correzioni gabriel #2

tesi/conv.py

@@ -7,7 +7,7 @@ try:
 except:
     allsymbols = json.load(open('../unicode-latex.json'))
     
-mysymbols = ['≡', '≠', '≼', '→', '←', '⊀', '⋠', '≺', '∀',  'ε', '₀', '₂', '₁', '₃', '₄', 'ₐ', 'ₖ', 'ₘ', 'ₙ', 'ᵢ', 'ⁱ', '⋮', 'ₛ', 'ₜ', '≃', '⇔', '∧', '∅', 'ℕ', 'ⱼ', 'ʲ', '⊥', 'π', 'α', 'β', '∞', 'σ', '≤', '⊈', '∧', '∨', '∃', '⇒', '∩', '∉', '⋃', 'ᵏ', 'ₗ', 'ˡ', 'ₒ', 'ᵣ', 'ᴵ', '≈', '⊆' ]
+mysymbols = ['≡', '≠', '≼', '→', '←', '⊀', '⋠', '≺', '∀',  'ε', '₀', '₂', '₁', '₃', '₄', 'ₐ', 'ₖ', 'ᵥ', 'ₘ', 'ₙ', 'ᵢ', 'ⁱ', '⋮', 'ₛ', 'ₜ', '≃', '⇔', '∧', '∅', 'ℕ', 'ⱼ', 'ʲ', '⊥', 'π', 'α', 'β', '∞', 'σ', '≤', '⊈', '∧', '∨', '∃', '⇒', '∩', '∉', '⋃', 'ᵏ', 'ₗ', 'ˡ', 'ₒ', 'ᵣ', 'ᴵ', '≈', '⊆' ]
 extrasymbols = {'〚': '\llbracket', '〛': r'\rrbracket', '̸': '\neg', '¬̸': '\neg', '∈': '\in ', 'ₛ': '_S', 'ₜ': '_T'}
 
 symbols = {s: allsymbols[s] for s in mysymbols}

tesi/drafts/partial-decomposition-proof.org

@@ -0,0 +1,139 @@
+# done
+Auxiliary operations:
+
+  index family of a constructor
+  Idx(K) := [0; arity(K)[
+
+  head of an ordered family (we write x for any object here, value, pattern etc.)
+  head((xᵢ)^{i ∈ I}) = x_min(I)
+
+  tail of an ordered family
+  tail((xᵢ)^{i ∈ I}) := (xᵢ)^{i ≠ min(I)}
+
+  head constructor of a value or pattern
+  constr(K(xᵢ)ⁱ) = K
+  constr(_) = ⊥
+  constr(x) = ⊥
+
+  first non-⊥ element of an ordered family
+  First((xᵢ)ⁱ) := ⊥                   if ∀i, xᵢ = ⊥
+  First((xᵢ)ⁱ) := x_min{i | xᵢ ≠ ⊥}   if ∃i, xᵢ ≠ ⊥
+# end
+
+Definition of running a pattern row against a (same-size) value vector  (pᵢ)ⁱ(vᵢ)ⁱ
+  ∅                   (∅)                  := []
+  (_, tail(pᵢ)ⁱ)      (vᵢ)                 := tail(pᵢ)ⁱ(tail(vᵢ)ⁱ)
+  (x, tail(pᵢ)ⁱ)      (vᵢ)                 := σ[x↦v₀]        if tail(pᵢ)ⁱ(tail(vᵢ)ⁱ) = σ
+  (K(qⱼ)ʲ, tail(pᵢ)ⁱ) (K(v'ⱼ)ʲ,tail(vⱼ)ʲ)  := ((qⱼ)ʲ +++ tail(pᵢ)ⁱ)((v'ⱼ)ʲ +++ tail(vᵢ)ⁱ)
+  (K(qⱼ)ʲ, tail(pᵢ)ⁱ) (K'(v'ₗ)ˡ,tail(vⱼ)ʲ) := ⊥              if K ≠ K'
+  (q₁|q₂, tail(pᵢ)ⁱ)  (vᵢ)ⁱ                := First((q₁,tail(pᵢ)ⁱ)(vᵢ)ⁱ, (q₂,tail(pᵢ)ⁱ)(vᵢ)ⁱ)
+
+Definition of group decomposition:
+  let constrs((pᵢ)^{i ∈ I}) = { K | K = constr(pᵢ), i ∈ I }
+  let Groups(m) where m = ((aᵢ)ⁱ ((pᵢⱼ)ⁱ → eⱼ)ⁱʲ) =
+    let (Kₖ)ᵏ = constrs(pᵢ₀)ⁱ in
+    ( Kₖ →
+        ((a₀.ₗ)ˡ +++ tail(aᵢ)ⁱ)
+        (
+         if pₒⱼ is Kₖ(qₗ) then
+            (qₗ)ˡ +++ tail(pᵢⱼ)ⁱ  → eⱼ
+         if pₒⱼ is _ then
+            (_)ˡ +++ tail(pᵢⱼ)ⁱ   → eⱼ
+         else ⊥
+        )ʲ
+    ), (
+      tail(aᵢ)ⁱ, (tail(pᵢⱼ)ⁱ  → eⱼ if p₀ⱼ is _ else ⊥)ʲ
+    )
+
+Groups(m) is an auxiliary function that decomposes a matrix m into
+submatrices, according to the head constructor of their first pattern.
+Groups(m) returns one submatrix m_r for each head constructor K that
+occurs on the first row of m, plus one "wildcard submatrix" m_{wild}
+that matches on all values that do not start with one of those head
+constructors.
+
+Intuitively, m is equivalent to its decomposition in the following
+sense: if the first pattern of an input vector (v_i)^i starts with one
+of the head constructors Kₖ, then running (v_i)^i against m is the same
+as running it against the submatrix m_{Kₖ}; otherwise (its head
+constructor is none of the Kₖ) it is equivalent to running it against
+the wildcard submatrix.
+
+We formalize this intuition as follows:
+
+Lemma (Groups):
+  Let $m$ be a matrix with $Groups(m) = (kᵣ → mᵣ)ᵏ, m_{wild}$.
+  For any value vector $(vᵢ)ᴵ$ such that $v₀ = K(v'ₗ)ˡ$ for some
+  constructor K,
+  we have:
+  \[
+    if K = Kₖ for some k then
+      m(vᵢ)ⁱ = mₖ((v'ₗ)ˡ +++ (vᵢ)^{i∈I\{0}})
+    otherwise
+      m(vᵢ)ⁱ = m_{wild}(vᵢ)^{i∈I\{0}}
+  \]
+
+Proof:
+
+  Let $m$ be a matrix ((aᵢ)ⁱ, ((pᵢⱼ)ⁱ → eⱼ)ʲ) with $Groups(m) = (Kₖ → mₖ)ᵏ, m_{wild}$.
+
+  Below we are going to assume that m is a simplified matrix:
+
+    the first row does not with an or-pattern or a variable
+
+  I think it would be simpler to not consider them at all in the
+  formalization. If we do want them, the discrepancy can be fixed
+  later by defining a simplification step that explodes
+  head-or-patterns and binds variable, and showing that it preserves
+  the matrix semantics.
+
+  Let (vᵢ)ⁱ be an input matrix with v₀ = Kᵥ(v'ₗ)ˡ for some constructor Kᵥ.
+
+  We have to show that:
+  - if Kₖ = Kᵥ for some Kₖ ∈ constrs(p₀ⱼ)ʲ, then
+      m(vᵢ)ⁱ = mₖ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ)
+  - otherwise
+      m(vᵢ)ⁱ = m_{wild}(tail(vᵢ)ⁱ)
+
+  Let us call (rₖⱼ) the j-th row of the submatrix mₖ, and rⱼ_{wild}
+  the j-th row of the wildcard submatrix m_{wild}.
+
+  Our goal contains same-behavior equalities between matrices, for
+  a fixed input vector (vᵢ)ⁱ. It suffices to show same-behavior
+  equalities between each row of the matrices for this input
+  vector. We show that for any j,
+
+  - if Kₖ = Kᵥ for some Kₖ ∈ constrs(p₀ⱼ)ʲ, then
+      (pᵢⱼ)ⁱ(vᵢ)ⁱ = rₖⱼ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ
+  - otherwise
+      (pᵢⱼ)ⁱ(vᵢ)ⁱ = rⱼ_{wild} tail(vᵢ)ⁱ
+
+  In the first case (Kᵥ is Kₖ for some Kₖ ∈ constrs(p₀ⱼ)ʲ), we
+  have to prove that
+
+    (pᵢⱼ)ⁱ(vᵢ)ⁱ = rₖⱼ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ
+
+  By definition of mₖ we know that rₖⱼ is equal to
+  
+    if pₒⱼ is Kₖ(qₗ) then
+       (qₗ)ˡ +++ tail(pᵢⱼ)ⁱ  → eⱼ
+    if pₒⱼ is _ then
+       (_)ˡ +++ tail(pᵢⱼ)ⁱ   → eⱼ
+    else ⊥
+
+  By definition of (pᵢ)ⁱ(vᵢ)ⁱ we know that (pᵢⱼ)ⁱ(vᵢ) is equal to
+
+    (K(qⱼ)ʲ, tail(pᵢⱼ)ⁱ) (K(v'ₗ)ˡ,tail(vᵢ)ⁱ)  := ((qⱼ)ʲ +++ tail(pᵢⱼ)ⁱ)((v'ₗ)ˡ +++ tail(vᵢ)ⁱ)
+    (_, tail(pᵢⱼ)ⁱ)      (vᵢ)                 := tail(pᵢⱼ)ⁱ(tail(vᵢ)ⁱ)
+    (K(qⱼ)ʲ, tail(pᵢⱼ)ⁱ) (K'(v'ₗ)ˡ,tail(vⱼ)ʲ) := ⊥              if K ≠ K'
+
+  We prove this first case by a second case analysis on p₀ⱼ.
+
+    TODO
+
+  In the second case (Kᵥ is distinct from Kₖ for all Kₖ ∈ constrs(pₒⱼ)ʲ),
+  we have to prove that
+
+      (pᵢⱼ)ⁱ(vᵢ)ⁱ = rⱼ_{wild} tail(vᵢ)ⁱ
+
+    TODO

tesi/tesi_unicode.org

@@ -336,7 +336,7 @@ for content equality.
 \end{mathpar}
 
 If the source decision tree (left hand side) is a terminal while the
-target decistion tree (right hand side) is not, the algorithm proceeds
+target decisiorn tree (right hand side) is not, the algorithm proceeds
 by \emph{explosion} of the right hand side. Explosion means that every
 child of the right hand side is tested for equality against the left
 hand side.
@@ -1316,42 +1316,53 @@ Base cases:
    the first expression and (Leaf bb)(v) := Match bb
 2. [| (aⱼ)ʲ, ∅ |] ≡ Failure 
 Regarding non base cases:
-Let's first define
+Let's first define some auxiliary functions
-| let Idx(k) := [0; arity(k)[
+- The index family of a constructor
-| let First(∅) := ⊥
+  | Idx(K) := [0; arity(K)[
-| let First((aⱼ)ʲ) := a_{min(j∈J≠∅)} 
+- head of an ordered family (we write x for any object here, value, pattern etc.)
-\[
+  | head((xᵢ)^{i ∈ I}) = x_min(I)
-m := ((a_i)^i ((p_{ij})^i \to e_j)^{ij})
+- tail of an ordered family
-\]
+  | tail((xᵢ)^{i ∈ I}) := (xᵢ)^{i ≠ min(I)}
-\[
+- head constructor of a value or pattern
- (k_k)^k := headconstructor(p_{i0})^i
+  | constr(K(xᵢ)ⁱ) = K
-\]
+  | constr(_) = ⊥
-| Groups(m) := (k_{k} \to  ((a)_{0.l})^{l  \in  Idx(k_{k})} +++ (a_{i})^{i\in{}I\DZ}), 
+  | constr(x) = ⊥
-|(
+- first non-⊥ element of an ordered family
-| \quad if p_{0j} is k(q_{l}) then 
+  | First((xᵢ)ⁱ) := ⊥ \quad \quad \quad if ∀i, xᵢ = ⊥
-| \quad \quad \quad (qₗ)^{l \in Idx(k_k)} +++ (p_{ij})^{i\in{}I\DZ} \to e_{j} 
+  | First((xᵢ)ⁱ) := x_min{i \vert{} xᵢ ≠ ⊥}  \quad  if ∃i, xᵢ ≠ ⊥
-| \quad    if p_{0j} is "_" then 
+- definition of group decomposition:
-| \quad \quad \quad ("_")^{l \in Idx(k_{k})} +++ (p_{ij})^{i\in{}I\DZ}   \to e_{j} 
+| let constrs((pᵢ)^{i ∈ I}) = { K \vert{} K = constr(pᵢ), i ∈ I }
-|  \quad   else \bot
+| let Groups(m) where m = ((aᵢ)ⁱ ((pᵢⱼ)ⁱ → eⱼ)ⁱʲ) =
-|)^{j ∈ J},
+| \quad let (Kₖ)ᵏ = constrs(pᵢ₀)ⁱ in
-|  ((a_{i})^{i\in{}I\DZ}, ((p_{ij})^{i\in{}I\DZ}  \to eⱼ if p_{0j} is \_ else \bot)^{j\in{}J}) 
+| \quad ( Kₖ →
-Groups(m) is an auxiliary function that source a matrix m into
+| \quad \quad ((a₀.ₗ)ˡ +++ tail(aᵢ)ⁱ)
+| \quad \quad (
+| \quad \quad if pₒⱼ is Kₖ(qₗ) then
+| \quad \quad \quad (qₗ)ˡ +++ tail(pᵢⱼ)ⁱ  → eⱼ
+| \quad \quad if pₒⱼ is _ then
+| \quad \quad \quad (_)ˡ +++ tail(pᵢⱼ)ⁱ   → eⱼ
+| \quad \quad else ⊥
+| \quad \quad )ʲ
+| \quad ), (
+| \quad \quad tail(aᵢ)ⁱ, (tail(pᵢⱼ)ⁱ  → eⱼ if p₀ⱼ is _ else ⊥)ʲ
+| \quad  )
+Groups(m) is an auxiliary function that decomposes a matrix m into
 submatrices, according to the head constructor of their first pattern.
-Groups(m) returns one submatrix m_r for each head constructor k that
+Groups(m) returns one submatrix m_r for each head constructor K that
 occurs on the first row of m, plus one "wildcard submatrix" m_{wild}
 that matches on all values that do not start with one of those head
 constructors.
-
 Intuitively, m is equivalent to its decomposition in the following
-sense: if the first pattern of an input vector (vᵢ)ⁱ starts with one
+sense: if the first pattern of an input vector (v_i)^i starts with one
-of the head constructors k, then running (vᵢ)ⁱ against m is the same
+of the head constructors Kₖ, then running (v_i)^i against m is the same
-as running it against the submatrix mₖ; otherwise (its head
+as running it against the submatrix m_{Kₖ}; otherwise (its head
-constructor is none of the k) it is equivalent to running it against
+constructor is none of the Kₖ) it is equivalent to running it against
 the wildcard submatrix.
 
-We formalize this intuition as follows:
+We formalize this intuition as follows
-Lemma (Groups):
+*** Lemma (Groups):
-Let /m/ be a matrix with \[Groups(m) = (k_r \to m_r)^k, m_{wild}\].
+    Let /m/ be a matrix with 
+    | Groups(m) = (k_r \to m_r)^k, m_{wild}
     For any value vector $(v_i)^l$ such that $v_0 = k(v'_l)^l$ for some
     constructor k,
     we have:
@@ -1365,43 +1376,60 @@ TODO: fix \{0}
 \end{comment}
 
 *** Proof:
-Let $m$ be a matrix with \[Group(m) = (k_r \to m_r)^k, m_{wild}\]
+Let /m/ be a matrix ((aᵢ)ⁱ, ((pᵢⱼ)ⁱ → eⱼ)ʲ) with 
-Let $(v_i)^i$ be an input matrix with $v_0 = k(v'_l)^l$ for some k.
+| Groups(m) = (Kₖ → mₖ)ᵏ, m_{wild}
-We proceed by case analysis:
+Below we are going to assume that m is a simplified matrix such that
-- either k is one of the kₖ for some k
+the first row does not contain an or-pattern or a binding to a
-- or k is none of the (kₖ)ᵏ
+variable.
 
-Both m(vᵢ)ⁱ and mₖ(vₖ)ᵏ are defined as the first matching result of
+Let (vᵢ)ⁱ be an input matrix with v₀ = Kᵥ(v'_{l})ˡ for some constructor Kᵥ.
-a family over each row rⱼ of a matrix 
+We have to show that:
+- if Kₖ = Kᵥ for some Kₖ ∈ constrs(p₀ⱼ)ʲ, then
+    m(vᵢ)ⁱ = mₖ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ)
+- otherwise
+    m(vᵢ)ⁱ = m_{wild}(tail(vᵢ)ⁱ)
+Let us call (rₖⱼ) the j-th row of the submatrix mₖ, and rⱼ_{wild}
+the j-th row of the wildcard submatrix m_{wild}.
 
-We know, from the definition of
+Our goal contains same-behavior equalities between matrices, for
-Groups(m), that mₖ is 
+a fixed input vector (vᵢ)ⁱ. It suffices to show same-behavior
-| ((a){0.l})^{l∈Idx(kₖ)} +++ (aᵢ)^{i∈I\DZ}),
+equalities between each row of the matrices for this input
-|    (
+vector. We show that for any j,
-|  \quad   if p_{0j} is k(qₗ) then
+- if Kₖ = Kᵥ for some Kₖ ∈ constrs(p₀ⱼ)ʲ, then
-| \quad   \quad    (qₗ)ˡ +++ (p_{ij})^{i∈I\DZ } → eⱼ 
+  | (pᵢⱼ)ⁱ(vᵢ)ⁱ = rₖⱼ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ
-|  \quad   if p_{0j} is _ then
+- otherwise
-|   \quad  \quad   (_)ˡ +++ (p_{ij})^{i∈I\DZ}   → eⱼ 
+  | (pᵢⱼ)ⁱ(vᵢ)ⁱ = rⱼ_{wild} tail(vᵢ)ⁱ
-| \quad    else ⊥
+In the first case (Kᵥ is Kₖ for some Kₖ ∈ constrs(p₀ⱼ)ʲ), we
-|    )^{j∈J}
+have to prove that
+| (pᵢⱼ)ⁱ(vᵢ)ⁱ = rₖⱼ((v'ₗ)ˡ +++ tail(vᵢ)ⁱ
+By definition of mₖ we know that rₖⱼ is equal to
+| if pₒⱼ is Kₖ(qₗ) then
+| \quad  (qₗ)ˡ +++ tail(pᵢⱼ)ⁱ  → eⱼ
+| if pₒⱼ is _ then
+| \quad  (_)ˡ +++ tail(pᵢⱼ)ⁱ   → eⱼ
+| else ⊥
+\begin{comment}
+Maybe better as a table?
+| pₒⱼ    | rₖⱼ                       |
+|--------+---------------------------|
+| Kₖ(qₗ) | (qₗ)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ |
+| _      | (_)ˡ +++ tail(pᵢⱼ)ⁱ → eⱼ  |
+| else   | ⊥                         |
+\end{comment}
+By definition of (pᵢ)ⁱ(vᵢ)ⁱ we know that (pᵢⱼ)ⁱ(vᵢ) is equal to
+| (K(qⱼ)ʲ, tail(pᵢⱼ)ⁱ) (K(v'ₗ)ˡ,tail(vᵢ)ⁱ)  := ((qⱼ)ʲ +++ tail(pᵢⱼ)ⁱ)((v'ₗ)ˡ +++ tail(vᵢ)ⁱ)
+| (_, tail(pᵢⱼ)ⁱ)      (vᵢ)                 := tail(pᵢⱼ)ⁱ(tail(vᵢ)ⁱ)
+| (K(qⱼ)ʲ, tail(pᵢⱼ)ⁱ) (K'(v'ₗ)ˡ,tail(vⱼ)ʲ) := ⊥ if K ≠ K'
 
-By definition, m(vᵢ)ⁱ is
+We prove this first case by a second case analysis on p₀ⱼ.
-| m(vᵢ)ⁱ = First(rⱼ(vᵢ)ⁱ)ʲ for m = ((aᵢ)ⁱ, (rⱼ)ʲ)
-| (pᵢ)ⁱ (vᵢ)ⁱ  =  {
-|  \quad if p₀ = k(qₗ)ˡ, v₀ = k'(v'ₖ)ᵏ, k=Idx(k') and l=Idx(k) 
-| \quad \quad  if k ≠ k' then ⊥ 
-|      if k = k' then ((qₗ)ˡ +++ (pᵢ)^{i∈I\DZ}) ((v'ₖ)ᵏ +++ (vᵢ)^{i∈I\DZ} )
-|    if p₀ = (q₁\vert{}q₂) then 
-|      First(  (q₁pᵢ^{i∈I\DZ}) vᵢ^{i∈I\DZ}, (q₂pᵢ^{i∈I \DZ}) vᵢ^{i∈I\DZ})}
 
-For this reason, if we can prove that 
+TODO
-| ∀j,  rⱼ(vᵢ)ⁱ = r'ⱼ((v'ₖ)ᵏ ++ (vᵢ)ⁱ)
-it follows that 
-| m(vᵢ)ⁱ = mₖ((v'ₖ)ᵏ ++ (vᵢ)ⁱ)
-from the above definition.
 
-We can also show that aᵢ = (a_{0.l})ˡ +++ a_{i∈I\DZ} because v(a₀) = K(v(a){0.l})ˡ)
+In the second case (Kᵥ is distinct from Kₖ for all Kₖ ∈ constrs(pₒⱼ)ʲ),
+we have to prove that
+| (pᵢⱼ)ⁱ(vᵢ)ⁱ = rⱼ_{wild} tail(vᵢ)ⁱ
 
+TODO
 
 
 ** Target translation
@@ -1597,7 +1625,7 @@ Similarly we say that a couple of decision trees in the presence of
 an input space /S/ are /not/ equivalent when:
 | equiv(S, Cₛ, Cₜ, gs) = No(vₛ,vₜ) ∧ covers(Cₜ, S) → vₛ≃vₜ ∈ S ∧ Cₛ(vₛ) ≠gs Cₜ(vₜ)
 Corollary: For a full input space /S/, that is the universe of the
-target program we say:
+target program:
 | equiv(S, 〚tₛ〛ₛ, 〚tₜ〛ₜ, ∅) = Yes  ⇔  tₛ ≃ tₜ