COP4020 Programming Languages

COP4020Programming Languages Introduction to Axiomatic Semantics Prof. Robert van Engelen

Assertions and Preconditions • Assertions are used by programmers to verify run-time execution • An assertion is a logical formula to verify the state of variables and data to ensure safe continuation • A failed assertion should stop the program • Assertions are placed by programmers explicitly in codeassert (len>0); mean = sum/len; • By contrast, preconditions state the initial conditions under which an algorithm has been proven correct COP4020 Fall 2006

Preconditions, Postconditions and Partial Correctness • We will place assertions before and after a commandC: { Precondition } C { Postcondition } • We say that the command C is partially correct with respect to the <precondition,postcondition> specification, provided that • The command C is executed with values that make the precondition true • If the command terminates, then the resulting values make the postcondition true • Total correctness requires termination COP4020 Fall 2006

Assignment Axiom • If we view assertions as predicates, the assignment axiom can be stated { P(E) } V := E { P(V) }that is, if we state a property of V after the assignment, then the property must hold for expression E before the assignment • We can use substitution to derive the precondition given a postcondition formula P: this is the assignment axiom: { P[VE] } V := E { P }where P[VE] denotes the substitution of V by E in P COP4020 Fall 2006

Examples for Assignments • { k = 5 } k := k + 1 { k = 6 }(k = 6)[kk+1]  (k+1 = 6)  (k = 5) • { j = 3 and k = 4 } j := j + k { j = 7 and k = 4 }(j = 7 and k = 4)[jj+k]  (j+k = 7 and k = 4)  (j = 3 and k = 4) • { true } x := 2 { x = 2 }(x = 2)[x2]  (2 = 2)  (true) • { a > 0 } a := a - 1 { a> 0 }(a> 0)[aa - 1]  (a - 1 > 0)  (a > 0) Assuming a is int ! • { false } y := 1 { y = 2 } No state can satisfy precondition ! = partially correct COP4020 Fall 2006

Validity of Assignment Axiom • At first glance it seems that working backward from a postcondition is more complicated than necessary and we could use { true } V := E { V = E } • However, consider { true } m := m + 1 { m = m + 1}and we find that { m = m + 1 }  { false } assignment equality COP4020 Fall 2006

Statement Composition: Sequence Axiom • The sequence axiom: { P } C1 { Q } C2 { R }Q is a postcondition of C1 and a precondition for C2 • Written as a rule of inference { P } C1 { Q } { Q } C2 { R } { P } C1 ; C2 { R } COP4020 Fall 2006

Example Sequencing • We usually write the sequencing vertical and insert the assertions between the statements:{ i> 0 }k := i + 1;{ k > 0 and j = j }  { k > 0 }i := j;{ k > 0 and i = j } • The rule of inference: { i> 0 } k := i + 1 { k > 0 } { k > 0 } i := j { k > 0 and i = j } { i> 0 } k := i + 1; i := j { k > 0 and i = j } (k > 0)[ki + 1] (k > 0 and i = j)[ij] COP4020 Fall 2006

Skip Axiom • The ‘skip’ statement is a no-op{ P } skip { P }pre- and postconditions are identical COP4020 Fall 2006

If-then-else Axiom • The if-then-else axiom written vertically:{ P }if B then { P and B }C1 { Q }else { P and not B }C2 { Q }end if{ Q } COP4020 Fall 2006

If-then-else Axiom • And as an inference rule: { P and B } C1 { Q } { P and not B } C2 { Q } { P } if B then C1 else C2 end if { Q } COP4020 Fall 2006

The if-then-else Weakest Precondition Rule • We can derive the weakest preconditionP of and if-then-else using:P (not B or P1) and (B or P2)where P1 is the precondition of C1 given postcondition Q and P2 is the precondition of C2 given postcondition Q • Example:{ ( x < 0 or x > 0) and (x > 0 or true) }  { true }if x > 0 then { x > 0 } y := xelse { 0 > 0 }  { true } y := 0end if{ y > 0 } Compute preconditionsP1 and P2 of C1 and C2 COP4020 Fall 2006

Precondition Strengthening • Logical implication ( or ) means stronger condition weaker condition(more restrictive) (less restrictive) • For example: • x = y and y = 0  y = 0 • x  0  x = 0 or x < 0 or x > 0 • x = 0  x> 0 • x = y  true • false x = y2 COP4020 Fall 2006

Using Precondition Strengthening • We can always make a precondition stronger than necessary to complete a proof • For example, suppose we know that x > 0 and y = 2 at the start of the program:{ x > 0 and y = 2} { x > 0}y := x{ y = x and y > 0 } (y = x and y > 0)[yx] (x = x and x > 0) COP4020 Fall 2006

Loops and Loop Invariants • A loop-invariant condition is a logical formula that is true before the loop, in the loop, and after the loop • An common example: grocery shopping • The invariant is:groceries needed = groceries on list + groceries in cart cart := empty;{ groceries needed = groceries on list + groceries in cart }{ groceries needed = groceries on list }while grocery list not empty do { groceries needed = groceries on list + groceries in cart and not empty list } add grocery to cart; take grocery off list; { groceries needed = groceries on list + groceries in cart }end do;{ groceries needed = groceries on list + groceries in cart and empty list } { groceries needed = groceries in cart } COP4020 Fall 2006

While-loop Axiom • The while-loop axiom uses a loop invariant I, which must be determined • Invariant cannot generally be automatically computed and must be “guessed” by an experienced programmer{ I }while B do { I and B } C { I }end do{ I and not B } COP4020 Fall 2006

While-loop Example (1) • Loop invariant I (f*k! = n! and k> 0){ n> 0 }k := n;f := 1;while k > 0 dof := f*k;k := k-1;end do{ f = n! } Proof that this algorithm is correct given precondition n>0 and postcondition f=n! COP4020 Fall 2006

While-loop Example (2) • Loop invariant I (f*k! = n! and k> 0){ n> 0 }k := n;f := 1;{ f*k! = n! and k> 0 }while k > 0 do{ f*k! = n! and k> 0 and k > 0 }f := f*k;k := k-1;{ f*k! = n! and k> 0 }end do{ f*k! = n! and k> 0 and k< 0 }{ f = n! } Add while-loop preconditions and postconditions based on the invariant COP4020 Fall 2006

While-loop Example (3) • Loop invariant I (f*k! = n! and k> 0){ n> 0 }k := n;{ 1*k! = n! and k> 0 }f := 1;{ f*k! = n! and k> 0 }while k > 0 do { f*k! = n! and k> 0 and k > 0 }f := f*k;{ f*(k-1)! = n! and k-1 > 0 }k := k-1; { f*k! = n! and k> 0 }end do{ f*k! = n! and k> 0 and k< 0 } { f = n! } Use assignment axioms COP4020 Fall 2006

While-loop Example (4) • Loop invariant I (f*k! = n! and k> 0){ n> 0 }{ n! = n! and n> 0 }k := n;{ 1*k! = n! and k> 0 }f := 1;{ f*k! = n! and k> 0 }while k > 0 do { f*k! = n! and k> 0 and k > 0 }{ f*k*(k-1)! = n! and k-1 > 0 }f := f*k;{ f*(k-1)! = n! and k-1 > 0 }k := k-1; { f*k! = n! and k> 0 }end do{ f*k! = n! and k> 0 and k< 0 } { f = n! } Use assignment axioms COP4020 Fall 2006

While-loop Example (5) • Loop invariant I (f*k! = n! and k> 0){ n> 0 } { n! = n! and n> 0 }k := n;{ 1*k! = n! and k> 0 }f := 1;{ f*k! = n! and k> 0 }while k > 0 do { f*k! = n! and k> 0 and k > 0 }  { f*k*(k-1)! = n! and k-1 > 0 } f := f*k; { f*(k-1)! = n! and k-1 > 0 } k := k-1; { f*k! = n! and k> 0 }end do{ f*k! = n! and k> 0 and k< 0 } { f = n! } Use precondition strengthening to prove the correctness of implications COP4020 Fall 2006

While-loop Example (6) • Loop invariant I (f*k! = n! and k> 0){ n> 0 } { n! = n! and n> 0 }k := n;{ 1*k! = n! and k> 0 }f := 1;{ f*k! = n! and k> 0 }while k > 0 do { f*k! = n! and k> 0 and k > 0 }  { f*k*(k-1)! = n! and k-1 > 0 } f := f*k; { f*(k-1)! = n! and k-1 > 0 } k := k-1; { f*k! = n! and k> 0 }end do{ f*k! = n! and k> 0 and k< 0 } { f*k! = n! and k = 0 } { f = n! } Use simplification and logical implications to complete the proof COP4020 Fall 2006

Specifications • A postcondition specification can by any logical formula • A specification that states the input-output requirements of an algorithm is needed to prove correctness • A specification that tests a violation can aid in debugging • For example (precondition strengthening is disallowed): { (n > 0 or false) and (n < 0 or n = 0) }  { false }if (n < 0) { false } p = 2;else { n = 0 } p = n+1;{ p = 1 }k = m / (p-1); if (n < 0) p = 2;else p = n+1;k = m / (p-1);// Error when p = 1 Means: never possible COP4020 Fall 2006

COP4020 Programming Languages