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## PowerPoint Slideshow about ' Axiomatic Semantics' - miette

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Axiomatic Semantics

Motivation Program

Input

Output

- 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.

- Transform input to output.

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- Sorting algorithms
- Bubble sort; Shell sort;
- Insertion sort; Selection sort;
- Merge sort; Quick sort;
- Heap sort;

To show 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 States : Variables -> Values Assertions : States -> Boolean

{ x = n and y = m }

t := x;

x := y;

y := t;

{ x = m and y = n}

- program variables vs ghost/logic variables

(= Powerset of States)

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Partial vs Total Correctness

{P} S {Q}

- S is partiallycorrect for P and Qif and only if whenever S is executed in a state satisfying Pand the execution terminates,then the resulting state satisfies Q.
- S is totallycorrect for P and Qif 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 Not totally correct, but partially correct Not partially correct

- 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 }

- {true} while true do; {x = 0}

- {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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Rule of Consequence Example

{P’} S {Q’} and P=>P’and Q’=>Q

{P} S {Q}

- Strengthening the antecedent
- Weakening the consequent

{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 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 Band 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

{Invand B} S {Inv}

{Inv} while B do S {Invand not B}

Proof of Correctness

{P} while B do S {Q}

= P => Invand {Inv} B {Inv}

and {Invand B} S {Inv}

and {Invandnot 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}

notB holds

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

B holdsnotBholds

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

B holds B holdsnotBholds

- Infinite loop if B never becomes false.

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Example1 : while-loop correctness Precondition implies invariant

{ 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)}

{ 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
- Variant : ( n - y )
y : 1 -> 2 -> … -> n

(n-y) : (n-1) -> (n-2) -> … -> 0

- Variant : ( n - y )

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Detailed Working

wp( y++; x:=x*y; , {x=y! and n>=y})

= wp(y++,{x*y=y! and n>=y})

= wp(y++,{x=y-1! and n>=y})

= wp(y++,{x=y-1! and n>=y})

= {x=y+1-1! and n>=y+1}

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GCD-LCM code

PRE: (x = n) and (y = m)

- u := x; v := y;
- while (x <> y) do
ASSERT: (** INVARIANT **)

- begin
- if x > y then x := x - y; u := u + v
- else y := y - x; v := v + u
- end;
POST: (x = gcd(n,m)) and (lcm (n,m) = (u+v) div 2)

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while-loop : Dijkstra’s Approach

wp( while B do S , Q)

= P0 or P1 or … or Pn or …

= there existsk >= 0 such that Pk

Pi: Set of states causingi-iterations ofwhile-loopbefore halting in a state inQ.

P0 = not B and Q

P1 = B and wp(S, P0)

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

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Example2 : while-loop correctness

P0 = { y >= n and x = n! }

Pk = B and wp(S,Pk-1)

P1 = { y<n andy+1>=n and x*(y+1) = n! }

Pk = y=n-kand x=(n-k)!

Weakest Precondition Assertion:

Wp = there existsk >= 0 such that

P0 or {y = n-kand x = (n-k)!}

Verification :

P = n>0 and x=1 and y=1

Fori = n-1: P => Wp

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Induction Proof

Hypothesis : Pk = {y=n-kand 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)

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