TP03: Curry-Howard & induction in HA


From last week

Using the Curry-Howard isomorphism, find λ-terms proving the following formulæ.
  • P ∧ (Q ∨ R) → (P ∧ Q) ∨ (P ∧ R)
  • ((P ∨ Q) → R) → (P → R) ∧ (Q → R)
  • (P → R) ∨ (Q → R) → P ∧ Q → R
Section CurryHoward.
Variables P Q R : Prop.

Prove the following statements in natural deduction (on paper). What are the corresponding λ-terms ?
  • ¬¬P → ¬¬Q → ¬¬(P ∧ Q)
  • (P → ¬¬Q) → ¬¬(P → Q)
  • (P → ¬¬Q) → ¬¬P → ¬¬Q

End CurryHoward.

On Induction in Arithmetic

The idea of this part is to prove some results about Heyting arithmetic. Since we want to prove everything without relying on the standrad library, we define our set of natural numbers Nat.

Axiomatisation of (some part of) Heyting arithmetic

Parameter Nat : Set.

Function and relation symbols
Parameter Succ : Nat -> Nat.
Parameter Zero : Nat.
Parameter Lt : Nat -> Nat -> Prop.
Notation "x < y" := (Lt x y).

Axioms
Axiom Induction : forall P,
  P Zero -> (forall x, P x -> P (Succ x)) -> forall x, P x.

Axiom NonConf : forall x, ~ Succ x = Zero.
Axiom SuccInj : forall x y, Succ x = Succ y -> x = y.

Axiom Not_Lt_Zero : forall x, ~ (x < Zero).
Axiom Lt_Succ_disj : forall x y, y < Succ x -> y < x \/ y = x.
Axiom Lt_Succ : forall x, x < Succ x.
Axiom Lt_trans : forall x y z, x < y -> y < z -> x < z.

Our goal is to prove:
  • decidability of equality
  • well-founded induction
  • equivalence of excluded middle and the minimum principle

Some basic results about Succ and <

As a warm-up, prove the following lemmata.
Lemma Zero_or_Succ : forall x, x = Zero \/ exists y, x = Succ y.

Lemma Succ_Not_Idempotent : forall x, ~ Succ x = x.

Lemma Lt_Zero_Succ : forall x, Zero < Succ x.

Decidability of equality

Find the statement expressing that equality in Heyting arithmetic is decidable (Remember: we are in intuitionistic logic). Prove it.
Theorem EqDec : ...

Well founded induction

Let us start with the most obvious proof sketch.
Theorem WF_Induction : forall P : Nat -> Prop,
  (forall x, (forall y, y < x -> P y) -> P x) -> forall x, P x.
Proof.
intros P HP. apply Induction.
  apply HP. intros y ?. exfalso. apply (Not_Lt_Zero y). assumption.
  intros x Hx.
Abort.

1) Where does the problem comes from? How can we strenghten the induction hypothesis to avoid it? Prove the lemma with the strengthened hypothesis.
Lemma WF_Pre_ind : ...

2) Conclude by proving well-founded induction for Nat.
Theorem wf_Nat_ind : forall P : Nat -> Prop,
  (forall x, (forall y, y < x -> P y) -> P x) -> forall x, P x.

In HA, the minimum principle is equivalent to the excluded middle


Definition MinP (P : Nat -> Prop) : Prop :=
  (exists x, P x) ->
  exists x, P x /\ forall y, y < x -> ~ P y.

1) Prove that excluded middle implies the minimum principle.
Theorem EM_MinP : (forall P, P \/ ~P) -> forall A, MinP A.

2) For the converse implication, we will use the minimum principle on the following proposition:
      Q x := x = Zero ∧ P ∨ x = Succ Zero ∧ ¬P ∨ x = Succ (Succ Zero)
Using this trick, prove the converse implication.
Theorem MinP_EM : (forall A, MinP A) -> forall P, P \/ ~ P.

Double negation of excluded middle and of the minimum principle

1) Prove the double negation of excluded middle.
Lemma NotNotEM : forall A, ~ ~ (A \/ ~ A).

For the double negation of the minimum principle, we will use an alternate fomulation of ¬¬ MinP:
      ∀P (x : Nat), P x → ¬¬(∃ x, P x /\ ∀ y, y < x → ¬P y).
    
Definition NotNotMinP_alt (P : Nat -> Prop) :=
  forall x, P x -> ~ ~ exists x, P x /\ forall y, y < x -> ~ P y.

2) Prove the following lemma and then that the alternate formulation implies ¬¬MinP.
Definition forallP_exP : forall T (A : T -> Prop) (P : Prop),
  (forall x, A x -> P) -> (exists x, A x) -> P.

3) To prove the alternate formulation, we note that on a doubly negated statement, we can use excluded middle.
Lemma NotNot_EM : forall P A, ((P \/ ~P) -> ~~A) -> ~~A.

4) Now prove the alternate formulation and conclude.
Theorem NotNotMinP_Shift : forall P, NotNotMinP_alt P.

Corollary NotNotMinP : forall P, ~~MinP P.

Definition of < by +

Finally, we prove that by defining
      x < y := exists z, Succ z + x = y
we get an order satisfying all the axioms we used do far.
Definition of + and its properties
Parameter Plus : Nat -> Nat -> Nat.
Notation "x + y" := (Plus x y).
Axiom PlusZero : forall x, Zero + x = x.
Axiom PlusSucc : forall x y, Succ x + y = Succ (x + y).
Notation "x < y" := (exists z, Succ z + x = y).

Prove the four lemmata.
Lemma Not_PLt_Zero : forall x, ~ (x < Zero).

Lemma PLt_Succ : forall x, x < Succ x.

Lemma PLt_Succ_disj : forall x y, y < Succ x -> y < x \/ y = x.

Lemma PLt_trans : forall x y z, x < y -> y < z -> x < z.
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