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Transcript of CSI 3125, Axiomatic Semantics, page 1 Axiomatic semantics The assignment statement Statement...
CSI 3125, Axiomatic Semantics, page 1
Axiomatic semantics
• The assignment statement• Statement composition• The "if-then-else" statement• The "while" statement• Narrowing and widening• Termination• Two diversions• The greatest common divisor• The "if-then" statement
Points to discuss:
CSI 3125, Axiomatic Semantics, page 2
Program verification
Program verification includes two steps.
1. Associate a formula with every meaningful step of the computation.
2. Show that the final formula logically follows from the initial one through all intermediate steps and formulae.
CSI 3125, Axiomatic Semantics, page 3
What is axiomatic semantics?
• Axiomatic semantics of assignments, compound statements, conditional statements, and iterative statements has been developed by Professor C. A. R. Hoare.
• The elementary building blocks are the formulae for assignments and conditions.
• The effects of other statements are described by inference rules that combine formulae for assignments (just as statements themselves are combinations of assignments and conditions).
CSI 3125, Axiomatic Semantics, page 4
The assignment statement
Let be a logical formula that contains variable v.
v e is a formula which we get from
when we replace all occurrences of
variable v with expression e.
CSI 3125, Axiomatic Semantics, page 5
Replacement, an example
Before replacement:
h >= 0 & h <= n & n > 0
h 0 0 >= 0 & 0 <= n & n > 0
after replacement
CSI 3125, Axiomatic Semantics, page 6
Another example
m == min( 1 <= i & i <= k–1: ai ) &
k–1 <= N
k k+1 m == min( 1 <= i & i <= (k+1) – 1: ai ) &
(k+1)–1 <= N
m == min( 1 <= i & i <= k: ai ) & k <= N
CSI 3125, Axiomatic Semantics, page 7
The axiom for the assignment statement
{v e } v = e {}
Example:
{ 0 >= 0 & 0 <= n & n > 0 }
x = 0;
{ x >= 0 & x <= n & n > 0 }
CSI 3125, Axiomatic Semantics, page 8
Two small puzzles
{ ??? } z = z + 1; { z <= N }
{ a > b } a = a – b; { ??? }
CSI 3125, Axiomatic Semantics, page 9
Statement composition
ASSUME THAT
{´ } S ´ {´´ }
and
{´´ } S ´´ {´´´ }
CONCLUDE THAT
{´ } S ´ S ´´ {´´´ }
In other words:
{´ } S ´ {´´ } S ´´ {´´´ }
CSI 3125, Axiomatic Semantics, page 10
A more complicated example
x = 0; f = 1;while (x != n) { x = x + 1; f = f * x;}
We want to prove that
{ f == x! }x = x + 1;f = f * x;
{ f == x! }
CSI 3125, Axiomatic Semantics, page 11
Let's apply the inference rule for composition.
´ is f == x!
´´´ is f == x!
S ´ is x = x + 1;
S ´´ is f = f * x;
The factorial
CSI 3125, Axiomatic Semantics, page 12
We need to find a ´´ for which we can prove:{ f == x! } x = x + 1;
{´´ }f = f * x; { f == x! }
Observe that f == x! f == ((x + 1) – 1)!
and therefore f == (x – 1)! x x + 1 f == x!That is:{ f == x! } x = x + 1; {f == (x – 1)! }
´ ´´S ´
The factorial (2)
CSI 3125, Axiomatic Semantics, page 13
Now, let us observe that
f == (x – 1)! f * x == (x – 1)! * x == x!
So, we have
f == x! f f * x f == (x – 1)!
That is,
{f == (x – 1)! } f = f * x; {f == x! }
´´ ´´´S ´´
The factorial (3)
QED
CSI 3125, Axiomatic Semantics, page 14
The "if-else" statement
ASSUME THAT
{ & } S´ {}
and
{ & } S´´ {}
CONCLUDE THAT
{} if ( ) S ´ else S ´´ {}
Both paths through the if-else statement establish the same fact . That is why the whole conditional statement establishes this fact.
CSI 3125, Axiomatic Semantics, page 15
The statementif ( a < 0 ) b = -a; else b = a;
makes the formula b == abs(a) true.Specifically, the following fact holds:
{true}if ( a < 0 ) b = -a; else b = a;{ b == abs(a) }
Here: is true is b == abs(a) is a < 0
Also: S´ is b = -a;S´´ is b = a;
"if-else", an example
CSI 3125, Axiomatic Semantics, page 16
We will consider cases. First, we assume that is true:
true & a < 0 a < 0 – a == abs(a)Therefore, by the assignment axiom:
{– a == abs(a)} b = -a; {b == abs(a)}
Similarly, when we assume , we get this:
true & a < 0 a 0 a == abs(a)Therefore:
{a == abs(a)} b = a; {b == abs(a)}
"if-else", an example (2)
CSI 3125, Axiomatic Semantics, page 17
This shows that both S´ and S´´ establish the same condition:
b == abs(a)Our fact has been proven:
{true}if ( a < 0 ) b = -a; else b = a;{ b == abs(a) }
In other words, our conditional statement computes abs(a). It does so without any preconditions: "true" means that there are no restrictions on the initial values of a and b.
"if-else", an example (3)
CSI 3125, Axiomatic Semantics, page 18
The "while" statement
A loop invariant is a condition that is true immediately before entering the loop, stays true during its execution, and is still true after the loop has terminated.
ASSUME THAT
{ & } S {}
[That is, S preserves .]CONCLUDE THAT
{ } while ( ) S { & }
provided that the loop terminates.
CSI 3125, Axiomatic Semantics, page 19
The factorial again...
x = 0; f = 1;while ( x != n ) {x = x + 1;f = f * x;
}
Assume for now that n ≥ 0. After computingx = 0; f = 1;
we have f == x! because it is true that 1 == 0!We showed earlier that{ f == x! } x = x + 1; f = f * x; { f == x! }
CSI 3125, Axiomatic Semantics, page 20
Now, is f == x! is x != n is x == n
Using the inference rule for "while" loops:
{ f == x! }while ( x != n ) { x = x + 1; f = f * x;}{ f == x! & x == n}
The factorial again... (2)
CSI 3125, Axiomatic Semantics, page 21
Notice that
f == x! & x == n f == n!
This means two things:{ true } x = 0; f = 1; { f == x! }
AND{ f == x! } while ( x != n ) {
x = x + 1; f = f * x;}
{ f == n!}
The factorial again... (3)
CSI 3125, Axiomatic Semantics, page 22
In other words, the program establishes f == n! without any preconditions on the initial values of f and n, assuming that we only deal with n ≥ 0.
The axiom for statement composition gives us:{ true } x = 0; f = 1;
while ( x != n ) { x = x + 1; f = f * x;}
{ f == n!}
So: this program does compute the factorial of n.
The factorial again... (4)
CSI 3125, Axiomatic Semantics, page 23
Our reasoning agrees with the intuition of loop invariants: we adjust some variables and make the invariant temporarily false, but we re-establish it by adjusting some other variables.
{ f == x! } x = x + 1; {f == (x – 1)! }the invariant is "almost true"
{f == (x – 1)! } f = f * x; {f == x! }the invariant is back to normal
This reasoning is not valid for infinite loops:the terminating condition & is never reached, and we know nothing of the situation following the loop.
The factorial again... (5)
CSI 3125, Axiomatic Semantics, page 24
Narrowing and widening
ASSUME THAT
´
and
{} S { }
CONCLUDE THAT
{´ } S { }
ASSUME THAT
{} S { }
and
´
CONCLUDE THAT
{ } S { ´ }
These rules can be used to narrow a precondition, or to widen a postcondition.
CSI 3125, Axiomatic Semantics, page 25
n! is computed, for any nonnegative n, with true as the precondition (it is always computed successfully);
So, n! will in particular must be computed successfully if initially n == 5.
Narrowing and widening, a small example
CSI 3125, Axiomatic Semantics, page 26
A larger example (in a more concise notation)
{ N >= 1 } { N >= 1 & 1 == 1 & a1 == a1 } i = 1; s = a1;
{ N >= 1 & i == 1 & s == a1 } { N >= 1 & s == a1 + … + ai } INVARIANTwhile ( i != N ) {
{ N >= 1 & s == a1 + … + ai & i != N }i = i + 1;
{ N >= 1 & s == a1 + … + ai–1 & i – 1 != N }s = s + ai;
{ N >= 1 & s == a1 + … + ai }}
{ N >= 1 & s == a1 + … + ai & i == N } { N >= 1 & s == a1 + … + aN }
CSI 3125, Axiomatic Semantics, page 27
• We have shown that this program computes the sum of a1, ..., aN.
• The precondition N >= 1 is only necessary to prove termination.
A larger example (2)
CSI 3125, Axiomatic Semantics, page 28
Termination
• Proofs like these show only partial correctness.– Everything is fine if the loop stops.– Otherwise we don't know (but the program
may be correct for most kinds of data).• A reliable proof must show that all loops in the
program are finite.• We can prove termination by showing how each
step brings us closer to the final condition.
CSI 3125, Axiomatic Semantics, page 29
Once again, the factorial…
• Initially, x == 0.
• Every step increases x by 1, so we go through the numbers 0, 1, 2, ...
• n >= 0 must be found among these numbers.
• Notice that this reasoning will not work forn < 0: the program loops.
CSI 3125, Axiomatic Semantics, page 30
A decreasing function
• A loop terminates when the value of some function of program variables goes down to 0 during the execution of the loop.
• For the factorial program, such a function could be n – x. Its value starts at n and decreases by 1 at every step.
• For summation, we can take N – i.
CSI 3125, Axiomatic Semantics, page 31
Multiplication by successive additions
{ B >= 0 & B == B & 0 == 0} FOR TERMINATIONb = B; p = 0;{ b == B & p == 0 } { p == A * (B – b) } INVARIANTwhile ( b != 0 ) {
p = p + A;{ p == A * (B – (b – 1)) }b = b - 1;{ p == A * (B – b) }
}{ p == A * (B – b) & b == 0} { p == A * B }The loop terminates, because the value of the variable b goes down to 0.
CSI 3125, Axiomatic Semantics, page 32
Two diversions
Prove that the sequencep = a; a = b; b = p;
exchanges the values of a and b :
{ a == A & b == B }p = a; a = b; b = p;
{ b == A & a == B }
The highlights of a proof:{ a == A & b == B } p = a;{ p == A & b == B } a = b;{ p == A & a == B } b = p;{ b == A & a == B }
CSI 3125, Axiomatic Semantics, page 33
x = x + y;
y = x - y;
x = x - y;
Two diversions (2)
Discover and PROVE the behaviour of the following sequence of statements for integer variables x, y:
CSI 3125, Axiomatic Semantics, page 34
{x == X & y == Y }
{x + y == X + Y & y == Y } x = x + y;
{x == X + Y & y == Y }
{x == X + Y & x - y == X } y = x - y;
{x == X + Y & y == X }
{ x - y == Y & y == X } x = x - y;
{ x == Y & y == X }
Two diversions (3)
CSI 3125, Axiomatic Semantics, page 35
The greatest common divisor
{ X > 0 & Y > 0 }a = X; b = Y;{ } what should the invariant be?while ( a != b ) { & a != b } {if ( a > b ) { & a != b & a > b }
a = a - b;else { & a != b & (a > b) }
b = b - a;}{ & (a != b) }{ GCD( X, Y ) == a }
CSI 3125, Axiomatic Semantics, page 36
We will need only a few properties of greatest common divisors:
GCD( n + m, m ) == GCD( n, m )GCD( n, m + n ) == GCD( n, m )
The first step (very formally):
{ X > 0 & Y > 0 } { X > 0 & Y > 0 & X == X & Y == Y }
a = X; b = Y;
{ a > 0 & b > 0 & a == X & b == Y }
GCD (2)
CSI 3125, Axiomatic Semantics, page 37
When the loop stops, we geta == b & GCD( a, b ) == a
We may want this condition in the invariant:a == b & GCD( X, Y ) == GCD( a, b )
At the beginning of the loop, we have:{ a > 0 & b > 0 & a == X & b == Y } {a > 0 & b > 0 & GCD( X, Y ) == GCD( a, b ) }
So, the invariant could be this:a > 0 & b > 0 & GCD( X, Y ) == GCD( a, b )
GCD (3)
CSI 3125, Axiomatic Semantics, page 38
We should be able to prove that{a > 0 & b > 0 & GCD(X, Y) == GCD(a, b) & a != b}while ......{a > 0 & b > 0 & GCD(X, Y) == GCD(a, b)}
The final condition will bea > 0 & b > 0 &GCD(X, Y) == GCD(a, b) & a == b
and this will implyGCD( X, Y ) == a
GCD (4)
CSI 3125, Axiomatic Semantics, page 39
The loop consists of one conditional statement.
Our proof will be complete if we show this:
{a > 0 & b > 0 & GCD(X, Y) == GCD(a, b) & a != b}
if ( a > b )a = a - b;
elseb = b - a;
{a > 0 & b > 0 & GCD(X, Y) == GCD(a, b)}
GCD (5)
CSI 3125, Axiomatic Semantics, page 40
Consider first the case of a > b.
{a > 0 & b > 0 &GCD(X, Y) == GCD(a, b) & a != b & a > b }
{a – b > 0 & b > 0 & GCD(X, Y) == GCD(a – b, b)}
a = a - b;
{a > 0 & b > 0 & GCD(X, Y) == GCD(a, b)}
GCD (6)
CSI 3125, Axiomatic Semantics, page 41
Now, the case of a > b.
{a > 0 & b > 0 & GCD(X, Y) == GCD(a, b) & a != b & (a > b) }
{a > 0 & b – a > 0 & GCD(X, Y) == GCD(a, b – a)}
b = b - a;
{a > 0 & b > 0 & GCD(X, Y) == GCD(a, b)}
GCD (7)
CSI 3125, Axiomatic Semantics, page 42
Both branches of if-else give the same final condition. We will complete the correctness proof when we show that the loop terminates.We show how the value of max( a, b ) decreases at each turn of the loop.
Let a == A, b == B at the beginning of a step. Assume first that a > b:
max( a, b ) == A,so a – b < A, b < A,
therefore max( a – b, b ) < A.
GCD (8)
CSI 3125, Axiomatic Semantics, page 43
Now assume that a < b:
max( a, b ) == B,
b – a < B, a < B,
therefore max( a, b – a ) < B.
Since a > 0 and b > 0, max( a, b ) > 0.
This means that decreasing the values of a, b cannot go forever.
QED
GCD (9)
CSI 3125, Axiomatic Semantics, page 44
The "if" statement
ASSUME THAT
{ & } S { }
and
&
CONCLUDE THAT
{ } if ( ) S { }
CSI 3125, Axiomatic Semantics, page 45
We will show the following:
{ N > 0 }
k = 1; m = a1;while ( k != N ) { k = k + 1;
if ( ak < m ) m = ak;}
{ m == min( 1 <= i & i <= N: ai ) }
An example with "if"
CSI 3125, Axiomatic Semantics, page 46
Loop termination is obvious:
the value of N – k goes down to zero.
Here is a good invariant: at the kth turn of the loop, when we have already looked at a1, ..., ak, we know that
m == min( 1 <= i & i <= k : ai ).
Initially, we have this:{ N > 0 } k = 1; m = a1;{ k == 1 & m == a1 } { k == 1 & m == min( 1 <= i & i <= k : ai ) }
Minimum
CSI 3125, Axiomatic Semantics, page 47
We must prove the following:
{ m == min( 1 <= i & i <= k : ai ) & k != N }k = k + 1;if ( ak < m ) m = ak;{ m == min( 1 <= i & i <= k : ai ) }
Minimum
CSI 3125, Axiomatic Semantics, page 48
{ m == min( 1 <= i & i <= k : ai ) & k != N }
{ m == min( 1 <= i & i <= (k + 1) – 1: ai ) & (k + 1) – 1 != N }
k = k + 1;
{ m == min( 1 <= i & i <= k – 1: ai ) & k – 1 != N }
Note that k – 1 != N ensures the existence of ak.
Minimum (2)
CSI 3125, Axiomatic Semantics, page 49
This remains to be shown:
{ m == min( 1 <= i & i <= k – 1: ai ) & k – 1 != N }
if ( ak < m ) m = ak;
{ m == min( 1 <= i & i <= k: ai ) }
The fact we will use is this:
min( 1 <= i & i <= k: ai ) ==min2( min( 1 <= i & i <= k – 1: ai ), ak )
Minimum (3)
CSI 3125, Axiomatic Semantics, page 50
We will consider two cases of the conditional statement.
First, (ak < m).
{m == min(1 <= i & i <= k – 1: ai ) &k – 1 != N & (ak < m)}
{m == min2(min( 1 <= i & i <= k – 1: ai ), ak )}
{m == min(1 <= i & i <= k: ai )}
Minimum (4)
CSI 3125, Axiomatic Semantics, page 51
Now, ak < m.
{m == min(1 <= i & i <= k – 1: ai ) &k – 1 != N & ak < m}
{ak == min2( min( 1 <= i & i <= k – 1: ai ), ak )}
{ak == min(1 <= i & i <= k: ai )}
m = ak;
{m == min(1 <= i & i <= k: ai )}
So, the body of the loop preserves the condition
m == min( 1 <= i & i <= k: ai )
Minimum (5)
CSI 3125, Axiomatic Semantics, page 52
Now, the whole loop works as follows:
{ m == min( 1 <= i & i <= k: ai ) }while ( k != N ) }
k = k + 1; if ( ak < m ) ak = m;}
{ m == min( 1 <= i & i <= k: ai ) & k == N } { m == min( 1 <= i & i <= N: ai ) }
All in all, we have shown that our program finds the minimum of N numbers, if only N > 0.
QED
Minimum (6)
CSI 3125, Axiomatic Semantics, page 53
Yet another "while" loop
{ B > 0 } FOR TERMINATIONb = 1; p = A;while ( b != B ) {
b = b + 1;p = p * A;
}{ ??? }
Examples
CSI 3125, Axiomatic Semantics, page 54
Yet another "while" loop (2)
{ B > 0 & 1 == 1 & A == A} FOR TERMINATIONb = 1; p = A;{ b == 1 & p == A } { p == A ** b } INVARIANTwhile ( b != B ) {
b = b + 1;{ p == A ** (b - 1) }p = p * A;{ p == A ** b }
}{ p == A ** b & b == B} { p == A ** B }The loop terminates: the value B - b goes down to 0.
Examples
CSI 3125, Axiomatic Semantics, page 55
{ N > 0 } FOR TERMINATIONk = 1;while ( k != N ) {
if ( Ak > Ak+1 )
{ p = Ak; Ak = Ak+1; Ak+1 = p; } k = k + 1; }{ ??? }
Another example with "if"Examples
CSI 3125, Axiomatic Semantics, page 56
{ N > 0 } FOR TERMINATIONk = 1;{ Ak == max( 1 <= i & i <= k: Ai ) } INVARIANTwhile ( k != N ) { { Ak == max( 1 <= i & i <= k: Ai ) & k != N } if ( Ak > Ak+1 ) { p = Ak; Ak = Ak+1; Ak+1 = p; } { Ak+1 == max( 1 <= i & i <= k+1: Ai ) } k = k + 1; { Ak == max( 1 <= i & i <= k: Ai ) }}
{Ak == max( 1 <= i & i <= k: Ai ) & k == N } {AN == max( 1 <= i & i <= N: Ai ) }
Another example with "if" (2)Examples
CSI 3125, Axiomatic Semantics, page 57
{Ak == max( 1 <= i & i <= k: Ai ) & k != N }
case 1: Ak > Ak+1
{ Ak == max( 1 <= i & i <= k: Ai ) & k != N & Ak > Ak+1}
p = Ak; { p > Ak+1 }
Ak = Ak+1; { p > Ak }
Ak+1 = p; { Ak+1 > Ak }
{ Ak+1 == max( 1 <= i & i <= k+1: Ai ) }
case 2: Ak <= Ak+1
{ Ak == max( 1 <= i & i <= k: Ai ) & k != N & Ak <= Ak+1 } { Ak+1 == max( 1 <= i & i <= k+1: Ai ) }
Another example with "if" (3)Examples