Prove that floating point addition is not associative. For illustration purposes, we will require one of the inputs to be a NaN. We have:

>>> prove $ assocPlus (0/0)
Falsifiable. Counter-example:
  s0 = 0.0 :: Float
  s1 = 0.0 :: Float

Indeed:

>>> let i = 0/0 :: Float
>>> i + (0.0 + 0.0)
NaN
>>> ((i + 0.0) + 0.0)
NaN

But keep in mind that NaN does not equal itself in the floating point world! We have:

>>> let nan = 0/0 :: Float in nan == nan
False

Prove that addition is not associative, even if we ignore NaN/Infinity values. To do this, we use the predicate fpIsPoint, which is true of a floating point number (SFloat or SDouble) if it is neither NaN nor Infinity. (That is, it's a representable point in the real-number line.)

We have:

>>> assocPlusRegular
Falsifiable. Counter-example:
  x =  3.7634227e-37 :: Float
  y = -3.7612938e-37 :: Float
  z = -1.1036833e-38 :: Float

Indeed, we have:

>>> let  x =  3.7634227e-37 :: Float
>>> let  y = -3.7612938e-37 :: Float
>>> let  z = -1.1036833e-38 :: Float
>>> x + (y + z)
-1.0823943e-38
>>> (x + y) + z
-1.0823947e-38

Note the difference between two additions!

Demonstrate that a+b = a does not necessarily mean b is 0 in the floating point world, even when we disallow the obvious solution when a and b are Infinity. We have:

>>> nonZeroAddition
Falsifiable. Counter-example:
  a =  -4.611686e18 :: Float
  b = 1.3552526e-20 :: Float

Indeed, we have:

>>> (-4.611686e18 + 1.3552526e-20) == (-4.611686e18 :: Float)
True

But:

>>> 1.3552526e-20 == (0 :: Float)
False

This example illustrates that a * (1/a) does not necessarily equal 1. Again, we protect against division by 0 and NaN/Infinity.

We have:

>>> multInverse
Falsifiable. Counter-example:
  a = 8.988465676670122e307 :: Double

Indeed, we have:

>>> let a = 8.988465676670122e307 :: Double
>>> a * (1/a)
1.0000000000000002

One interesting aspect of floating-point is that the chosen rounding-mode can effect the results of a computation if the exact result cannot be precisely represented. SBV exports the functions fpAdd, fpSub, fpMul, fpDiv, fpFMA and fpSqrt which allows users to specify the IEEE supported RoundingMode for the operation. This example illustrates how SBV can be used to find rounding-modes where, for instance, addition can produce different results. We have:

>>> roundingAdd
Satisfiable. Model:
  rm = RoundTowardPositive :: RoundingMode
  x  =           255.96877 :: Float
  y  =             255.875 :: Float

(Note that depending on your version of Z3, you might get a different result.) Unfortunately we can't directly validate this result at the Haskell level, as Haskell only supports RoundNearestTiesToEven. We have:

>>> 255.96877 + 255.875 :: Float
511.84375

While we cannot directly see the result when the mode is RoundTowardPositive in Haskell, we can use SBV to provide us with that result thusly:

>>> sat $ \z -> z .== fpAdd sRoundTowardPositive 255.96877 (255.875 :: SFloat)
Satisfiable. Model:
  s0 = 511.84378 :: Float

We can see why these two resuls are indeed different: The RoundTowardPositive (which rounds towards positive-infinity) produces a larger result. Indeed, if we treat these numbers as Double values, we get:

>>> 255.96877 + 255.875 :: Double
511.84377

we see that the "more precise" result is larger than what the Float value is, justifying the larger value with RoundTowardPositive. A more detailed study is beyond our current scope, so we'll merely -- note that floating point representation and semantics is indeed a thorny subject, and point to http://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/02Numerics/Double/paper.pdf as an excellent guide.