-- {{{ Logic/general definitions

-- import basic logic definitions from <logic.pg>

@include logic.pg

-- }}}

-- {{{ Axioms

-- Now we can define the Peano axioms. Unlike with equality, perga is not
-- powerful enough to construct the natural numbers (or at least to prove the
-- Peano axioms as theorems from a definition constructible in perga). However,
-- working with axioms is extremely common in math. As such, perga has a system
-- for doing just that, namely the ★axiom★ system.
-- 
-- In a definition, rather than give a value for the term, we can give it the
-- value axiom, in which case a type ascription is mandatory. Perga will then
-- trust our type ascription, and assume going forward that the identifier we
-- defined is a value of the asserted type. For example, we will use the axiom
-- system to assert the existence of a type of natural numbers

axiom nat : ★;

-- As you can imagine, this can be risky. For instance, there's nothing stopping
-- us from saying
--     uh_oh : false := axiom;
-- or stipulating more subtly contradictory axioms. As such, as in mathematics,
-- axioms should be used with care.
-- 
-- There's another problem with axioms, namely that perga cannot do any
-- computations with axioms. The more you can define within perga natively, the
-- better, as computations done without axioms can be utilized by perga. For
-- example, in <computation.pg>, we define the natural numbers as Church
-- numerals entirely within perga. There, the proof that 1 + 1 = 2 is just
-- eq_refl, since they reduce to the same thing. Here, 1 + 1 = 2 will
-- require a proof, since perga is unable to do computations with things defined
-- as axioms.
-- 
-- With these warnings in place, the Peano axioms are proven to be consistent,
-- so we should be fine. I'm formalizing the second order axioms given in the
-- wikipedia article on the Peano axioms
-- (https://en.wikipedia.org/wiki/Peano_axioms).

-- axiom 1: 0 is a natural number
axiom zero : nat;

-- axiom 6: For every n, S n is a natural number.
axiom suc (n : nat) : nat;

-- axiom 7: If S n = S m, then n = m.
axiom suc_inj : forall (n m : nat), eq nat (suc n) (suc m) → eq nat n m;

-- axiom 8: No successor of any natural number is zero.
axiom suc_nonzero : forall (n : nat), not (eq nat (suc n) zero);

-- axiom 9: Induction! For any proposition P on natural numbers, if P(0) holds,
-- and if for every natural number n, P(n) ⇒ P(S n), then P holds for all n.
axiom nat_ind : forall (P : nat → ★), P zero → (forall (n : nat), P n → P (suc n))
    → forall (n : nat), P n;

-- }}}

-- {{{ Basic Definitions

-- Now that we have stipulated these axioms, we are free to use them to make
-- definitions, prove theorems, etc.
-- 
-- Our first theorem, as a warm up, is to prove that every natural number is
-- either 0 or the successor of another natural number.
-- 
-- First, we will make a bunch of abbreviations, since these terms get really
-- long and complicated really quickly.

-- Some abbreviations for common numbers.
def one   : nat := suc zero;
def two   : nat := suc one;
def three : nat := suc two;
def four  : nat := suc three;
def five  : nat := suc four;

-- A nice infix abbreviation for equality on natural numbers
def = (n m : nat) := eq nat n m;
infixl 11 =;

-- First, the predecessor of n is m if n = suc m.
def pred (n m : nat) : ★ := n = suc m;

-- Our claim is a disjunction, whose first option is that n = 0.
def szc_l (n : nat) := n = zero;

-- The second option is that n has a predecessor.
def szc_r (n : nat) := exists nat (pred n);

-- So the claim we are trying to prove is that either one of the above options
-- holds for every n.
def szc (n : nat) := szc_l n ∨ szc_r n;

-- And here's our proof!
def suc_or_zero : forall (n : nat), szc n :=
    -- We will prove this by induction.
    nat_ind szc
        -- For the base case, the first option holds, i.e. 0 = 0
        (or_intro_l (szc_l zero) (szc_r zero) (eq_refl nat zero))

        -- For the inductive case, suppose the theorem holds for n.
        (fun (n : nat) (_ : szc n) => 
            -- Then the right option holds for suc n, since suc n is the
            -- successor of n
            or_intro_r (szc_l (suc n)) (szc_r (suc n))
                (exists_intro nat (pred (suc n)) n (eq_refl nat (suc n))));

-- }}}

-- {{{ Recursive Definitions

-- The next step would normally be to define addition and prove properties of
-- addition. However, we need to take a very long and difficult detour in order
-- to be able to define addition in the first place. The normal way addition is
-- defined is by the following two equations:
--
-- 1)       n + 0 = n
-- 2) n + (suc m) = suc (n + m)
--
-- It is clear that this definition is ok, since m gets smaller with every
-- application of equation 2, until m reaches zero, whereupon we can use
-- equation 1. This argument, while perfectly reasonable, is deceptively
-- difficult to formalize. It turns out that this structure of recursively
-- defining a function by two equations similar to 1 and 2 is extremely common.
-- So, we will take the time to prove the following theorem.
--
-- Theorem (recursive definitions):
-- For every type A, element z : A, and function fS : nat → A → A, there
-- exists a unique function f : nat → A satisfying
-- 1)                       f 0 = z
-- 2) forall n : nat, f (suc n) = fS n (f n)
--
-- Once we've proved this theorem, we can easily define addition such that,
-- i.e. that for any fixed n : nat, "n+" is the unique function satisfying
-- 1)                       n + 0 = n
-- 2) forall m : nat, n + (suc m) = suc (n + m)
-- Notice that this is the exact form we need in order to apply the theorem.
--
-- However, proving this theorem is *extremely long and difficult*. I would
-- recommend skipping this section and coming back to it after you are much
-- more used to perga. As such, I will go into a lot less detail on each of
-- these proofs than in the later sections.
--
-- Here's our game plan. We will define a relation R : nat → A → ★ such that
-- R x y iff for every relation Q : nat → A → ★ satisfying
-- 1) Q 0 z and
-- 2) forall (x : nat) (y : A), Q x y → Q (suc x) (fS x y),
-- Q x y.
-- In more mathy lingo, we take R to be the intersection of every relation
-- satisfying 1 and 2. From there we will, with much effort, prove that R is
-- actually a function satisfying the equations we want it to.

section RecursiveDefs

-- First, fix an A, z : A, and fS : nat → A → A
variable (A : ★) (z : A) (fS : nat → A → A);

-- {{{ Defining R
-- Here is condition 1 formally expressed in perga.
def cond1 (Q : nat → A → ★) := Q zero z;

-- Likewise for condition 2.
def cond2 (Q : nat → A → ★) :=
    forall (n : nat) (y : A), Q n y → Q (suc n) (fS n y);

-- From here we can define R.
def rec_rel (x : nat) (y : A) :=
    forall (Q : nat → A → ★), cond1 Q → cond2 Q → Q x y;
-- }}}

-- {{{ R is total
def total (A B : ★) (R : A → B → ★) := forall (a : A), exists B (R a);

def rec_rel_cond1 : cond1 rec_rel :=
    fun (Q : nat → A → ★) (h1 : cond1 Q) (h2 : cond2 Q) => h1;

def rec_rel_cond2 : cond2 rec_rel :=
    fun (n : nat) (y : A) (h : rec_rel n y)
        (Q : nat → A → ★) (h1 : cond1 Q) (h2 : cond2 Q) =>
        h2 n y (h Q h1 h2);

def rec_rel_total : total nat A rec_rel :=
    let (P (x : nat) := exists A (rec_rel x)) in
    nat_ind P
    (exists_intro A (rec_rel zero) z rec_rel_cond1)
    (fun (n : nat) (IH : P n) =>
        exists_elim A (P (suc n)) (rec_rel n) IH
        (fun (y0 : A) (hy : rec_rel n y0) =>
            exists_intro A (rec_rel (suc n)) (fS n y0) (rec_rel_cond2 n y0 hy)))
    end;

-- }}}

-- {{{ Defining R2
def alt_cond1 (x : nat) (y : A) :=
    x = zero ∧ eq A y z;

def cond_y2 (x : nat) (y : A)
    (x2 : nat) (y2 : A) :=
        (eq A y (fS x2 y2)) ∧ (rec_rel x2 y2);

def cond_x2 (x : nat) (y : A) (x2 : nat) :=
    pred x x2 ∧ exists A (cond_y2 x y x2);

def alt_cond2 (x : nat) (y : A) := exists nat (cond_x2 x y);

def rec_rel_alt (x : nat) (y : A) := alt_cond1 x y ∨ alt_cond2 x y;
-- }}}

-- {{{ R = R2

-- {{{ R2 ⊆ R
def R2_sub_R_case1 (x : nat) (y : A) : alt_cond1 x y → rec_rel x y :=
    fun (case1 : alt_cond1 x y) =>
        let (x0 := and_elim_l (x = zero) (eq A y z) case1)
            (yz := and_elim_r (x = zero) (eq A y z) case1)
            (a1 := (eq_sym A y z yz) (rec_rel zero) rec_rel_cond1)
        in
            (eq_sym nat x zero x0) (fun (n : nat) => rec_rel n y) a1
        end;

def R2_sub_R_case2 (x : nat) (y : A) : alt_cond2 x y → rec_rel x y :=
    fun (case2 : alt_cond2 x y) =>
        let (h1 := cond_x2 x y)
            (h2 := cond_y2 x y)
        in
        exists_elim nat (rec_rel x y) h1 case2
        (fun (x2 : nat) (hx2 : h1 x2) =>
            let (hpred := and_elim_l (pred x x2) (exists A (h2 x2)) hx2)
                (hex   := and_elim_r (pred x x2) (exists A (h2 x2)) hx2)
            in
            exists_elim A (rec_rel x y) (h2 x2) hex
            (fun (y2 : A) (hy2 : h2 x2 y2) =>
                let (hpreim := and_elim_l (eq A y (fS x2 y2)) (rec_rel x2 y2) hy2)
                    (hR     := and_elim_r (eq A y (fS x2 y2)) (rec_rel x2 y2) hy2)
                    (a1     := rec_rel_cond2 x2 y2 hR)
                    (a2     := (eq_sym A y (fS x2 y2) hpreim) (rec_rel (suc x2)) a1)
                in
                    (eq_sym nat x (suc x2) hpred) (fun (n : nat) => rec_rel n y) a2
                end)
            end)
        end;

def R2_sub_R (x : nat) (y : A) : rec_rel_alt x y → rec_rel x y :=
    fun (h : rec_rel_alt x y) =>
        or_elim (alt_cond1 x y) (alt_cond2 x y) (rec_rel x y) h
        (R2_sub_R_case1 x y)
        (R2_sub_R_case2 x y);
-- }}}

-- {{{ R ⊆ R2
def R2_cond1 : cond1 rec_rel_alt :=
    or_intro_l (alt_cond1 zero z) (alt_cond2 zero z)
    (and_intro (zero = zero) (eq A z z)
        (eq_refl nat zero)
        (eq_refl A z));

def R2_cond2 : cond2 rec_rel_alt :=
    fun (x2 : nat) (y2 : A) (h : rec_rel_alt x2 y2) =>
    let (x   := suc x2)
        (y   := fS x2 y2)
        (cx2 := cond_x2 x y)
        (cy2 := cond_y2 x y x2)
    in
        or_intro_r (alt_cond1 x y) (alt_cond2 x y)
        (exists_intro nat cx2 x2
            (and_intro (pred x x2) (exists A cy2)
                (eq_refl nat x)
                (exists_intro A cy2 y2
                    (and_intro (eq A y y) (rec_rel x2 y2)
                        (eq_refl A y)
                        (R2_sub_R x2 y2 h)))))
    end;


def R_sub_R2 (x : nat) (y : A) : rec_rel x y → rec_rel_alt x y :=
    fun (h : rec_rel x y) => h rec_rel_alt R2_cond1 R2_cond2;
-- }}}

-- }}}

-- {{{ R2 (hence R) is functional

def fl_in (A B : ★) (R : A → B → ★) (x : A) := forall (y1 y2 : B),
    R x y1 → R x y2 → eq B y1 y2;

def fl (A B : ★) (R : A → B → ★) := forall (x : A), fl_in A B R x;

def R2_zero (y : A) : rec_rel_alt zero y → eq A y z :=
    let (cx2 := cond_x2 zero y)
        (cy2 := cond_y2 zero y)
    in fun (h : rec_rel_alt zero y) =>
        or_elim (alt_cond1 zero y) (alt_cond2 zero y) (eq A y z) h
        (fun (case1 : alt_cond1 zero y) => and_elim_r (zero = zero) (eq A y z) case1)
        (fun (case2 : alt_cond2 zero y) =>
            exists_elim nat (eq A y z) cx2 case2
            (fun (x2 : nat) (h2 : cx2 x2) =>
                suc_nonzero x2
                (eq_sym nat zero (suc x2) (and_elim_l (pred zero x2) (exists A (cy2 x2)) h2))
                (eq A y z)))
    end;

def R2_suc (x2 : nat) (y : A) : rec_rel_alt (suc x2) y → exists A (cond_y2 (suc x2) y x2) :=
    let (x    := suc x2)
        (cx2  := cond_x2 x y)
        (cy2  := cond_y2 x y)
        (goal := exists A (cy2 x2))
    in
        fun (h : rec_rel_alt x y) =>
        or_elim (alt_cond1 x y) (alt_cond2 x y) goal h
        (fun (case1 : alt_cond1 x y) => suc_nonzero x2 (and_elim_l (x = zero) (eq A y z) case1) goal)
        (fun (case2 : alt_cond2 x y) =>
            exists_elim nat goal cx2 case2
            (fun (x22 : nat) (hx22 : cx2 x22) =>
                let (hpred  := and_elim_l (pred x x22) (exists A (cy2 x22)) hx22)
                    (hgoal  := and_elim_r (pred x x22) (exists A (cy2 x22)) hx22)
                    (x2_x22 := suc_inj x2 x22 hpred)
                in
                    (eq_sym nat x2 x22 x2_x22)
                    (fun (n : nat) => exists A (cy2 n))
                    hgoal
                end))
    end;

def R2_functional_base_case : fl_in nat A rec_rel_alt zero :=
    fun (y1 y2 : A) (h1 : rec_rel_alt zero y1) (h2 : rec_rel_alt zero y2) =>
        eq_trans A y1 z y2
        (R2_zero y1 h1)
        (eq_sym A y2 z (R2_zero y2 h2));

def R2_functional_inductive_step (x2 : nat) (IH : fl_in nat A rec_rel_alt x2) : fl_in nat A rec_rel_alt (suc x2) :=
    fun (y1 y2 : A) (h1 : rec_rel_alt (suc x2) y1) (h2 : rec_rel_alt (suc x2) y2) =>
        let (x    := suc x2)
            (cy1  := cond_y2 x y1 x2)
            (cy2  := cond_y2 x y2 x2)
            (goal := eq A y1 y2)
        in
            exists_elim A goal cy1 (R2_suc x2 y1 h1)
            (fun (y12 : A) (h12 : cy1 y12) =>
                exists_elim A goal cy2 (R2_suc x2 y2 h2)
                (fun (y22 : A) (h22 : cy2 y22) =>
                    let (y1_preim  := and_elim_l (eq A y1 (fS x2 y12)) (rec_rel x2 y12) h12)
                        (y2_preim  := and_elim_l (eq A y2 (fS x2 y22)) (rec_rel x2 y22) h22)
                        (R_x2_y12  := and_elim_r (eq A y1 (fS x2 y12)) (rec_rel x2 y12) h12)
                        (R_x2_y22  := and_elim_r (eq A y2 (fS x2 y22)) (rec_rel x2 y22) h22)
                        (R2_x2_y12 := R_sub_R2 x2 y12 R_x2_y12)
                        (R2_x2_y22 := R_sub_R2 x2 y22 R_x2_y22)
                        (y12_y22   := IH y12 y22 R2_x2_y12 R2_x2_y22)
                    in
                        eq_trans A y1 (fS x2 y12) y2
                        y1_preim
                        (eq_trans A (fS x2 y12) (fS x2 y22) y2
                            (eq_cong A A y12 y22 (fun (y : A) => fS x2 y) y12_y22)
                            (eq_sym A y2 (fS x2 y22) y2_preim))
                    end))
        end;

def R2_functional : fl nat A rec_rel_alt :=
    nat_ind (fl_in nat A rec_rel_alt) R2_functional_base_case R2_functional_inductive_step;

def R_functional : fl nat A rec_rel :=
    fun (n : nat) (y1 y2 : A) (h1 : rec_rel n y1) (h2 : rec_rel n y2) =>
        R2_functional n y1 y2 (R_sub_R2 n y1 h1) (R_sub_R2 n y2 h2);

-- }}}

-- {{{ Actually defining the function

def rec_def (x : nat) : A :=
    exists_elim A A (rec_rel x) (rec_rel_total x) (fun (y : A) (_ : rec_rel x y) => y);

-- }}}

-- {{{ It satisfies the properties we want it to

-- Kind of stupidly, we still need one more axiom. Due to how existentials are
-- defined, even though rec_def n is defined to be the y such that R n y, we
-- can't actually conclude that R n (rec_def n).

-- We need to assert that, even if you "forget" that a value came from an
-- existential, it still satisfies the property it definitionally is supposed
-- to satisfy. This annoying problem would be subverted with proper Σ-types,
-- provided they had η-equality.
axiom definite_description (A : ★) (P : A → ★) (h : exists A P) :
    P (exists_elim A A P h (fun (x : A) (_ : P x) => x));

-- Now we can use this axiom to prove that R n (rec_def n).
def rec_def_sat (x : nat) : rec_rel x (rec_def x) :=
    definite_description A (rec_rel x) (rec_rel_total x);

def eq1 (f : nat → A) := eq A (f zero) z;

def eq2 (f : nat → A) := forall (n : nat), eq A (f (suc n)) (fS n (f n));

-- f zero = z
def rec_def_sat_zero : eq1 rec_def := R_functional zero (rec_def zero) z (rec_def_sat zero) rec_rel_cond1;

-- f n = y → f (suc n) = fS n y
def rec_def_sat_suc : eq2 rec_def := fun (n : nat) =>
    R_functional (suc n) (rec_def (suc n)) (fS n (rec_def n)) (rec_def_sat (suc n)) (rec_rel_cond2 n (rec_def n) (rec_def_sat n));

-- }}}

-- {{{ The function satisfying these equations is unique

def rec_def_unique (f g : nat → A) (h1f : eq1 f) (h2f : eq2 f) (h1g : eq1 g) (h2g : eq2 g)
    : forall (n : nat), eq A (f n) (g n) :=
        nat_ind (fun (n : nat) => eq A (f n) (g n))
        -- base case: f 0 = g 0
        (eq_trans A (f zero) z (g zero) h1f (eq_sym A (g zero) z h1g))

        -- Inductive step
        (fun (n : nat) (IH : eq A (f n) (g n)) =>
            -- f (suc n) = fS n (f n)
            --           = fS n (g n)
            --           = g (suc n)
            eq_trans A (f (suc n)) (fS n (f n)) (g (suc n))
            -- f (suc n) = fS n (f n)
            (h2f n)
            -- fS n (f n) = g (suc n)
            (eq_trans A (fS n (f n)) (fS n (g n)) (g (suc n))
                -- fS n (f n) = fS n (g n)
                (eq_cong A A (f n) (g n) (fS n) IH)
                -- fS n (g n) = g (suc n)
                (eq_sym A (g (suc n)) (fS n (g n)) (h2g n))));

-- }}}

end RecursiveDefs

-- }}}

-- Now we can safely define addition.

-- First, here's the RHS of equation 2 as a function, since it will show up
-- multiple times.
def psuc (_ r : nat) := suc r;

-- And here's plus!
def plus (n : nat) : nat → nat := rec_def nat n psuc;

-- And here's an infix version of it
def + (n m : nat) : nat := plus n m;
infixl 20 +;

-- The first equation manifests itself as the familiar
--      n + 0 = 0.
def plus_0_r (n : nat) : n + zero = n :=
  rec_def_sat_zero nat n psuc;

-- The second equation, after a bit of massaging, manifests itself as the
-- likewise familiar
--      n + suc m = suc (n + m).
def plus_s_r (n m : nat) : n + suc m = suc (n + m) :=
  rec_def_sat_suc nat n psuc m;

-- -- We can now prove 1 + 1 = 2!
def one_plus_one_two : one + one = two :=
  -- 1 + (suc zero) = suc (1 + zero) = suc one
  eq_trans nat (one + one) (suc (one + zero)) two

  -- 1 + (suc zero) = suc (1 + zero)
  (plus_s_r one zero)

  -- suc (1 + zero) = suc one
  (eq_cong nat nat (one + zero) one suc (plus_0_r one));

-- We have successfully defined addition! Note that evaluating 1 + 1 to 2
-- requires a proof, unfortunately, since this computation isn't visible to
-- perga.
--
-- We will now prove a couple standard properties of addition.

-- First, associativity, namely that n + (m + p) = (n + m) + p.
def plus_assoc : forall (n m p : nat), n + (m + p) = n + m + p
  := [n m : nat]
  -- We prove this via induction on p for any fixed n and m.
  nat_ind
    (fun (p : nat) => n + (m + p) = n + m + p)

    -- Base case: p = 0
    -- WTS n + (m + 0) = (n + m) + 0
    -- n + (m + 0) = n + m = (n + m) + 0
    (eq_trans nat (n + (m + zero)) (n + m) (n + m + zero)
      -- n + (m + 0) = n + m
      (eq_cong nat nat (m + zero) m (fun (p : nat) => n + p) (plus_0_r m))

      -- n + m = (n + m) + 0
      (eq_sym nat (n + m + zero) (n + m) (plus_0_r (n + m))))

    -- Inductive step: IH = n + (m + p) = (n + m) + p
    (fun (p : nat) (IH : n + (m + p) = n + m + p) =>
      -- WTS n + (m + suc p) = (n + m) + suc p
      -- n + (m + suc p) = n + suc (m + p)
      --                 = suc (n + (m + p))
      --                 = suc ((n + m) + p)
      --                 = (n + m) + suc p
      eq_trans nat (n + (m + suc p)) (n + (suc (m + p))) (n + m + suc p)

      -- n + (m + suc p) = n + suc (m + p)
      (eq_cong nat nat (m + suc p) (suc (m + p)) (fun (a : nat) => (n + a)) (plus_s_r m p))

      -- n + suc (m + p) = (n + m) + suc p
      (eq_trans nat (n + suc (m + p)) (suc (n + (m + p))) (n + m + suc p)
        -- n + suc (m + p) = suc (n + (m + p))
        (plus_s_r n (m + p))

        -- suc (n + (m + p)) = (n + m) + suc p
        (eq_trans nat (suc (n + (m + p))) (suc (n + m + p)) (n + m + suc p)
          -- suc (n + (m + p)) = suc ((n + m) + p)
          (eq_cong nat nat (n + (m + p)) (n + m + p) suc IH)

          -- suc ((n + m) + p) = (n + m) + suc p
          (eq_sym nat (n + m + suc p) (suc (n + m + p))
            (plus_s_r (n + m) p)))));

-- Up next is commutativity, but we will need a couple lemmas first.

-- First, we will show that 0 + n = n.
def plus_0_l : forall (n : nat), zero + n = n :=
  -- We prove this by induction on n.
  nat_ind ([n : nat] zero + n = n)
    -- base case: WTS 0 + 0 = 0
    -- This is just plus_0_r 0
    (plus_0_r zero)

    -- inductive case
    (fun (n : nat) (IH : zero + n = n) =>
      -- WTS 0 + (suc n) = suc n
      -- 0 + suc n = suc (0 + n) = suc n
      eq_trans nat (zero + suc n) (suc (zero + n)) (suc n)
      -- 0 + suc n = suc (0 + n)
      (plus_s_r zero n)

      -- suc (0 + n) = suc n
      (eq_cong nat nat (zero + n) n suc IH));

-- Next, we will show that (suc n) + m = suc (n + m).
def plus_s_l (n : nat) : forall (m : nat), suc n + m = suc (n + m) :=
  -- We proceed by induction on m.
  nat_ind ([m : nat] suc n + m = suc (n + m))
    -- base case: (suc n) + 0 = suc (n + 0)
    -- suc n + 0 = suc n = suc (n + 0)
    (eq_trans nat (suc n + zero) (suc n) (suc (n + zero))
      -- suc n + 0 = suc n
      (plus_0_r (suc n))

      -- suc n = suc (n + 0)
      (eq_cong nat nat n (n + zero) suc
        -- n = n + 0
        (eq_sym nat (n + zero) n (plus_0_r n))))

    -- inductive case
    -- IH = suc n + m = suc (n + m)
    (fun (m : nat) (IH : suc n + m = suc (n + m)) =>
        -- WTS suc n + suc m = suc (n + suc m)
        -- suc n + suc m = suc (suc n + m)
        --               = suc (suc (n + m))
        --               = suc (n + suc m)
        (eq_trans nat (suc n + suc m) (suc (suc n + m)) (suc (n + suc m))
          -- suc n + suc m = suc (suc n + m)
          (plus_s_r (suc n) m)

          -- suc (suc n + m) = suc (n + suc m)
          (eq_trans nat (suc (suc n + m)) (suc (suc (n + m))) (suc (n + suc m))
            -- suc (suc n + m) = suc (suc (n + m))
            (eq_cong nat nat (suc n + m) (suc (n + m)) suc IH)

            -- suc (suc (n + m)) = suc (n + suc m)
            (eq_cong nat nat (suc (n + m)) (n + suc m) suc
              -- suc (n + m) = n + suc m
              (eq_sym nat (n + suc m) (suc (n + m)) (plus_s_r n m))))));

-- Finally, we can prove commutativity.
def plus_comm (n : nat) : forall (m : nat), n + m = m + n :=
  -- As usual, we proceed by induction.
  nat_ind ([m : nat] n + m = m + n)

  -- Base case: WTS n + 0 = 0 + n
  -- n + 0 = n = 0 + n
  (eq_trans nat (n + zero) n (zero + n)
    -- n + 0 = n
    (plus_0_r n)

    -- n = 0 + n
    (eq_sym nat (zero + n) n (plus_0_l n)))

  -- Inductive step:
  (fun (m : nat) (IH : n + m = m + n) =>
    -- WTS n + suc m = suc m + n
    -- n + suc m = suc (n + m)
    --           = suc (m + n)
    --           = suc m + n
    (eq_trans nat (n + suc m) (suc (n + m)) (suc m + n)
      -- n + suc m = suc (n + m)
      (plus_s_r n m)

      -- suc (n + m) = suc m + n
      (eq_trans nat (suc (n + m)) (suc (m + n)) (suc m + n)
        -- suc (n + m) = suc (m + n)
        (eq_cong nat nat (n + m) (m + n) suc IH)

        -- suc (m + n) = suc m + n
        (eq_sym nat (suc m + n) (suc (m + n)) (plus_s_l m n)))));