continuing with proof
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Francesco Mecca
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@ -960,16 +960,104 @@ Are K and Here clear here?
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We define the decomposition matrix /mₛ/ as
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We define the decomposition matrix /mₛ/ as
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| SMatrix mₛ := (aⱼ)^{j∈J}, ((pᵢⱼ)^{j∈J} → bbᵢ)^{i∈I}
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| SMatrix mₛ := (aⱼ)^{j∈J}, ((pᵢⱼ)^{j∈J} → bbᵢ)^{i∈I}
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Given that the decomposition matrix is computed from source terms, we
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\begin{comment}
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need to prove that it respects:
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Correggi prendendo in considerazione l'accessor
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| ∀(vⱼ)^{j∈J}, mₛ(vⱼ)ʲ = 〚mₛi 〛(vⱼ)ʲ
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\end{comment}
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v(Here) = v
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We define the constraint tree of source programs
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K(vᵢ)ⁱ(k.a) = vₖ(a) if k ∈ [0;n[
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[|tₛ|]
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in terms of the constraint tree of pattern matrices
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[|mₛ|]
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by the following:
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〚((pᵢ → bbᵢ)^{i∈I}〛 := 〚(Root), (pᵢ → bbᵢ)^{i∈I} 〛 # i ∈ I
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We also said that
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Constraint tree computed from pattern matrices respect the following invariant:
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(Leaf bb)(v) := Match bb
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| ∀v (vᵢ)^{i∈I} = v(aᵢ)^{i∈I} → 〚m〛(v) = m(vᵢ)^{i∈I} for m = ((aᵢ)^{i∈I}, (rᵢ)^{i∈I})
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(Node(a, (kᵢ → cᵢ)ⁱ, c_{fallback}))(v)
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where
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let v(a) be K(vⱼ)ʲ
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| v(Here) = v
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if k ∈ {Kᵢ}ⁱ then C_{min{i}|k=kᵢ}(v)
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| K(vᵢ)ⁱ(k.a) = vₖ(a) if k ∈ [0;n[
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else c_{fallback}(v)
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\begin{comment}
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TODO: EXPLAIN
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\end{comment}
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We proceed to show the correctness of the invariant by a case
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analysys.
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Base cases:
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1. [| ∅, (∅ → bbᵢ)ⁱ |] := Leaf bbᵢ where i := min(I), that is a
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constraint tree [|ms|] defined by an empty accessor and empty
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patterns pointing to blackboxes /bbᵢ/. This respects the invariant
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because a decomposition matrix in the case of empty rows returns
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the first expression and we known that (Leaf bb)(v) := Match bb
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2. [| (aⱼ)ʲ, ∅ |] := Failure
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Regarding non base cases:
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We define
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| let Idx(k) = [0; arity(k)[
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| let First(∅) = ⊥
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| let First((aⱼ)ʲ) = a_{min(j)} ## where j ∈ J ≠ ∅
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** Proof of equivalence checking
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The equivalence checking algorithm takes as parameters an input space
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/S/, a source constraint tree /Cₛ/ and a target constraint tree /Cₜ/:
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| equiv(S, Cₛ, Cₜ) → Yes | No(vₛ, vₜ)
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When the algorithm returns Yes and the input space is covered by /Cₛ/
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we can say that the couple of constraint trees are the same for
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every couple of source value /vₛ/ and target value /vₜ/ that are equivalent.
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\begin{comment}
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Define "covered"
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Is it correct to say the same? How to correctly distinguish in words ≊ and = ?
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\end{comment}
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| equiv(S, Cₛ, Cₜ) = Yes and cover(Cₜ, S) → ∀ vₛ ≊ vₜ∈S ∧ Cₛ(vₛ) = Cₜ(vₜ)
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In the case where the algorithm returns No we have at least a couple
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of counter example values vₛ and vₜ for which the two constraint trees
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outputs a different result.
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| equiv(S, Cₛ, Cₜ) = No(vₛ,vₜ) and cover(Cₜ, S) → ∀ vₛ ≊ vₜ∈S ∧ Cₛ(vₛ) ≠ Cₜ(vₜ)
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We define the following
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| Forall(Yes) = Yes
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| Forall(Yes::l) = Forall(l)
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| Forall(No(vₛ,vₜ)::_) = No(vₛ,vₜ)
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There exists and are injective:
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| int(k)∈ℕ (ar(k) = 0)
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| tag(k)∈ℕ (ar(k) > 0)
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| π(k) = {n|int(k) = n} x {n|tag{k} = n}
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where k is a constructor.
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\begin{comment}
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TODO: explain:
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∀v∈a→π, C_{/a→π}(v) = C(v)
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\end{comment}
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We proceed by case analysis:
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1. equiv(∅, Cₛ, Cₜ) := Yes
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\begin{comment}
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Devo spiegarlo?
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\end{comment}
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In the other subcases S is always non-empty.
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2. equiv(S, Failure, Failure) := Yes
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the statement holds because of equality between Failure nodes in
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the case of every possible value /v/.
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3. The result of the subcase where we have a source constraint tree
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/Cₛ/ that is either a Leaf terminal or a Failure terminal and a
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target constraint tree defined by an accessor /a/ and a positive
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number of couples constraint πᵢ and children nodes Cₜᵢ. The output
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the output of the algorithm is:
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| equiv(S, (Leaf bbₛ|Failure) as Cₛ, Node(a, (πᵢ → Cₜᵢ)ⁱ)) := Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)ⁱ)
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The statement holds because defined let Sᵢ := S∩(a→πᵢ)
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either the algorithm is true for every sub-input space Sᵢ and
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subtree Cₜᵢ
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| equiv(Sᵢ, Cₛ, Cₜᵢ) = Yes ∀i
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or we have a counter example vₛ, vₜ for which
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| vₛ≊vₜ∈Sₖ ∧ cₛ(vₛ) ≠ Cₜₖ(vₜ)
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then because
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| vₜ∈(a→πₖ) ⇒ Cₜ(vₜ) = Cₜₖ(vₜ)
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then
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| vₛ≊vₜ∈S ∧ Cₛ(vₛ)≠Cₜ(vₜ)
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and the result of the algorithm is
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| equiv(Sᵢ, Cₛ, Cₜᵢ) = No(vₛ, vₜ) for some minimal k∈I
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