139 lines
5.2 KiB
Org Mode
139 lines
5.2 KiB
Org Mode
# done
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Auxiliary operations:
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index family of a constructor
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Idx(K) := [0; arity(K)[
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head of an ordered family (we write x for any object here, value, pattern etc.)
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head((xᵢ)^{i ∈ I}) = x_min(I)
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tail of an ordered family
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tail((xᵢ)^{i ∈ I}) := (xᵢ)^{i ≠ min(I)}
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head constructor of a value or pattern
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constr(K(xᵢ)ⁱ) = K
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constr(_) = ⊥
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constr(x) = ⊥
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first non-⊥ element of an ordered family
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First((xᵢ)ⁱ) := ⊥ if ∀i, xᵢ = ⊥
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First((xᵢ)ⁱ) := x_min{i | xᵢ ≠ ⊥} if ∃i, xᵢ ≠ ⊥
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# end
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Definition of running a pattern row against a (same-size) value vector (pᵢ)ⁱ(vᵢ)ⁱ
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∅ (∅) := []
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(_, tail(pᵢ)ⁱ) (vᵢ) := tail(pᵢ)ⁱ(tail(vᵢ)ⁱ)
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(x, tail(pᵢ)ⁱ) (vᵢ) := σ[x↦v₀] if tail(pᵢ)ⁱ(tail(vᵢ)ⁱ) = σ
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(K(qⱼ)ʲ, tail(pᵢ)ⁱ) (K(v'ⱼ)ʲ,tail(vⱼ)ʲ) := ((qⱼ)ʲ +++ tail(pᵢ)ⁱ)((v'ⱼ)ʲ +++ tail(vᵢ)ⁱ)
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(K(qⱼ)ʲ, tail(pᵢ)ⁱ) (K'(v'ₗ)ˡ,tail(vⱼ)ʲ) := ⊥ if K ≠ K'
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(q₁|q₂, tail(pᵢ)ⁱ) (vᵢ)ⁱ := First((q₁,tail(pᵢ)ⁱ)(vᵢ)ⁱ, (q₂,tail(pᵢ)ⁱ)(vᵢ)ⁱ)
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Definition of group decomposition:
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let constrs((pᵢ)^{i ∈ I}) = { K | K = constr(pᵢ), i ∈ I }
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let Groups(m) where m = ((aᵢ)ⁱ ((pᵢⱼ)ⁱ → eⱼ)ⁱʲ) =
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let (Kₖ)ᵏ = constrs(pᵢ₀)ⁱ in
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( Kₖ →
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((a₀.ₗ)ˡ +++ tail(aᵢ)ⁱ)
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(
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if pₒⱼ is Kₖ(qₗ) then
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(qₗ)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ
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if pₒⱼ is _ then
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(_)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ
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else ⊥
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)ʲ
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), (
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tail(aᵢ)ⁱ, (tail(pᵢⱼ)ⁱ → eⱼ if p₀ⱼ is _ else ⊥)ʲ
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)
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Groups(m) is an auxiliary function that decomposes a matrix m into
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submatrices, according to the head constructor of their first pattern.
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Groups(m) returns one submatrix m_r for each head constructor K that
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occurs on the first row of m, plus one "wildcard submatrix" m_{wild}
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that matches on all values that do not start with one of those head
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constructors.
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Intuitively, m is equivalent to its decomposition in the following
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sense: if the first pattern of an input vector (v_i)^i starts with one
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of the head constructors Kₖ, then running (v_i)^i against m is the same
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as running it against the submatrix m_{Kₖ}; otherwise (its head
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constructor is none of the Kₖ) it is equivalent to running it against
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the wildcard submatrix.
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We formalize this intuition as follows:
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Lemma (Groups):
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Let $m$ be a matrix with $Groups(m) = (kᵣ → mᵣ)ᵏ, m_{wild}$.
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For any value vector $(vᵢ)ᴵ$ such that $v₀ = K(v'ₗ)ˡ$ for some
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constructor K,
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we have:
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\[
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if K = Kₖ for some k then
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m(vᵢ)ⁱ = mₖ((v'ₗ)ˡ +++ (vᵢ)^{i∈I\{0}})
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otherwise
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m(vᵢ)ⁱ = m_{wild}(vᵢ)^{i∈I\{0}}
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\]
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Proof:
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Let $m$ be a matrix ((aᵢ)ⁱ, ((pᵢⱼ)ⁱ → eⱼ)ʲ) with $Groups(m) = (Kₖ → mₖ)ᵏ, m_{wild}$.
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Below we are going to assume that m is a simplified matrix:
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the first row does not with an or-pattern or a variable
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I think it would be simpler to not consider them at all in the
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formalization. If we do want them, the discrepancy can be fixed
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later by defining a simplification step that explodes
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head-or-patterns and binds variable, and showing that it preserves
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the matrix semantics.
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Let (vᵢ)ⁱ be an input matrix with v₀ = Kᵥ(v'ₗ)ˡ for some constructor Kᵥ.
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We have to show that:
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- if Kₖ = Kᵥ for some Kₖ ∈ constrs(p₀ⱼ)ʲ, then
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m(vᵢ)ⁱ = mₖ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ)
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- otherwise
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m(vᵢ)ⁱ = m_{wild}(tail(vᵢ)ⁱ)
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Let us call (rₖⱼ) the j-th row of the submatrix mₖ, and rⱼ_{wild}
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the j-th row of the wildcard submatrix m_{wild}.
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Our goal contains same-behavior equalities between matrices, for
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a fixed input vector (vᵢ)ⁱ. It suffices to show same-behavior
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equalities between each row of the matrices for this input
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vector. We show that for any j,
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- if Kₖ = Kᵥ for some Kₖ ∈ constrs(p₀ⱼ)ʲ, then
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(pᵢⱼ)ⁱ(vᵢ)ⁱ = rₖⱼ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ
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- otherwise
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(pᵢⱼ)ⁱ(vᵢ)ⁱ = rⱼ_{wild} tail(vᵢ)ⁱ
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In the first case (Kᵥ is Kₖ for some Kₖ ∈ constrs(p₀ⱼ)ʲ), we
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have to prove that
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(pᵢⱼ)ⁱ(vᵢ)ⁱ = rₖⱼ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ
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By definition of mₖ we know that rₖⱼ is equal to
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if pₒⱼ is Kₖ(qₗ) then
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(qₗ)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ
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if pₒⱼ is _ then
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(_)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ
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else ⊥
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By definition of (pᵢ)ⁱ(vᵢ)ⁱ we know that (pᵢⱼ)ⁱ(vᵢ) is equal to
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(K(qⱼ)ʲ, tail(pᵢⱼ)ⁱ) (K(v'ₗ)ˡ,tail(vᵢ)ⁱ) := ((qⱼ)ʲ +++ tail(pᵢⱼ)ⁱ)((v'ₗ)ˡ +++ tail(vᵢ)ⁱ)
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(_, tail(pᵢⱼ)ⁱ) (vᵢ) := tail(pᵢⱼ)ⁱ(tail(vᵢ)ⁱ)
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(K(qⱼ)ʲ, tail(pᵢⱼ)ⁱ) (K'(v'ₗ)ˡ,tail(vⱼ)ʲ) := ⊥ if K ≠ K'
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We prove this first case by a second case analysis on p₀ⱼ.
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TODO
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In the second case (Kᵥ is distinct from Kₖ for all Kₖ ∈ constrs(pₒⱼ)ʲ),
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we have to prove that
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(pᵢⱼ)ⁱ(vᵢ)ⁱ = rⱼ_{wild} tail(vᵢ)ⁱ
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TODO
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