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Classical logic is somewhat surprising! Prove that given two arbitrary propositions a and b, either a implies b or the converse

Enrico is a lazy student. When asked to reverse and filter a list he comes up with the following ugly code.

The teacher asks him to prove that his ugly code is equivalent to the nicer code [seq x <- rev l | p x] he could have written by reusing the seq library.

Such proof is not trivial:

filtrev has an accumulator, rev (at least apparently) does not.
which is the invariant linking the accumulator acc and p in the code of filtrev? Depending on how you expose the accumulator on the right hand side of the goal, such invariant may need to be taken into account by the induction.

Relevant keywords for Search are: rev cons cat filter rcons Hint: it is perfectly fine to state intermediate lemmas

Prove the equivalence of these two sums. E.g. (n=8)

    1 + 3 + 5 + 7 = 7-0 + 7-2 + 7-4 + 7-6

Even the Mathematical Components library misses some theorems, for example the following one.

Would you help us improve the library with this lemma?

Hint: check out the theory of iota

Now, some algebra.

Big operations and zmodType. Now that ring_scope is open we are in an algebraic setting. Think about the meaning of + now, and find the right injectivity lemma.

We remind what Gauss integer are and recall some theory about it.

Definition gaussInteger := [qualify a x | ('Re x \in Cint) && ('Im x \in Cint)].
Axiom Cint_GI : forall (x : algC), x \in Cint -> x \is a gaussInteger.
Axiom GI_subring : subring_closed gaussInteger.

Fact GI_key : pred_key gaussInteger. Proof. by []. Qed.
Canonical GI_keyed := KeyedQualifier GI_key.
Canonical GI_opprPred := OpprPred GI_subring.
Canonical GI_addrPred := AddrPred GI_subring.
Canonical GI_mulrPred := MulrPred GI_subring.
Canonical GI_zmodPred := ZmodPred GI_subring.
Canonical GI_semiringPred := SemiringPred GI_subring.
Canonical GI_smulrPred := SmulrPred GI_subring.
Canonical GI_subringPred := SubringPred GI_subring.

Record GI := GIof {algGI : algC; algGIP : algGI \is a gaussInteger }.
Hint Resolve algGIP.

Canonical GI_subType := [subType for algGI].
Definition GI_eqMixin := [eqMixin of GI by <:].
Canonical GI_eqType := EqType GI GI_eqMixin.
Definition GI_choiceMixin := [choiceMixin of GI by <:].
Canonical GI_choiceType := ChoiceType GI GI_choiceMixin.
Definition GI_countMixin := [countMixin of GI by <:].
Canonical GI_countType := CountType GI GI_countMixin.
Definition GI_zmodMixin := [zmodMixin of GI by <:].
Canonical GI_zmodType := ZmodType GI GI_zmodMixin.
Definition GI_ringMixin := [ringMixin of GI by <:].
Canonical GI_ringType := RingType GI GI_ringMixin.
Definition GI_comRingMixin := [comRingMixin of GI by <:].
Canonical GI_comRingType := ComRingType GI GI_comRingMixin.

Lemma conjGIE x : (x^* \is a gaussInteger) = (x \is a gaussInteger).
Proof. by rewrite ![_ \is a _]qualifE algRe_conj algIm_conj rpredN. Qed.

Fact conjGI_subproof (x : GI) : (val x)^* \is a gaussInteger.
Proof. by rewrite conjGIE. Qed.

Canonical conjGI x := GIof (conjGI_subproof x).

Definition gaussNorm (x : algC) := x * x^*.

Axiom gaussNormE : forall x, gaussNorm x = `|x| ^+ 2.
Axiom gaussNormCnat : forall (x : GI), gaussNorm (val x) \in Cnat.

Prove these two facts. (Hint: conjugation is a morphism)

Question: Prove that GI euclidean for the stasm gaussNorm.

i.e. ∀ (a, b) ∈ GI × GI*, ∃ (q, r) ∈ GI² s.t. a = q b + r and φ(r) < φ(b)
Suggested strategy: sketch the proof on a paper first, don't let Coq divert you from your proofsketch
We first sketch the paper proof here and then do it in Coq:
- take a / b = x + i y
- take u the closest integer to x, and v the closest integer to y
- satisfy the existential with q = u + i v and r = a - q b, which are both Gauss integers.
- We want to show that |a - q b|² < |b|².
- It suffices to show |a / b - q|² < 1
- But |a / b - q|² = (u - x)² + (v - x)² ≤ ‌½² + ½² < 1
Now we give a Coq proof with holes, fill in the holes.

Lemma euclideanGI (a b : GI) : b != 0 ->
  exists2 qr : GI * GI, a = qr.1 * b + qr.2
                      & (gaussNorm (val qr.2) < gaussNorm (val b)).
Proof.
move=> b_neq0.
(* Trivial preliminaries *)
have oneV2 : 1 = 2%:R^-1 + 2%:R^-1 :> algC.
  by rewrite -mulr2n -[_ *+ 2]mulr_natr mulVf ?pnatr_eq0.
have V2ge0 : 0 <= 2%:R^-1 :> algC by admit.
have V2real : (2%:R^-1 : algC) \is Num.real by admit.
(* Closest integer to x, when x is real *)
pose approx (x : algC) : int :=
  floorC x + (if `|x - (floorC x)%:~R| <= 2%:R^-1 then 0 else 1).
have approxP x : x \is Creal -> `|x - (approx x)%:~R| <= 2%:R^-1.
  rewrite /approx => x_real; have /andP [x_ge x_le] := floorC_itv x_real.
  have [] // := @real_lerP _  `|_ - (floorC _)%:~R| _;
    first by rewrite addr0.
  rewrite [`|_ - (_ + 1)%:~R|]distrC !ger0_norm ?subr_ge0 //=;
     last by rewrite ltrW.
  move=> Dx1_gtV2; rewrite real_lerNgt ?rpredB // ?Creal_Cint ?Cint_int //.
  apply/negP=> /ltr_add /(_ Dx1_gtV2); rewrite -oneV2 !addrA addrNK.
  by rewrite [_ + 1]addrC rmorphD /= addrK ltrr.
have approxP2 x (_ : x \is Creal) : `|x - (approx x)%:~R| ^+ 2 < 2%:R^-1.
  rewrite (@ler_lt_trans _ (2%:R^-1 ^+ 2)) // ?ler_expn2r ?qualifE ?approxP //.
  by rewrite exprVn -natrX ltf_pinv ?qualifE ?ltr_nat ?ltr0n.
(* Proper proof *)
pose u := 'Re (val a / val b); pose v := 'Im (val a / val b).
have qGI : (approx u)%:~R + algCi * (approx v)%:~R \is a gaussInteger.
  (* Hint: use qualifE, alg*_rect and lemmas about Creal, Cint and _*)
  admit.
pose q := GIof qGI.
exists (q, a - q * b); first by rewrite addrC addrNK.
rewrite !gaussNormE /=.
rewrite -(@ltr_pmul2r _ (`|val b| ^-2)) ?invr_gt0 ?exprn_gt0 ?normr_gt0 //.
rewrite mulfV ?expf_eq0 /= ?normr_eq0 // -exprVn -exprMn.
rewrite -normfV -normrM mulrBl mulfK //.
suff uvP : `|u - (approx u)%:~R + 'i * (v - (approx v)%:~R)| ^+ 2 < 1.
  (* Hint: use algCrect and algebraic transformations *)
  admit.
set Du := _ - _; set Dv := _ - _.
have /andP [DuReal DvReal] : (Du \is Creal) && (Dv \is Creal).
(* Hint: use rpred*, Creal_*, normC2_rect, ream_normK and approxP2 *)
admit.
Admitted.

End GaussIntegers.
End AlgebraicHierarchy.

Section PolynomialsLagrange.

Open Scope ring_scope.
Import GRing.Theory Num.Theory.

Definition and properties of Lagrange polynomials.

Prove only one of the following lemmas

Taylor formula for polynomials

Endomorphisms u such that Ker u ⊕ Im u = E.

The endomorphisms of a space E of finite dimension n, such that u o v = 0 and v + u is invertible are exactly the endomorphisms such that Ker u ⊕ Im u = E.

Assume u o v = 0 and v + u is invertible,
- we have rank (v + u) = n
- we have rank v + rank u = n and we have Im v ⊂ Ker u
- Hence we have Im v = Ker u
- We deduce that Ker u ⊕ Im u = Im v ⊕ Im u = E

Variables (F : fieldType) (n' : nat).
Let n := n'.+1.

Lemma ex_6_13 (u : 'M[F]_n):
  reflect (exists2 v : 'M_n, v * u = 0 & v + u \is a GRing.unit)
          ((kermx u + u == 1)%MS && mxdirect (kermx u + u)).
Proof.
(* Hint: use mxrank* lemmas and search on addsmx, submx and eqmx
sometimes. Don't forget about mxdirect_addsP and sub_kermxP. *)
apply: (iffP idP) => [|[v vMu vDu]]; last first.
  have rkvDu: \rank (v + u)%R = n by admit.
  have /eqP rkvDrku : (\rank v + \rank u)%N == n.
    by rewrite eqn_leq; admit.
  have sub_v_ku : (v <= kermx u)%MS by admit.
  have /eqmxP/eqmx_sym eq_vu: (v == kermx u)%MS.
    rewrite -(geq_leqif (mxrank_leqif_eq _)) //.
    admit.
  rewrite submx1 sub1mx -col_leq_rank mxdirectEgeq /=.
  (* use adds_eqmx to lift eq_vu to a sum *)
  (* Warning: - (u + v)MS is a sum of spaces
     use addmx_sub_adds and mxrankS
     to compare the rank (u + v) with dim (Im u + Im v) *)
  (* finish using hypothesis *)
  admit.
move=> /andP [/eqmxP kuDu_eq1 /mxdirect_addsP kvDu_direct].
pose v := proj_mx (kermx u) u; exists v.
  admit.
rewrite -row_free_unit -kermx_eq0.
apply/negP => /negP /rowV0Pn [x /sub_kermxP]; rewrite mulmxDr.
move=> /(canRL (addrK _)); rewrite sub0r => eq_xv_Nxu.
apply/negP; rewrite negbK; apply/eqP.
have : (x *m v <= kermx u :&: u)%MS.
(* Hint: use sub_*, proj_mx*, eqmx_*, mxdirect_addsP, *)
  admit.
Admitted.

End LinearAlgebra.