sbv-3.5: SMT Based Verification: Symbolic Haskell theorem prover using SMT solving.

Copyright(c) Levent Erkok
Safe HaskellNone



Illustrates the use of sBranch, as a means of dealing with certain cases of the symbolic-termination problem.



bitCount :: SWord32 -> SWord8 Source

A fast implementation of population-count. Note that SBV already provides this functionality via sbvPopCount, using simple expansion and counting algorithm. sbvPopCount is linear in the size of the input, i.e., a 32-bit word would take 32 additions. This implementation here is faster in the sense that it takes as many additions as there are set-bits in the given word.

Of course, the issue is that this definition is recursive, and the usual definition via ite would never symbolically terminate: Recursion is done on the input argument: In each recursive call, we reduce the value n to n .&. (n-1). This eliminates one set-bit in the input. However, this claim is far from obvious. By the use of sBranch we tell SBV to call the SMT solver in each test to ensure we only evaluate the branches we need, thus avoiding the symbolic-termination issue. In a sense, the SMT solvers proves that the implementation terminates for all valid inputs.

Note that replacing sBranch in this implementation with ite would cause symbolic-termination to loop forever. Of course, this does not mean that sBranch is fast: It is costly to make external calls to the solver for each branch, so use with care.

prop :: IO ThmResult Source

Prove that the bitCount function implemented here is equivalent to the internal "slower" implementation. We have:

>>> prop

path :: SWord8 -> SWord8 Source

Illustrates the use of path-conditions in avoiding infeasible paths in symbolic simulation. If we used ite instead of sBranch in the else-branch of the implementation of path symbolic simulation would have encountered the error call, and hence would have failed. But sBranch keeps track of the path condition, and can successfully determine that this path will never be taken, and hence avoids the problem. Note that we can freely mix/match ite and sBranch calls; path conditions will be tracked in both cases. In fact, use of ite is advisable if we know for a fact that both branches are feasible, as it avoids the external call. sBranch will have the same result, albeit it'll cost more.

pathCheck :: IO ThmResult Source

Prove that path always produces either 10 or 20, i.e., symbolic simulation will not fail due to the error call. We have:

>>> pathCheck

Were we to use ite instead of sBranch in the implementation of path, this expression would have caused an exception to be raised at symbolic simulation time.