Background
I am learning assistance, Coq, on my own. So far, I have completed reading Yves Bertot's Coq in a Hurry. Now, my goal is to prove some basic results concerning the natural numbers, culminating with the so-called division algorithm. However, I have encountered some setbacks on my way towards that goal. In particular, the two following results have proved (pun intended) to be more difficult to prove in Coq than I initially imagined. In fact, I have, after many fruitless attempts, resorted to prove them by hand (as shown below). This is clearly not helping me become more proficient in handling Coq; which is why I turn to this forum. My hope is that someone on this site is able and willing to help me translate my proofs below into a proof that Coq accepts. All help is sincerely appreciated!
Theorem A
For all $x,y \in N$ \begin{equation} x < S(y) \subset x < y \lor \text{I}(N,x,y) \end{equation} Proof:
Suppose $x < S(y)$. Hence there is a $z \in N$ with \begin{equation} \text{I}(N,x+S(z),S(y)) \tag{*}\end{equation} Hence by (Peano 1b and 3) \begin{equation} \text{I}(N,x+z,y) \end{equation}
Define a predicate \begin{equation} Q(u):=(\text{I}(N,x+u,y) \subset x<y \lor \text{I}(N,x,y) \end{equation}
It is sufficient to show $Q(z)$. We prove this by induction on $z$. To see $Q(0)$, not ethat if $\text{I}(N,x+0,y)$ holds then $\text{I}(N,x,y)$ is true by Peano 1a. Thus, $x<y \lor \text{I}(n,x,y)$. Now, we prove $Q(S(v))$: Suppose $\text{I}(N,x+S(v),y)$. From this definition we have $x<y$ and thus $x<y \lor \text{I}(N,x,y)$ also in this case. Finally, Peano's fifth axiom gives $Q(z)$ and by $(*)$ we get $x < y \lor \text{I}(N,x,y)$. \begin{equation} \tag{$\square$} \end{equation}
Theorem B
For all $x,y \in N$ \begin{equation} x <y \lor \text{I}(N,x,y) \lor y<x \end{equation} Proof:
If $x<y$ then $\neg \text{I}(N,x,y)$ by definition, and if $x>y$ then $\neg \text{I}(N,x,y)$ also by definition. If $x>y$ and $y>x$ then by transitivity and reflexivity, we have $\text{I}(N,x,y)$, which is a contradiction. Consequently, no more than one of the statements is true.
We keep $y$ fixed and induct on $x$. When $\text{I}(N,0,y)$ we have $0 < y \lor \text{I}(N,0,y)$ for all $y$, which proves the base case. Next, suppose the theorem holds for $x$; now we want to prove the theorem for $S(x)$. From the trichotomy for $x$, there are three cases: $x<y,\text{I}(N,x,y)$, and $x>y$. If $x>y$, then clearly $S(x) >y$. If $\text{I}(N,x,y)$, then $S(x) >y$ (as $S(x) >x$ for all $x\in \mathbb{N}$). Finally, suppose $x <y$ Then, by theorem A we have $S(x) < y$ or $\text{I}(N,S(x),y)$, and in either case we are done. \begin{equation} \tag{$\square$} \end{equation}
The theorems that I wish to prove, can be expressed as follows in Coq.
Lemma less_lem (x y:N) : less x (succ y) -> or (less x y) (I N x y).
Theorem Ntrichotomy: (forall x y:N, or (less x y) (or (I N x y) (less y x))).
Useful results
Here, I have gathered some of the results that I have defined, and proved up to this point. These are the ones that I refer to above. *This is the code that I have managed to write so far, note that most consists of definitions. *
(* Sigma types *)
Inductive Sigma (A:Set)(B:A -> Set) :Set :=
Spair: forall a:A, forall b : B a,Sigma A B.
Definition E (A:Set)(B:A -> Set)
(C: Sigma A B -> Set)
(c: Sigma A B)
(d: (forall x:A, forall y:B x,
C (Spair A B x y))): C c :=
match c as c0 return (C c0) with
| Spair a b => d a b
end.
(* Binary sum type *)
Inductive sum' (A B:Set):Set :=
inl': A -> sum' A B | inr': B -> sum' A B.
Print sum'_rect.
Definition D (A B : Set)(C: sum' A B -> Set)
(c: sum' A B)
(d: (forall x:A, C (inl' A B x)))
(e: (forall y:B, C (inr' A B y))): C c :=
match c as c0 return C c0 with
| inl' x => d x
| inr' y => e y
end.
(* Three useful finite sets *)
Inductive N_0: Set :=.
Definition R_0
(C:N_0 -> Set)
(c: N_0): C c :=
match c as c0 return (C c0) with
end.
Inductive N_1: Set := zero_1:N_1.
Definition R_1
(C:N_1 -> Set)
(c: N_1)
(d_zero: C zero_1): C c :=
match c as c0 return (C c0) with
| zero_1 => d_zero
end.
Inductive N_2: Set := zero_2:N_2 | one_2:N_2.
Definition R_2
(C:N_2 -> Set)
(c: N_2)
(d_zero: C zero_2)
(d_one: C one_2): C c :=
match c as c0 return (C c0) with
| zero_2 => d_zero
| one_2 => d_one
end.
(* Natural numbers *)
Inductive N:Set :=
zero: N | succ : N -> N.
Print N.
Print N_rect.
Definition R
(C:N -> Set)
(d: C zero)
(e: (forall x:N, C x -> C (succ x))):
(forall n:N, C n) :=
fix F (n: N): C n :=
match n as n0 return (C n0) with
| zero => d
| succ n0 => e n0 (F n0)
end.
(* Boolean to truth-value converter *)
Definition Tr (c:N_2) : Set :=
match c as c0 with
| zero_2 => N_0
| one_2 => N_1
end.
(* Identity type *)
Inductive I (A: Set)(x: A) : A -> Set :=
r : I A x x.
Print I_rect.
Theorem J
(A:Set)
(C: (forall x y:A,
forall z: I A x y, Set))
(d: (forall x:A, C x x (r A x)))
(a:A)(b:A)(c:I A a b): C a b c.
induction c.
apply d.
Defined.
(* functions are extensional wrt
identity types *)
Theorem I_I_extensionality (A B: Set)(f: A -> B):
(forall x y:A, I A x y -> I B (f x) (f y)).
Proof.
intros x y P.
induction P.
apply r.
Defined.
(* addition *)
Definition add (m n:N) : N
:= R (fun z=> N) m (fun x y => succ y) n.
(* multiplication *)
Definition mul (m n:N) : N
:= R (fun z=> N) zero (fun x y => add y m) n.
(* Axioms of Peano verified *)
Theorem P1a: (forall x: N, I N (add x zero) x).
intro x.
(* force use of definitional equality
by applying reflexivity *)
apply r.
Defined.
Theorem P1b: (forall x y: N,
I N (add x (succ y)) (succ (add x y))).
intros.
apply r.
Defined.
Theorem P2a: (forall x: N, I N (mul x zero) zero).
intros.
apply r.
Defined.
Theorem P2b: (forall x y: N,
I N (mul x (succ y)) (add (mul x y) x)).
intros.
apply r.
Defined.
Definition pd (n: N): N :=
R (fun _=> N) zero (fun x y=> x) n.
(* alternatively
Definition pd (x: N): N :=
match x as x0 with
| zero => zero
| succ n0 => n0
end.
*)
Theorem P3: (forall x y:N,
I N (succ x) (succ y) -> I N x y).
intros x y p.
apply (I_I_extensionality N N pd (succ x) (succ y)).
apply p.
Defined.
Definition not (A:Set): Set:= (A -> N_0).
Definition isnonzero (n: N): N_2:=
R (fun _ => N_2) zero_2 (fun x y => one_2) n.
Theorem P4 : (forall x:N,
not (I N (succ x) zero)).
intro x.
intro p.
apply (J N (fun x y z =>
Tr (isnonzero x) -> Tr (isnonzero y))
(fun x => (fun t => t)) (succ x) zero)
.
apply p.
simpl.
apply zero_1.
Defined.
Theorem P5 (P:N -> Set):
P zero -> (forall x:N, P x -> P (succ x))
-> (forall x:N, P x).
intros base step n.
apply R.
apply base.
apply step.
Defined.
(* I(A,-,-) is an equivalence relation *)
Lemma Ireflexive (A:Set): (forall x:A, I A x x).
intro x.
apply r.
Defined.
Lemma Isymmetric (A:Set): (forall x y:A, I A x y -> I A y x).
intros x y P.
induction P.
apply r.
Defined.
Lemma Itransitive (A:Set):
(forall x y z:A, I A x y -> I A y z -> I A x z).
intros x y z P Q.
induction P.
assumption.
Defined.
Lemma succ_cong : (forall m n:N, I N m n -> I N (succ m) (succ n)).
intros m n H.
induction H.
apply r.
Defined.
Lemma zeroadd: (forall n:N, I N (add zero n) n).
intro n.
induction n.
simpl.
apply r.
apply succ_cong.
auto.
Defined.
Lemma succadd: (forall m n:N, I N (add (succ m) n) (succ (add m n))).
intros.
induction n.
simpl.
apply r.
simpl.
apply succ_cong.
auto.
Defined.
Lemma commutative_add: (forall m n:N, I N (add m n) (add n m)).
intros n m; elim n.
apply zeroadd.
intros y H; elim (succadd m y).
simpl.
rewrite succadd.
apply succ_cong.
assumption.
Defined.
Lemma associative_add: (forall m n k:N,
I N (add (add m n) k) (add m (add n k))).
intros m n k.
induction k.
simpl.
apply Ireflexive.
simpl.
apply succ_cong.
assumption.
Defined.
Definition or (A B : Set):= sum' A B.
Definition less (m n: N) :=
Sigma N (fun z => I N (add m (succ z)) n).
Lemma less_lem (x y:N) :
less x (succ y) -> or (less x y) (I N x y).
intro.
destruct H.
right.
(* Here is where I'm working right now *)
Defined.
Theorem Ntrichotomy: (forall x y:N,
or (less x y) (or (I N x y) (less y x))).
