Axiomatic semantics Points to discuss: 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

Program verification Program verification includes two steps. Associate a formula with every meaningful step of the computation. Show that the final formula logically follows from the initial one through all intermediate steps and formulae.

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).

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.

Replacement, an example Before replacement:   h >= 0 & h <= n & n > 0 h  0  0 >= 0 & 0 <= n & n > 0 after replacement

  m == min( 1 <= i & i <= k–1: ai ) & 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

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 }

Two small puzzles { ??? } z = z + 1; { z <= N } { a > b } a = a – b; { ??? }

Statement composition ASSUME THAT {´ } S ´ {´´ } and {´´ } S ´´ {´´´ } CONCLUDE THAT {´ } S ´ S ´´ {´´´ } In other words: {´ } S ´ {´´ } S ´´ {´´´ }

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! }

´´´ is f == x! The factorial 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 (2) 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 (3) 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 ´´ ´´´ QED

The "if-else" statement ASSUME THAT { &  } S´ {} and 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.

"if-else", an example The statement if ( 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 (2) 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 (3) 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.

The "while" statement ASSUME THAT { &  } S {} 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.

The factorial again... x = 0; f = 1; while ( x != n ) { x = x + 1; f = f * x; } Assume that n ≥ 0. After computing we have f == x! because it is true that 1 == 0!

The factorial again... (2) We showed earlier that { f == x! } x = x + 1; f = f * x; { f == x! } The reasoning will not change if we add the condition n ≥ 0, because the transitions do not depend on n. So, we can write { n ≥ 0 & f == x! } x = x + 1; f = f * x;

The factorial again... (3) Now,  is f == x!  is x != n   is x == n Using the inference rule for "while" loops: { n ≥ 0 & f == x! } while ( x != n ) { x = x + 1; f = f * x; } { n ≥ 0 & f == x! & x == n }

The factorial again... (4) Notice that f == x! & x == n  f == n! This means two things: { n ≥ 0 } x = 0; f = 1; { n ≥ 0 & f == x! } AND { n ≥ 0 & f == x! } while ( x != n ) { x = x + 1; f = f * x; } { n ≥ 0 & f == n! }

The factorial again... (5) 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: { n ≥ 0 } 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... (6) { f == x! } x = x + 1; { f == (x – 1)! } 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.

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.

Narrowing and widening, a small example 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.

A larger example (in a more concise notation) { N >= 1 & 1 == 1 & a1 == a1 } i = 1; s = a1; { N >= 1 & i == 1 & s == a1 }  { N >= 1 & s == a1 + … + ai }  INVARIANT while ( 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 }

A larger example (2) We have shown that this program computes the sum of a1, ..., aN. The precondition N >= 1 is only necessary to prove termination.

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.

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 for n < 0: the program loops.

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.

Multiplication by successive additions { B >= 0 & B == B & 0 == 0}  FOR TERMINATION b = B; p = 0; { b == B & p == 0 }  { p == A * (B – b) }  INVARIANT while ( 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.

Two diversions Prove that the sequence p = a; a = b; b = p; exchanges the values of a and b : { a == A & b == B } { 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 }

Two diversions (2) Discover and PROVE the behaviour of the following sequence of statements for integer variables x, y: x = x + y; y = x - y; x = x - y;

Two diversions (3) {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 }

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 }

GCD (2) 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 (3) When the loop stops, we get a == 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 (4) 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 be a > 0 & b > 0 & GCD(X, Y) == GCD(a, b) & a == b and this will imply GCD( X, Y ) == a

GCD (5) 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; else b = b - a; {a > 0 & b > 0 & GCD(X, Y) == GCD(a, b)}

GCD (6) 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 (7) 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 (8) 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 (9) 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

The "if" statement  &     ASSUME THAT {  &  } S {  } and  &     CONCLUDE THAT { } if (  ) S { }

An example with "if" 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 ) }

Minimum 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 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 (2) { m == min( 1 <= i & i <= k : ai ) & k != 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 (3) 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 (4) 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 (5) 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 (6) 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

Examples Yet another "while" loop { B > 0 }  FOR TERMINATION b = 1; p = A; while ( b != B ) { b = b + 1; p = p * A; } { ??? }

Yet another "while" loop (2) Examples Yet another "while" loop (2) { B > 0 & 1 == 1 & A == A}  FOR TERMINATION b = 1; p = A; { b == 1 & p == A }  { p == A ** b }  INVARIANT while ( 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.

Another example with "if" Examples Another example with "if" { N > 0 }  FOR TERMINATION k = 1; while ( k != N ) { if ( Ak > Ak+1 ) { p = Ak; Ak = Ak+1; Ak+1 = p; } k = k + 1; } { ??? }

Another example with "if" (2) Examples Another example with "if" (2) { N > 0 }  FOR TERMINATION k = 1; { Ak == max( 1 <= i & i <= k: Ai ) }  INVARIANT while ( 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" (3) Examples Another example with "if" (3) {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 } 