```-----------------------------------------------------------------------------
-- |
-- Module    : Documentation.SBV.Examples.ProofTools.Fibonacci
-- Copyright : (c) Levent Erkok
-- Maintainer: erkokl@gmail.com
-- Stability : experimental
--
-- Example inductive proof to show partial correctness of the for-loop
-- based fibonacci algorithm:
--
-- @
--     i = 0
--     k = 1
--     m = 0
--     while i < n:
--        m, k = k, m + k
--        i++
-- @
--
-- We do the proof against an axiomatized fibonacci implementation using an
-- uninterpreted function.
-----------------------------------------------------------------------------

{-# LANGUAGE DeriveAnyClass        #-}
{-# LANGUAGE DeriveFoldable        #-}
{-# LANGUAGE DeriveGeneric         #-}
{-# LANGUAGE DeriveTraversable     #-}
{-# LANGUAGE FlexibleInstances     #-}
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE NamedFieldPuns        #-}

module Documentation.SBV.Examples.ProofTools.Fibonacci where

import Data.SBV
import Data.SBV.Tools.Induction
import Data.SBV.Control

import GHC.Generics hiding (S)

-- * System state

-- | System state. We simply have two components, parameterized
-- over the type so we can put in both concrete and symbolic values.
data S a = S { i :: a, k :: a, m :: a, n :: a }
deriving (Show, Mergeable, Generic, Functor, Foldable, Traversable)

-- | 'Fresh' instance for our state
instance Fresh IO (S SInteger) where
fresh = S <\$> freshVar_ <*> freshVar_ <*> freshVar_ <*> freshVar_

-- | Encoding partial correctness of the sum algorithm. We have:
--
-- >>> fibCorrect
-- Q.E.D.
--
-- NB. In my experiments, I found that this proof is quite fragile due
-- to the use of quantifiers: If you make a mistake in your algorithm
-- or the coding, z3 pretty much spins forever without finding a counter-example.
-- However, with the correct coding, the proof is almost instantaneous!
fibCorrect :: IO (InductionResult (S Integer))
fibCorrect = induct chatty setup initial trans strengthenings inv goal
where -- Set this to True for SBV to print steps as it proceeds
-- through the inductive proof
chatty :: Bool
chatty = False

-- Declare fib as un uninterpreted function:
fib :: SInteger -> SInteger
fib = uninterpret "fib"

-- We setup to axiomatize the textbook definition of fib in SMT-Lib
setup :: Symbolic ()
setup = do constrain \$ fib 0 .== 0
constrain \$ fib 1 .== 1

-- This is unfortunate; but SBV currently does not support
-- adding quantified constraints in the query mode. So we
-- have to write this axiom in SMT-Lib. Note also how carefully
-- we've chosen this axiom to work with our proof!
addAxiom "fib_n" [ "(assert (forall ((x Int))"
, "                (= (fib (+ x 2)) (+ (fib (+ x 1)) (fib x)))))"
]

-- Initialize variables
initial :: S SInteger -> SBool
initial S{i, k, m, n} = i .== 0 .&& k .== 1 .&& m .== 0 .&& n .>= 0

-- We code the algorithm almost literally in SBV notation:
trans :: S SInteger -> [S SInteger]
trans st@S{i, k, m, n} = [ite (i .< n)
st { i = i + 1, k = m + k, m = k }
st
]

-- No strengthenings needed for this problem!
strengthenings :: [(String, S SInteger -> SBool)]
strengthenings = []

-- Loop invariant: @i@ remains at most @n@, @k@ is @fib (i+1)@
-- and @m@ is fib(i)@:
inv :: S SInteger -> SBool
inv S{i, k, m, n} =    i .<= n
.&& k .== fib (i+1)
.&& m .== fib i

-- Final goal. When the termination condition holds, the value @m@
-- holds the @n@th fibonacc number. Note that SBV does not prove the
-- termination condition; it simply is the indication that the loop
-- has ended as specified by the user.
goal :: S SInteger -> (SBool, SBool)
goal S{i, m, n} = (i .== n, m .== fib n)
```