Merge Sort with Leftovers, Correctly

Merge Sort with Leftovers

This is the fourth blog post in a series about developing correct implementations of basic data structures and algorithms using the Deduce language and proof checker.

In this blog post we study a fast sorting algorithm, Merge Sort. This classic algorithm splits the input list in half, recursively sorts each half, and then merges the two results back into a single sorted list.

The specification of Merge Sort is the same as Insertion Sort.

Specification: The merge_sort(xs) function returns a list that contains the same elements as xs but the elements in the result are in sorted order.

We follow the write-test-prove approach to develop a correct implementation of merge_sort.

Write the merge_sort function

The classic implementation of merge_sort would be something like the following.

function merge_sort(List<Nat>) -> List<Nat> {
  merge_sort(empty) = empty
  merge_sort(node(x,xs')) =
    let p = split(node(x,xs'))
    merge(merge_sort(first(p)), merge_sort(second(p)))
}

Unfortunately, Deduce rejects the above function definition because Deduce uses a very simple restriction to ensure the termination of recursive function, which is that a recursive call may only be made on a part of the input. In this case, the recursive call may only be applied to the sublist xs', not first(p) or second(p).

How can we work around this restriction? There’s an old trick that goes by many names (gas, fuel, etc.), which is to add another parameter of type Nat and use that for termination. Let us use the name msort for the following, and then we define merge_sort in terms of msort.

function msort(Nat, List<Nat>) -> List<Nat> {
  msort(0, xs) = xs
  msort(suc(n'), xs) =
    let p = split(xs)
    merge(msort(n', first(p)), msort(n', second(p)))
}

define merge_sort : fn List<Nat> -> List<Nat>
  = λxs{ msort(log(length(xs)), xs) }

In the above definition of merge_sort, we need to suppply enough gas so that msort won’t prematurely run out. Here we use the logarithm (base 2, rounding up) defined in Log.pf.

This definition of merge_sort and msort is fine, it has O(n log(n)) time complexity, so it is efficient. However, the use of split rubs me the wrong way because it requires traversing half of the input list. The use of split is necessary if one wanted to use parallelism to speed up the code, performing the two recursive calls in parallel. However, we are currently only interested in a single-threaded implementation.

Suppose you just finished baking a pie and intend to eat half now and half tomorrow night. One approach would be to split it in half and then eat one of the halves. Another approach is to just start eating the pie and stop when half of it is gone. That’s the approach that we will take with the next version of msort.

Specification The msort(n,xs) function sorts the first min(2ⁿ,length(xs)) many elements of xs and returns a pair containing (1) the sorted list and (2) the leftovers that were not yet sorted.

function msort(Nat, List<Nat>) -> Pair< List<Nat>, List<Nat> > {
  msort(0, xs) =
    switch xs {
      case empty { pair(empty, empty) }
      case node(x, xs') { pair(node(x, empty), xs') }
    }
  msort(suc(n'), xs) =
    let p1 = msort(n', xs)
    let p2 = msort(n', second(p1))
    let ys = first(p1)
    let zs = first(p2)
    pair(merge(length(ys) + length(zs), ys, zs), second(p2))
}

In the above case for suc(n'), the first recursive call to msort produces the pair p1 that includes a sorted list and the leftovers. We sort the leftovers with the second recursive call to msort. We return (1) the merge of the two sorted sublists and (2) the leftovers from the second recursive call to msort.

With the code for msort complete, we can turn to merge_sort. Similar to the previous version, we involke msort with the input list xs and use the logarithm of list length for the gas. This msort returns a pair, with the sorted results in the first component. The second component of the pair is an empty list because we supplied enough gas.

define merge_sort : fn List<Nat> -> List<Nat>
    = λxs{ first(msort(log(length(xs)), xs)) }

So far, we have neglected the implementation of merge. Here’s its specification.

Specification: The merge(xs,ys) function takes two sorted lists and returns a sorted list that contains just the elements from the two input lists.

Here’s the classic implementation of merge. The idea is to compare the two elements at the front of each list and use the lower of the two as the first element of the output. Then do the recursive call with the two lists, minus the element that was chosen. Again, we use an extra gas parameter to ensure termination. To ensure that we have enough gas, we will choose the sum of the lengths of the two input lists.

function merge(Nat, List<Nat>, List<Nat>) -> List<Nat> {
  merge(0, xs, ys) = empty
  merge(suc(n), xs, ys) =
    switch xs {
      case empty { ys }
      case node(x, xs') {
        switch ys {
          case empty {
            node(x, xs')
          }
          case node(y, ys') {
            if x ≤ y then
              node(x, merge(n, xs', node(y, ys')))
            else
              node(y, merge(n, node(x, xs'), ys'))
          }
        }
     }
   }
}

Test

We have three functions to test, merge, msort and merge_sort.

Test merge

We test that the result of merge is sorted and that it contains all the elements from the two input lists, which we check using count.

define L_1337 = node(1, node(3, node(3, node(7, empty))))
define L_2348 = node(2, node(3, node(4, node(8, empty))))
define L_12333478 = merge(length(L_1337) + length(L_2348), L_1337, L_2348)
assert sorted(L_12333478)
assert all_elements(append(L_1337, L_2348),
  λx{count(L_1337)(x) + count(L_2348)(x) = count(L_12333478)(x) })

Test msort

In the following tests, we vary the gas from 0 to 3, varying how much of the input list L18 gets sorted in the call to msort. The take(n,xs) function returns the first n elements of xs and drop(n,xs) drops the first n elements of xs and returns the remaining portion of xs.

define L18 = append(L_1337, L_2348)

define p0 = msort(0, L18)
define t0 = take(pow2(0), L18)
define d0 = drop(pow2(0), L18)
assert sorted(first(p0))
assert all_elements(t0, λx{count(t0)(x) = count(first(p0))(x) })
assert all_elements(d0, λx{count(d0)(x) = count(second(p0))(x) })

define p1 = msort(1, L18)
define t1 = take(pow2(1), L18)
define d1 = drop(pow2(1), L18)
assert sorted(first(p1))
assert all_elements(t1, λx{count(t1)(x) = count(first(p1))(x) })
assert all_elements(d1, λx{count(d1)(x) = count(second(p1))(x) })

define p2 = msort(2, L18)
define t2 = take(pow2(2), L18)
define d2 = drop(pow2(2), L18)
assert sorted(first(p2))
assert all_elements(t2, λx{count(t2)(x) = count(first(p2))(x) })
assert all_elements(d2, λx{count(d2)(x) = count(second(p2))(x) })

define p3 = msort(3, L18)
define t3 = take(pow2(3), L18)
define d3 = drop(pow2(3), L18)
assert sorted(first(p3))
assert all_elements(t3, λx{count(t3)(x) = count(first(p3))(x) })
assert all_elements(d3, λx{count(d3)(x) = count(second(p3))(x) })

Test merge_sort

Next we test that merge_sort returns a sorted list that contains the same elements as the input list. For input, we reuse the list L18 from above.

define s_L18 = merge_sort(L18)
assert sorted(s_L18)
assert all_elements(t0, λx{count(L18)(x) = count(s_L18)(x) })

We can bundle several tests, with varying-length inputs, into one assert by using all_elements and interval.

assert all_elements(interval(3, 0),
    λn{ let xs = reverse(interval(n, 0))
        let ls = merge_sort(xs)
        sorted(ls) and
        all_elements(xs, λx{count(xs)(x) = count(ls)(x)})
    })

Prove

Compared to the proof of correctness for insertion_sort, we have considerably more work to do for merge_sort. Instead of two functions, we have three functions to consider: merge, msort, and merge_sort. Furthermore, these functions are more complex than insert and insertion_sort. Nevertheless, we are up to the challenge!

Prove correctness of merge

The specificaiton of merge has two parts, one part saying that the elements of the output must be the elements of the two input lists, and the another part saying that the output must be sorted, provided the two input lists are sorted.

Here is how we state the theorem for the first part.

theorem mset_of_merge: all n:Nat. all xs:List<Nat>, ys:List<Nat>.
  if length(xs) + length(ys) = n
  then mset_of(merge(n, xs, ys)) = mset_of(xs) ⨄ mset_of(ys)

Here is the theorem stating that the output of merge is sort.

theorem merge_sorted: all n:Nat. all xs:List<Nat>, ys:List<Nat>.
  if sorted(xs) and sorted(ys) and
     length(xs) + length(ys) = n
  then sorted(merge(n, xs, ys))

Prove the mset_of_merge theorem

We begin with the proof of mset_of_merge. Because merge(n, xs, ys) is recursive on the natural number n, we proceed by induction on Nat.

  induction Nat
  case 0 {
    arbitrary xs:List<Nat>, ys:List<Nat>
    suppose prem: length(xs) + length(ys) = 0
    ?
  }
  case suc(n') suppose IH {
    ?
  }

In the case for n = 0, we need to prove

  mset_of(merge(0,xs,ys)) = mset_of(xs) ⨄ mset_of(ys)

and merge(0,xs,ys) returns empty, so we need to show that mset_of(xs) ⨄ mset_of(ys) is the empty multiset. From the premise prem, both xs and ys must be empty.

  have lxs_lys_z: length(xs) = 0 and length(ys) = 0
    by apply add_to_zero[length(xs)][length(ys)] to prem
  have xs_mt: xs = empty
    by apply length_zero_empty[Nat,xs] to lxs_lys_z
  have ys_mt: ys = empty
    by apply length_zero_empty[Nat,ys] to lxs_lys_z

After rewriting with those equalities and applying the definition of merge and mset_of:

  rewrite xs_mt | ys_mt
  definition {merge, mset_of}

it remains to prove m_fun(λ{0}) = m_fun(λ{0}) ⨄ m_fun(λ{0}) (the sum of two empty multisets is the empty multiset), which we prove with the theorem m_sum_empty from MultiSet.pf.

  symmetric m_sum_empty[Nat, m_fun(λx{0}) :MultiSet<Nat>]

In the case for n = suc(n'), we need to prove

  mset_of(merge(suc(n'),xs,ys)) = mset_of(xs) ⨄ mset_of(ys)

Looking a the suc clause of merge, there is a switch on xs and then on ys. So our proof will be structured analogously.

  switch xs {
    case empty {
      ?
    }
    case node(x, xs') suppose xs_xxs {
      ?
    }
  }

In the case for xs = empty, we conclude simply by use of the definitions of merge and mset_of and the fact that combining mset_of(ys) with the empty multiset produces mset_of(ys).

  case empty {
    definition {merge, mset_of}
    conclude mset_of(ys) = m_fun(λx{0}) ⨄ mset_of(ys)
      by symmetric empty_m_sum[Nat, mset_of(ys)]
  }

In the case for xs = node(x, xs'), merge performs a switch on ys, so our proof does too.

  switch ys {
    case empty {
      ?
    }
    case node(y, ys') suppose ys_yys {
      ?
    }

The case for ys = empty, is similar to the case for xs = empty. We conclude by use of the definitions of merge and mset_of and the fact that combining mset_of(ys) with the empty multiset produces mset_of(ys).

  definition {merge, mset_of}
  conclude m_one(x) ⨄ mset_of(xs')
         = m_one(x) ⨄ mset_of(xs') ⨄ m_fun(λ{0})
    by rewrite m_sum_empty[Nat, m_one(x) ⨄ mset_of(xs')].

In the case for ys = node(y, ys'), we continue to follow the structure of merge and switch on x ≤ y.

  definition merge
  switch x ≤ y {
    case true suppose xy_true {
      ?
    }
    case false suppose xy_false {
      ?
    }
  }

In the case for (x ≤ y) = true, the goal becomes

m_one(x) ⨄ mset_of(merge(n',xs',node(y,ys'))) 
= m_one(x) ⨄ mset_of(xs') ⨄ m_one(y) ⨄ mset_of(ys')

Which follows from the induction hypothesis instantiated with xs' and node(y,ys').

  mset_of(merge(n',xs',node(y,ys')))
= mset_of(xs') ⨄ mset_of(node(y, ys'))

Filling in the details, we prove this case as follows.

  case true suppose xy_true {
    definition mset_of
    have sxs_sys_sn: suc(length(xs')) + suc(length(ys')) = suc(n')
      by enable length rewrite xs_xxs | ys_yys in prem
    have len_xs_yys: length(xs') + length(node(y,ys')) = n'
      by enable {operator +,length}
         injective suc sxs_sys_sn
    have IH': mset_of(merge(n',xs',node(y,ys')))
            = mset_of(xs') ⨄ mset_of(node(y, ys'))
      by apply IH[xs', node(y, ys')] to len_xs_yys
    rewrite IH'
    definition mset_of
    rewrite m_sum_assoc[Nat, m_one(x), mset_of(xs'),
                        (m_one(y) ⨄ mset_of(ys'))].
  }

In the case for (x ≤ y) = false, the goal becomes

  m_one(y) ⨄ mset_of(merge(n',node(x,xs'),ys')) 
= m_one(x) ⨄ mset_of(xs') ⨄ m_one(y) ⨄ mset_of(ys')

The induction hypothesis instantiated with node(x,xs') and ys' is

  mset_of(merge(n',node(x,xs'),ys'))
= mset_of(node(x,xs')) ⨄ mset_of(ys')

So the goal follows from the fact that multiset sum is associative and commutative.

theorem mset_of_merge: all n:Nat. all xs:List<Nat>, ys:List<Nat>.
  if length(xs) + length(ys) = n
  then mset_of(merge(n, xs, ys)) = mset_of(xs) ⨄ mset_of(ys)
proof
  induction Nat
  case 0 {
    arbitrary xs:List<Nat>, ys:List<Nat>
    suppose prem: length(xs) + length(ys) = 0
    have lxs_lys_z: length(xs) = 0 and length(ys) = 0
      by apply add_to_zero[length(xs)][length(ys)] to prem
    have xs_mt: xs = empty
      by apply length_zero_empty[Nat,xs] to lxs_lys_z
    have ys_mt: ys = empty
      by apply length_zero_empty[Nat,ys] to lxs_lys_z
    rewrite xs_mt | ys_mt
    definition {merge, mset_of}
    symmetric m_sum_empty[Nat, m_fun(λx{0}) :MultiSet<Nat>]
  }
  case suc(n') suppose IH {
    arbitrary xs:List<Nat>, ys:List<Nat>
    suppose prem: length(xs) + length(ys) = suc(n')
    switch xs {
      case empty {
        definition {merge, mset_of}
        conclude mset_of(ys) = m_fun(λx{0}) ⨄ mset_of(ys)
          by symmetric empty_m_sum[Nat, mset_of(ys)]
      }
      case node(x, xs') suppose xs_xxs {
        switch ys {
          case empty {
            definition {merge, mset_of}
            conclude m_one(x) ⨄ mset_of(xs')
                   = m_one(x) ⨄ mset_of(xs') ⨄ m_fun(λ{0})
              by rewrite m_sum_empty[Nat, m_one(x) ⨄ mset_of(xs')].
          }
          case node(y, ys') suppose ys_yys {
            definition merge
            switch x ≤ y {
              case true suppose xy_true {
                definition mset_of
                have sxs_sys_sn: suc(length(xs')) + suc(length(ys')) = suc(n')
                  by enable length rewrite xs_xxs | ys_yys in prem
                have len_xs_yys: length(xs') + length(node(y,ys')) = n'
                  by enable {operator +,length}
                     injective suc sxs_sys_sn
                have IH': mset_of(merge(n',xs',node(y,ys')))
                        = mset_of(xs') ⨄ mset_of(node(y, ys'))
                  by apply IH[xs', node(y, ys')] to len_xs_yys
                rewrite IH'
                definition mset_of
                rewrite m_sum_assoc[Nat, m_one(x), mset_of(xs'),
                                    (m_one(y) ⨄ mset_of(ys'))].
              }
              case false suppose xy_false {
                definition mset_of
                have sxs_sys_sn: suc(length(xs')) + suc(length(ys')) = suc(n')
                  by enable length rewrite xs_xxs | ys_yys in prem
                have len_xxs_ys: length(node(x,xs')) + length(ys') = n'
                  by enable {operator +,length}
                     injective suc
                     rewrite add_suc[length(xs')][length(ys')] in
                     sxs_sys_sn
                have IH': mset_of(merge(n',node(x,xs'),ys'))
                        = mset_of(node(x,xs')) ⨄ mset_of(ys')
                  by apply IH[node(x,xs'), ys'] to len_xxs_ys
                equations
                        m_one(y) ⨄ mset_of(merge(n',node(x,xs'),ys'))
                      = m_one(y) ⨄ ((m_one(x) ⨄ mset_of(xs')) ⨄ mset_of(ys'))
                      by rewrite IH' definition mset_of.
                  ... = m_one(y) ⨄ (m_one(x) ⨄ (mset_of(xs') ⨄ mset_of(ys')))
                      by rewrite m_sum_assoc[Nat, m_one(x), mset_of(xs'),
                                             mset_of(ys')].
                  ... = (m_one(y) ⨄ m_one(x)) ⨄ (mset_of(xs') ⨄ mset_of(ys'))
                      by rewrite m_sum_assoc[Nat, m_one(y), m_one(x),
                               (mset_of(xs') ⨄ mset_of(ys'))].
                  ... = (m_one(x) ⨄ m_one(y)) ⨄ (mset_of(xs') ⨄ mset_of(ys'))
                      by rewrite m_sum_commutes[Nat, m_one(x), m_one(y)].
                  ... = m_one(x) ⨄ (m_one(y) ⨄ (mset_of(xs') ⨄ mset_of(ys')))
                      by rewrite m_sum_assoc[Nat, m_one(x), m_one(y),
                          (mset_of(xs') ⨄ mset_of(ys'))].
                  ... = m_one(x) ⨄ ((m_one(y) ⨄ mset_of(xs')) ⨄ mset_of(ys'))
                      by rewrite m_sum_assoc[Nat, m_one(y), mset_of(xs'),
                          mset_of(ys')].
                  ... = m_one(x) ⨄ ((mset_of(xs') ⨄ m_one(y)) ⨄ mset_of(ys'))
                      by rewrite m_sum_commutes[Nat, m_one(y), mset_of(xs')].
                  ... = m_one(x) ⨄ (mset_of(xs') ⨄ (m_one(y) ⨄ mset_of(ys')))
                      by rewrite m_sum_assoc[Nat, mset_of(xs'), m_one(y),
                         mset_of(ys')].
                  ... = (m_one(x) ⨄ mset_of(xs')) ⨄ (m_one(y) ⨄ mset_of(ys'))
                      by rewrite m_sum_assoc[Nat, m_one(x), mset_of(xs'),
                          (m_one(y) ⨄ mset_of(ys'))].
              }
            }
          }
        }
      }
    }
  }
end

The mset_of_merge theorem also holds for sets, using the set_of function. We prove the following set_of_merge theorem as a corollary of mset_of_merge.

theorem set_of_merge: all xs:List<Nat>, ys:List<Nat>.
  set_of(merge(length(xs) + length(ys), xs, ys)) = set_of(xs) ∪ set_of(ys)
proof
  arbitrary xs:List<Nat>, ys:List<Nat>
  have mset_of_merge: mset_of(merge(length(xs) + length(ys), xs, ys))
                    = mset_of(xs) ⨄ mset_of(ys)
    by apply mset_of_merge[length(xs) + length(ys)][xs, ys] to .
  equations
    set_of(merge(length(xs) + length(ys), xs, ys))
        = set_of_mset(mset_of(merge(length(xs) + length(ys), xs, ys)))
          by symmetric som_mset_eq_set[Nat]
                             [merge(length(xs) + length(ys), xs, ys)]
    ... = set_of_mset(mset_of(xs)) ∪ set_of_mset(mset_of(ys))
          by rewrite mset_of_merge  som_union[Nat,mset_of(xs),mset_of(ys)]
    ... = set_of(xs) ∪ set_of(ys)
          by rewrite som_mset_eq_set[Nat][xs] | som_mset_eq_set[Nat][ys].
end

Prove the merge_sorted theorem

Next up is the merge_sorted theorem. The structure of the proof will be similar to the one for mset_of_merge, because they both follow the structure of merge. So begin with induction on Nat.

theorem merge_sorted: all n:Nat. all xs:List<Nat>, ys:List<Nat>.
  if sorted(xs) and sorted(ys) and length(xs) + length(ys) = n
  then sorted(merge(n, xs, ys))
proof
  induction Nat
  case 0 {
    ?
  }
  case suc(n') suppose IH {
    arbitrary xs:List<Nat>, ys:List<Nat>
    suppose prem
    definition merge
    switch xs {
      case empty {
        ?
      }
      case node(x, xs') suppose xs_xxs {
        switch ys {
          case empty {
            ?
          }
          case node(y, ys') suppose ys_yys {
            switch x ≤ y {
              case true suppose xy_true {
                ?
              }
              case false suppose xy_false {
                ?
              }
            }
          }
        }
      }
    }
  }
end

In the case for n = 0, we need to prove sorted(merge(0, xs, ys)). But merge(0, xs, ys) = empty, and sorted(empty) is trivially true. So we conclude the case for n = 0 as follows.

  case 0 {
    arbitrary xs:List<Nat>, ys:List<Nat>
    suppose _
    definition merge
    conclude sorted(empty) by definition sorted.
  }

We move on to the case for n = suc(n') and xs = empty. Here merge returns ys, and we already know that ys is sorted from the premise.

    case empty {
      conclude sorted(ys) by prem
    }

In the case for xs = node(x, xs') and ys = empty, the merge function returns node(x, xs') (aka. xs), and we already know that xs is sorted from the premise.

  case empty {
    conclude sorted(node(x,xs'))  by rewrite xs_xxs in prem
  }

In the case for ys = node(y, ys') and (x ≤ y) = true, the merge function returns node(x, merge(n',xs',node(y,ys'))). So we need to prove the following.

  sorted(merge(n',xs',node(y,ys'))) and
  all_elements(merge(n',xs',node(y,ys')),λb{x ≤ b})

To prove the first, we invoke the induction hypothesis intantiated to xs' and node(y,ys') as follows.

  have s_xs: sorted(xs')
    by enable sorted rewrite xs_xxs in prem
  have s_yys: sorted(node(y,ys'))
    by rewrite ys_yys in prem
  have len_xs_yys: length(xs') + length(node(y,ys')) = n'
    by enable {operator +,length}
       have sxs: suc(length(xs')) + suc(length(ys')) = suc(n')
          by rewrite xs_xxs | ys_yys in prem
       injective suc sxs
  have IH_xs_yys: sorted(merge(n',xs',node(y,ys')))
    by apply IH[xs',node(y,ys')]
       to s_xs, s_yys, len_xs_yys

It remains to prove that x is less-or-equal to to all the elements in the rest of the output list:

  all_elements(merge(n',xs',node(y,ys')),λb{x ≤ b})

The theorem all_elements_eq_member in List.pf says

  all_elements(xs,P) = (all x:T. if x ∈ set_of(xs) then P(x))

which combined with the set_of_merge corollary above, simplifies our goal to

  all z:Nat. (if z ∈ set_of(xs') ∪ set_of(node(y,ys')) then x ≤ z)

So we have a few cases to consider and need to prove x ≤ z in each one. Consider the case where z ∈ set_of(xs'). Then we can deduce x ≤ z from the fact that node(x, xs') is sorted.

  have x_le_xs: all_elements(xs', λb{x ≤ b})
    by definition sorted in rewrite xs_xxs in prem
  conclude x ≤ z by
    apply all_elements_member[Nat][xs'][z, λb{x ≤ b}]
    to x_le_xs, z_in_xs

Next, consider the case where y = z. Then we can immediately conclude because x ≤ y.

Finally, consider when z ∈ set_of(ys'). Because node(y,ys') is sorted, we know y ≤ z. Then combined with x ≤ y, we conclude that x ≤ z by transitivity.

  have y_le_ys: all_elements(ys', λb{y ≤ b})
    by definition sorted in rewrite ys_yys in prem
  have y_z: y ≤ z
    by apply all_elements_member[Nat][ys'][z,λb{y ≤ b}]
       to y_le_ys, z_in_ys
  have x_y: x ≤ y by rewrite xy_true.
  conclude x ≤ z
    by apply less_equal_trans[x][y,z] to x_y, y_z

The last case to consider is for ys = node(y, ys') and (x ≤ y) = false. The reasoning is similar to the case for (x ≤ y) = true, so we’ll skip the detailed explanation.

Here’s the completed proof of merge_sorted.

theorem merge_sorted: all n:Nat. all xs:List<Nat>, ys:List<Nat>.
  if sorted(xs) and sorted(ys) and length(xs) + length(ys) = n
  then sorted(merge(n, xs, ys))
proof
  induction Nat
  case 0 {
    arbitrary xs:List<Nat>, ys:List<Nat>
    suppose _
    definition merge
    conclude sorted(empty) by definition sorted.
  }
  case suc(n') suppose IH {
    arbitrary xs:List<Nat>, ys:List<Nat>
    suppose prem
    definition merge
    switch xs {
      case empty {
        conclude sorted(ys) by prem
      }
      case node(x, xs') suppose xs_xxs {
        switch ys {
          case empty {
            conclude sorted(node(x,xs'))  by rewrite xs_xxs in prem
          }
          case node(y, ys') suppose ys_yys {
            /* Apply the induction hypothesis
             * to prove sorted(merge(n',xs',node(y,ys')))
             */
            have s_xs: sorted(xs')
              by enable sorted rewrite xs_xxs in prem
            have s_yys: sorted(node(y,ys'))
              by rewrite ys_yys in prem
            have len_xs_yys: length(xs') + length(node(y,ys')) = n'
              by enable {operator +,length}
                 have sxs: suc(length(xs')) + suc(length(ys')) = suc(n')
                    by rewrite xs_xxs | ys_yys in prem
                 injective suc sxs
            have IH_xs_yys: sorted(merge(n',xs',node(y,ys')))
              by apply IH[xs',node(y,ys')]
                 to s_xs, s_yys, len_xs_yys

            /* Apply the induction hypothesis
             * to prove sorted(merge(n',node(x,xs'),ys'))
             */
            have len_xxs_ys: length(node(x,xs')) + length(ys') = n'
              by definition {operator +,length}
                 rewrite symmetric len_xs_yys
                 definition length
                 rewrite add_suc[length(xs')][length(ys')].
            have s_xxs: sorted(node(x, xs'))
              by enable sorted rewrite xs_xxs in prem
            have s_ys: sorted(ys')
              by definition sorted in rewrite ys_yys in prem
            have IH_xxs_ys: sorted(merge(n',node(x,xs'),ys'))
              by apply IH[node(x,xs'),ys']
                 to s_xxs, s_ys, len_xxs_ys

            have x_le_xs: all_elements(xs', λb{x ≤ b})
              by definition sorted in rewrite xs_xxs in prem
            have y_le_ys: all_elements(ys', λb{y ≤ b})
              by definition sorted in rewrite ys_yys in prem
            
            switch x ≤ y {
              case true suppose xy_true {
                definition sorted
                suffices sorted(merge(n',xs',node(y,ys'))) and
                         all_elements(merge(n',xs',node(y,ys')), λb{x ≤ b})
                IH_xs_yys, 
                conclude all_elements(merge(n',xs',node(y,ys')),λb{x ≤ b})  by
                  rewrite all_elements_eq_member
                     [Nat,merge(n',xs',node(y,ys')),λb{x ≤ b}]
                  rewrite symmetric len_xs_yys
                  rewrite set_of_merge[xs',node(y,ys')]
                  arbitrary z:Nat
                  suppose z_in_xs_yys: z ∈ set_of(xs') ∪ set_of(node(y,ys'))
                  suffices x ≤ z
                  cases apply member_union
                               [Nat,z,set_of(xs'),set_of(node(y,ys'))]
                        to z_in_xs_yys
                  case z_in_xs: z ∈ set_of(xs') {
                    conclude x ≤ z by
                      apply all_elements_member[Nat][xs'][z, λb{x ≤ b}]
                      to x_le_xs, z_in_xs
                  }
                  case z_in_ys: z ∈ set_of(node(y,ys')) {
                    cases apply member_union[Nat,z,single(y),set_of(ys')]
                          to definition set_of in z_in_ys
                    case z_sy: z ∈ single(y) {
                      have y_z: y = z
                          by definition {operator ∈, single, rep} in z_sy
                      conclude x ≤ z by rewrite symmetric y_z | xy_true.
                    }
                    case z_in_ys: z ∈ set_of(ys') {
                      have y_z: y ≤ z
                        by apply all_elements_member[Nat][ys'][z,λb{y ≤ b}]
                           to y_le_ys, z_in_ys
                      have x_y: x ≤ y by rewrite xy_true.
                      conclude x ≤ z
                          by apply less_equal_trans[x][y,z] to x_y, y_z
                    }
                  }
              }
              case false suppose xy_false {
                have not_x_y: not (x ≤ y)
                  by suppose xs rewrite xy_false in xs
                have y_x: y ≤ x
                  by apply less_implies_less_equal[y][x] to
                     (apply not_less_equal_greater[x,y] to not_x_y)
                definition sorted
                suffices sorted(merge(n',node(x,xs'),ys')) and
                         all_elements(merge(n',node(x,xs'),ys'),λb{y ≤ b})
                IH_xxs_ys, 
                conclude all_elements(merge(n',node(x,xs'),ys'),λb{y ≤ b}) by
                  rewrite all_elements_eq_member
                     [Nat,merge(n',node(x,xs'),ys'),λb{y ≤ b}]
                  rewrite symmetric len_xxs_ys
                  rewrite set_of_merge[node(x,xs'),ys']
                  arbitrary z:Nat
                  suppose z_in_xxs_ys: z ∈ set_of(node(x,xs')) ∪ set_of(ys')
                  suffices y ≤ z
                  cases apply member_union
                               [Nat,z,set_of(node(x,xs')),set_of(ys')]
                        to z_in_xxs_ys
                  case z_in_xxs: z ∈ set_of(node(x,xs')) {
                    have z_in_sx_or_xs: z ∈ single(x) or z ∈ set_of(xs')
                      by apply member_union[Nat,z,single(x),set_of(xs')]
                         to definition set_of in z_in_xxs
                    cases z_in_sx_or_xs
                    case z_in_sx: z ∈ single(x) {
                      have x_z: x = z
                          by definition {operator ∈, single, rep} in z_in_sx
                      conclude y ≤ z  by rewrite symmetric x_z  y_x
                    }
                    case z_in_xs: z ∈ set_of(xs') {
                      have x_z: x ≤ z
                        by apply all_elements_member[Nat][xs'][z,λb{x ≤ b}]
                           to x_le_xs, z_in_xs
                      conclude y ≤ z 
                         by apply less_equal_trans[y][x,z] to y_x, x_z
                    }
                  }
                  case z_in_ys: z ∈ set_of(ys') {
                    conclude y ≤ z by
                      apply all_elements_member[Nat][ys'][z,λb{y ≤ b}]
                      to y_le_ys, z_in_ys
                  }
              }
            }
          }
        }
      }
    }
  }
end

Prove correctness of msort

First we show that the two lists produced by msort contain the same elements as the input list.

theorem mset_of_msort: all n:Nat. all xs:List<Nat>.
  mset_of(first(msort(n, xs)))  ⨄  mset_of(second(msort(n, xs))) = mset_of(xs)
proof
  induction Nat
  case 0 {
    arbitrary xs:List<Nat>
    definition msort
    switch xs {
      case empty {
        definition {first, second}
        suffices mset_of(empty) ⨄ mset_of(empty) = mset_of(empty)
        definition {mset_of}
        rewrite m_sum_empty[Nat,m_fun(λx{0})].
      }
      case node(x, xs') {
        definition {first, second, mset_of}
        suffices m_one(x) ⨄ mset_of(empty) ⨄ mset_of(xs')
               = m_one(x) ⨄ mset_of(xs')
        definition {mset_of}
        rewrite m_sum_empty[Nat,m_one(x)].
      }
    }
  }
  case suc(n') suppose IH {
    arbitrary xs:List<Nat>
    definition {msort, first, second}
    
    let ys = first(msort(n',xs))
    let ls = second(msort(n',xs))
    rewrite have first(msort(n',xs)) = ys  by definition ys.
    rewrite have second(msort(n',xs)) = ls  by definition ls.
    
    let zs = first(msort(n', ls))
    let ms = second(msort(n', ls))
    rewrite have first(msort(n', ls)) = zs by definition zs.
    rewrite have second(msort(n', ls)) = ms by definition ms.

    equations
          mset_of(merge(length(ys) + length(zs),ys,zs)) ⨄ mset_of(ms)
        = (mset_of(ys) ⨄ mset_of(zs)) ⨄ mset_of(ms)
          by rewrite (apply mset_of_merge[length(ys) + length(zs)][ys,zs] to .).
    ... = mset_of(ys) ⨄ (mset_of(zs) ⨄ mset_of(ms))
          by rewrite m_sum_assoc[Nat, mset_of(ys), mset_of(zs), mset_of(ms)].
    ... = mset_of(ys) ⨄ mset_of(ls)
          by rewrite have mset_of(zs) ⨄ mset_of(ms) = mset_of(ls)
                     by definition {zs, ms} IH[ls].
    ... = mset_of(xs)
          by definition {ys, ls} IH[xs]
  }
end

Next, we prove that the first output list is sorted. We make use of the merge_sorted theorem in this proof.

theorem msort_sorted: all n:Nat. all xs:List<Nat>. 
  sorted(first(msort(n, xs)))
proof
  induction Nat
  case 0 {
    arbitrary xs:List<Nat>
    switch xs {
      case empty {
        definition {msort, first}
        conclude sorted(empty)  by definition sorted.
      }
      case node(x, xs') {
        definition {msort, first}
        conclude sorted(node(x,empty))
            by definition {sorted, sorted, all_elements}.
      }
    }
  }
  case suc(n') suppose IH {
    arbitrary xs:List<Nat>
    let ys = first(msort(n',xs))
    let zs = first(msort(n',second(msort(n',xs))))
    have IH1: sorted(ys)  by definition ys IH[xs]
    have IH2: sorted(zs)  by definition zs IH[second(msort(n',xs))]
    definition {msort, first}
    definition {ys, zs} in
    apply merge_sorted[length(ys) + length(zs)][ys, zs] to IH1, IH2
  }
end

It remains to show that first output of msort is of length min(2ⁿ,length(xs)). Instead of using min, I separated the proof into a couple cases depending on whether 2ⁿ ≤ length(xs). However, I first needed to prove the lengths of the two output lists adds up to the length of the input list.

theorem msort_length: all n:Nat. all xs:List<Nat>.
  length(first(msort(n, xs)))  +  length(second(msort(n, xs))) = length(xs)

The proof of msort_length required a theorem that the length of the output of merge is the sum of the lengths of the inputs.

theorem merge_length: all n:Nat. all xs:List<Nat>, ys:List<Nat>.
  if length(xs) + length(ys) = n
  then length(merge(n, xs, ys)) = n

So in the case when the length of the input list is greater than 2ⁿ, the first output of msort is of length 2ⁿ.

theorem msort_length_less_equal: all n:Nat. all xs:List<Nat>.
  if pow2(n) ≤ length(xs)
  then length(first( msort(n, xs) )) = pow2(n)
proof
  induction Nat
  case 0 {
    arbitrary xs:List<Nat>
    suppose prem
    switch xs {
      case empty suppose xs_mt {
        conclude false
            by definition {pow2, length, operator≤} in
               rewrite xs_mt in prem
      }
      case node(x, xs') suppose xs_xxs {
        definition {msort,first}
        conclude length(node(x,empty)) = pow2(0)
            by definition {length, length, pow2}.
      }
    }
  }
  case suc(n') suppose IH {
    arbitrary xs:List<Nat>
    suppose prem
    have len_xs: pow2(n') + pow2(n') ≤ length(xs)
      by rewrite add_zero[pow2(n')] in
         definition {pow2, operator*, operator*,operator*} in prem
    definition {pow2, msort, first}

    let ys = first(msort(n',xs))
    let ls = second(msort(n',xs))
    have ys_def: first(msort(n',xs)) = ys  by definition ys.
    have ls_def: second(msort(n',xs)) = ls  by definition ls.
    rewrite ys_def | ls_def
    
    let zs = first(msort(n', ls))
    let ms = second(msort(n', ls))
    have zs_def: first(msort(n', ls)) = zs by definition zs.
    have ms_def: second(msort(n', ls)) = ms by definition ms.
    rewrite zs_def | ms_def

    have p2n_le_xs: pow2(n') ≤ length(xs)
      by have p2n_le_2p2n: pow2(n') ≤ pow2(n') + pow2(n')
           by less_equal_add[pow2(n')][pow2(n')]
         apply less_equal_trans[pow2(n')][pow2(n') + pow2(n'), length(xs)]
         to p2n_le_2p2n, len_xs

    have len_ys: length(ys) = pow2(n')
      by rewrite ys_def in apply IH[xs] to p2n_le_xs
      
    have len_ys_ls_eq_xs: length(ys) + length(ls) = length(xs)
      by rewrite ys_def | ls_def in msort_length[n'][xs]

    have p2n_le_ls: pow2(n') ≤ length(ls)
      by have pp_pl: pow2(n') + pow2(n') ≤ pow2(n') + length(ls)
           by rewrite symmetric len_ys_ls_eq_xs | len_ys in len_xs
         apply less_equal_left_cancel[pow2(n')][pow2(n'), length(ls)] to pp_pl
            
    have len_zs: length(zs) = pow2(n')
      by rewrite zs_def in apply IH[ls] to p2n_le_ls

    have len_ys_zs: length(ys) + length(zs) = 2 * pow2(n')
      by rewrite len_ys | len_zs
         definition {operator*,operator*,operator*}
         rewrite add_zero[pow2(n')].

    conclude length(merge(length(ys) + length(zs),ys,zs)) = 2 * pow2(n')
      by rewrite len_ys_zs
         apply merge_length[2 * pow2(n')][ys, zs] to len_ys_zs
  }
end

When the length of the input list is less than 2ⁿ, the length of the first output is the same as the length of the input.

theorem msort_length_less: all n:Nat. all xs:List<Nat>.
  if length(xs) < pow2(n)
  then length(first( msort(n, xs) )) = length(xs)
proof
  induction Nat
  case 0 {
    arbitrary xs:List<Nat>
    suppose prem
    switch xs {
      case empty suppose xs_mt {
        definition {msort, length, first}.
      }
      case node(x, xs') suppose xs_xxs {
        definition {msort,first, length, length}
        have xs_0: length(xs') = 0
          by definition {operator ≤, length, operator<, pow2} in 
             rewrite xs_xxs in prem
        rewrite xs_0.
      }
    }
  }
  case suc(n') suppose IH {
    arbitrary xs:List<Nat>
    suppose prem
    definition{msort, first}

    let ys = first(msort(n',xs))
    let ls = second(msort(n',xs))
    have ys_def: first(msort(n',xs)) = ys  by definition ys.
    have ls_def: second(msort(n',xs)) = ls  by definition ls.
    rewrite ys_def | ls_def
    
    let zs = first(msort(n', ls))
    let ms = second(msort(n', ls))
    have zs_def: first(msort(n', ls)) = zs by definition zs.
    have ms_def: second(msort(n', ls)) = ms by definition ms.
    rewrite zs_def | ms_def

    have xs_le_two_p2n: length(xs) < pow2(n') + pow2(n')
      by rewrite add_zero[pow2(n')] in
         definition {pow2, operator*,operator*,operator*} in prem

    have ys_ls_eq_xs: length(ys) + length(ls) = length(xs)
      by rewrite ys_def | ls_def in msort_length[n'][xs]

    have pn_xs_or_xs_pn: pow2(n') ≤ length(xs) or length(xs) < pow2(n')
      by dichotomy[pow2(n'), length(xs)]
    cases pn_xs_or_xs_pn
    case pn_xs: pow2(n') ≤ length(xs) {
    
      have ys_pn: length(ys) = pow2(n')
          by rewrite ys_def in apply msort_length_less_equal[n'][xs] to pn_xs

      have ls_l_pn: length(ls) < pow2(n')
          by have pn_ls_l_2pn: pow2(n') + length(ls) < pow2(n') + pow2(n')
               by rewrite symmetric ys_ls_eq_xs | ys_pn in xs_le_two_p2n
             apply less_left_cancel[pow2(n'), length(ls), pow2(n')] to pn_ls_l_2pn

      have len_zs: length(zs) = length(ls)
          by rewrite zs_def in apply IH[ls] to ls_l_pn

      equations
        length(merge(length(ys) + length(zs),ys,zs))
            = length(ys) + length(zs)
              by apply merge_length[length(ys) + length(zs)][ys,zs] to .
        ... = length(ys) + length(ls)
              by rewrite len_zs.
        ... = length(xs)
              by ys_ls_eq_xs
    }
    case xs_pn: length(xs) < pow2(n') {
    
      have len_ys: length(ys) = length(xs)
        by rewrite ys_def in apply IH[xs] to xs_pn

      have len_ls: length(ls) = 0
        by apply left_cancel[length(ys)][length(ls), 0] to
           rewrite add_zero[length(ys)] | len_ys
           rewrite len_ys in ys_ls_eq_xs

      have ls_l_pn: length(ls) < pow2(n')
        by rewrite len_ls  pow_positive[n'] 
      
      have len_zs: length(zs) = 0
        by rewrite zs_def | len_ls in apply IH[ls] to ls_l_pn

      equations
        length(merge(length(ys) + length(zs),ys,zs))
          = length(ys) + length(zs)
            by apply merge_length[length(ys) + length(zs)][ys, zs] to .
      ... = length(xs)
            by rewrite len_zs | add_zero[length(ys)] | len_ys.
    }
  }
end

Prove correctness of merge_sort

The proof that merge_sort produces a sorted list is a straightforward corollary of the msort_sorted theorem.

theorem merge_sort_sorted: all xs:List<Nat>.
  sorted(merge_sort(xs))
proof
  arbitrary xs:List<Nat>
  definition merge_sort
  msort_sorted[log(length(xs))][xs]
end

The proof that the contents of the output of merge_sort are the same as the input is a bit more involved. So if we use the definitoin of merge_sort, we then need to show that

mset_of(first(msort(log(length(xs)),xs))) = mset_of(xs)

which means we need to show that all the elements in xs end up in the first output and that there are not any leftovers. Let ys be the first output of msort and ls be the leftovers. The theorem less_equal_pow_log in Log.pf tells us that length(xs) ≤ pow2(log(length(xs))). So in the case where they are equal, we can use the msort_length_less_equal theorem to show that length(ys) = length(xs). In the case where length(xs) is strictly smaller, we use the msort_length_less theorem to prove that length(ys) = length(xs). Finally, we show that the length of ls is zero by use of msort_length and some properties of arithmetic like left_cancel (in Nat.pf).

Here is the proof of mset_of_merge_sort in full.

theorem mset_of_merge_sort: all xs:List<Nat>.
  mset_of(merge_sort(xs)) = mset_of(xs)
proof
  arbitrary xs:List<Nat>
  definition merge_sort
  let n = log(length(xs))
  have n_def: log(length(xs)) = n  by definition n.
  let ys = first(msort(n,xs))
  have ys_def: first(msort(n,xs)) = ys  by definition ys.
  let ls = second(msort(n,xs))
  have ls_def: second(msort(n,xs)) = ls  by definition ls.

  have len_xs: length(xs) ≤ pow2(n)
    by rewrite symmetric n_def
       less_equal_pow_log[length(xs)]
  have len_ys: length(ys) = length(xs)
    by cases apply less_equal_implies_less_or_equal[length(xs)][pow2(n)]
             to len_xs
       case len_xs_less {
         rewrite ys_def in apply msort_length_less[n][xs] to len_xs_less
       }
       case len_xs_equal {
         have pn_le_xs: pow2(n) ≤ length(xs)
           by rewrite len_xs_equal  less_equal_refl[pow2(n)]
         have len_ys_pow2: length(ys) = pow2(n)
           by rewrite symmetric ys_def
              apply msort_length_less_equal[n][xs] to pn_le_xs
         transitive len_ys_pow2 (symmetric len_xs_equal)
       }
  have len_ys_ls_eq_xs: length(ys) + length(ls) = length(xs)
    by rewrite ys_def | ls_def in msort_length[n][xs]
  have len_ls: length(ls) = 0
    by apply left_cancel[length(ys)][length(ls), 0] to
       rewrite add_zero[length(ys)] | len_ys
       rewrite len_ys in len_ys_ls_eq_xs
  have ls_mt: ls = empty
    by apply length_zero_empty[Nat, ls] to len_ls

  have ys_ls_eq_xs: mset_of(ys)  ⨄  mset_of(ls) = mset_of(xs)
    by rewrite ys_def | ls_def in mset_of_msort[n][xs]

  rewrite n_def
  rewrite ys_def
  equations
    mset_of(ys)
        = mset_of(ys)  ⨄  m_fun(λx{0})
          by rewrite m_sum_empty[Nat, mset_of(ys)].
    ... = mset_of(ys)  ⨄  mset_of(ls)
          by rewrite ls_mt definition mset_of.
    ... = mset_of(xs)
          by ys_ls_eq_xs
end

Exercise: merge_length and msort_length

Prove the following theorems.

theorem merge_length: all n:Nat. all xs:List<Nat>, ys:List<Nat>.
  if length(xs) + length(ys) = n
  then length(merge(n, xs, ys)) = n

theorem msort_length: all n:Nat. all xs:List<Nat>.
  length(first(msort(n, xs)))  +  length(second(msort(n, xs))) = length(xs)

Exercise: classic Merge Sort

Test and prove the correctness of the classic definition of merge_sort, which we repeat here.

function msort(Nat, List<Nat>) -> List<Nat> {
  msort(0, xs) = xs
  msort(suc(n'), xs) =
    let p = split(xs)
    merge(msort(n', first(p)), msort(n', second(p)))
}

define merge_sort : fn List<Nat> -> List<Nat>
  = λxs{ msort(log(length(xs)), xs) }

You will need define the split function.