Coherence of Heyting arithmetic in Coq

Terms

Terms:

inductively defined over the Peano signature $(0, S, +, ×)$ as

• variable or constants

• $f(t_1, \ldots, t_n)$ where $f$ is a function of arity $n$ and the $t_i$ are terms

Fixpoint tlift n t k :=
  match t with
    | Tvar i => Tvar (if k <=? i then n+i else i)
    | Tzero => Tzero
    | Tsucc u => Tsucc (tlift n u k)
    | Tplus u v => Tplus (tlift n u k) (tlift n v k)
    | Tmult u v => Tmult (tlift n u k) (tlift n v k)
  end.

Lifting

Lifting:

tlift n t k increases by $n$ all the variables $≥ k$ in the term $t$ (recall that we’re using De Bruijn indices)

Lifting properties

• lifting by $n$ over $k$ then $n'$ over $k'$ $⟺$ lifting by $n+n'$ over $k$ provided that $k ≤ k' ≤ k+n$

  Lemma tlift_1 : forall t n n' k k', k <= k' -> k' <= k + n ->
    tlift n' (tlift n t k) k' = tlift (n' + n) t k.
  

• lifting by $n$ over $k$ then $n'$ over $k'$ $⟺$ lifting by $n$ over $n'+k$ then lifting by $n'$ over $k'$ provided that $k' ≤ k$

  Lemma tlift_2 : forall t n n' k k', k' <= k ->
    tlift n' (tlift n t k) k' = tlift n (tlift n' t k') (n' + k).
  

⟹ Straightforward proofs:

induction t; intros; simpl; f_equal; repeat break; auto.

relying on f_equal and the fact that inductively defined terms are of the form $f(t_1, \ldots, t_n)$

Substitutions

Substitution:

tsubst x t' t

• replaces x in t by t' where all the variables have been lifted by x

• and decrement by one all the variables greater than x in t

  this decrementing step stems from the $∀$-elimination and $∃$-introduction natural deduction rules:

    | Rforall_e Γ B t : Γ:-(Fforall B) -> Γ:-(fsubst 0 t B)
    | Rexists_i Γ B t : Γ:-(fsubst 0 t B) -> Γ:-(Fexists B)
  

Substitution-related Lemmas

• Substituting in a lifted term:

  Lemma tsubst_1 : forall t x t' n k, k <= x -> x <= k + n ->
    tsubst x t' (tlift (S n) t k) = tlift n t k.
  

• Substituting, then lifting $⟺$ Lifting then Substituting

  Lemma tsubst_2 : forall t x t' n k, k <= x ->
    tlift n (tsubst x t' t) k = tsubst (n + x) t' (tlift n t k).
  
  Lemma tsubst_3 : forall t x t' n k,
    tlift n (tsubst x t' t) (x + k) =
    tsubst x (tlift n t' k) (tlift n t (x + S k)).
  

• Substituting two times in a row:

  Lemma tsubst_4 : forall t x t' y u',
    tsubst (x + y) u' (tsubst x t' t) =
    tsubst x (tsubst y u' t') (tsubst (x + S y) u' t).
  

Bounded terms

Terms whose variables are $< n$:

Inductive cterm n : term -> Prop

cterm-related lemmas

• Monotony:

  Lemma cterm_1 : forall n t, cterm n t -> forall n', n <= n' -> cterm n' t.
  

• Lifting above the upper bound does nothing:

  Lemma cterm_2 : forall n t, cterm n t -> forall k, tlift k t n = t.
  

• Susbstituting above the upper bound either:

  Lemma cterm_3 : forall n t, cterm n t -> forall t' j, n <= j -> tsubst j t' t = t.
  

• Substituting the last variable by a closed term decrements the upper bound by one:

  Lemma cterm_4 : forall n t, cterm (S n) t ->
    forall t', cterm 0 t' -> cterm n (tsubst n t' t).
  

Valuations

Semantics: Interpretation of terms and formulas via valuations

Definition valuation := list nat.

Fixpoint tinterp (v:valuation) t :=
  match t with
    | Tvar j => nth j v O
    | Tzero => O
    | Tsucc t => S (tinterp v t)
    | Tplus t t' => tinterp v t + tinterp v t'
    | Tmult t t' => tinterp v t * tinterp v t'
  end.

Fixpoint finterp v A :=
  match A with
    | Fequal t t' => tinterp v t = tinterp v t'
    | Ffalse => False
    | Fand B C => finterp v B /\ finterp v C
    | For B C => finterp v B \/ finterp v C
    | Fimplies B C => finterp v B -> finterp v C
    | Fexists B => exists n, finterp (n::v) B
    | Fforall B => forall n, finterp (n::v) B
  end.

tinterp-related lemmas

tinterp:

• for a tlifted term
• for a tsubstituted one

Lemma tinterp_1 : forall t v0 v1 v2,
  tinterp (v0++v1++v2) (tlift (length v1) t (length v0)) =
  tinterp (v0++v2) t.

Lemma tinterp_2 : forall t' t v1 v2,
  tinterp (v1 ++ v2) (tsubst (length v1) t' t) =
  tinterp (v1 ++ (tinterp v2 t') :: v2) t.

Then: we lift these lemmas to formulas