2 } (k = 2)– { n = 5 and x = 5! And y = 6} (no iteration)