gabriel draft

tesi/tesi_unicode.org

@@ -1,4 +1,6 @@
 \begin{comment}
+TODO: neg is parsed incorrectly
+TODO: chiedi a Gabriel se T e S vanno bene, ma prima controlla che siano coerenti
 * TODO Scaletta [1/6]
   - [X] Introduction
   - [-] Background [80%]
@@ -1396,7 +1398,6 @@ m := ((a_i)^i ((p_{ij})^i \to e_j)^{ij})
  (k_k)^k := headconstructor(p_{i0})^i
 \]
 \begin{equation}
-\begin{align}
  Groups(m) := ( k_k \to  ((a)_{0.l})^{l  \in  Idx(k_k)} +++ (a_i)^{i \in I\backslash \{0\} }), \\
     ( if p_{0j} is k(q_l) then \\
         (qₗ)^{l \in Idx(k_k)} +++ (p_{ij})^{i \in I\backslash \{0\}}  \to e_j \\
@@ -1404,7 +1405,6 @@ m := ((a_i)^i ((p_{ij})^i \to e_j)^{ij})
         (\_)^{l \in Idx(k_k)} +++ (p_{ij})^{i \in I\backslash \{0\}}   \to e_j \\
      else \bot  )^j ), \\
   ((a_i)^{i \in I\backslash \{0\}}, ((p_{ij})^{i \in I\backslash \{0\}}  \to eⱼ if p_{0j} is \_ else \bot)^{j \in J}) 
-\end{align}
 \end{equation}
 
 Groups(m) is an auxiliary function that decomposes a matrix m into
@@ -1516,7 +1516,7 @@ an accessor → π relation (In other words???)
 Should I swap π and π' 
 \end{comment}
 
-\subsubsection Proof of equivalence checking
+\subsubsection{Equivalence checking}
 The equivalence checking algorithm takes as parameters an input space
 /S/, a source decision tree /Cₛ/ and a target decision tree /Cₜ/:
 | equiv(S, Cₛ, Cₜ) → Yes | No(vₛ, vₜ)
@@ -1539,9 +1539,9 @@ We define  the following
 | Forall(Yes::l) = Forall(l)
 | Forall(No(vₛ,vₜ)::_) = No(vₛ,vₜ)
 There exists and are injective:
-|  int(k)∈ℕ (ar(k) = 0) 
-|  tag(k)∈ℕ (ar(k) > 0) 
-|  π(k) = {n|int(k) = n} x {n|tag{k} = n}
+| int(k) ∈ ℕ (arity(k) = 0) 
+| tag(k) ∈ ℕ (arity(k) > 0) 
+| π(k) = {n\vert int(k) = n} x {n\vert tag(k) = n}
 where k is a constructor.
 
 \begin{comment}
@@ -1554,51 +1554,52 @@ We proceed by case analysis:
 I start numbering from zero to leave the numbers as they were on the blackboard, were we skipped some things
 I think the unreachable case should go at the end.
 \end{comment}
-0. in case of unreachable: Cₛ(vₛ) = Absurd(Unreachable) ≠ Cₜ(vₜ) ∀vₛ,vₜ
+0. in case of unreachable: 
+| Cₛ(vₛ) = Absurd(Unreachable) ≠ Cₜ(vₜ) ∀vₛ,vₜ
 1. In the case of an empty input space
    | equiv(∅, Cₛ, Cₜ) := Yes
    and that is trivial to prove because there is no pair of values (vₛ, vₜ) that could be
    tested against the decision trees.
    In the other subcases S is always non-empty.
-2. equiv(S, Failure, Failure) := Yes
-   the statement holds because of equality between Failure nodes in
-   the case of every possible value /v/.
-3. The result of the subcase where we have a source decision tree
-   /Cₛ/ that is either a Leaf terminal or a Failure terminal and a
-   target decision tree defined by an accessor /a/ and a positive
-   number of couples constraint πᵢ and children nodes Cₜᵢ. The output
-   the output of the algorithm is:
-   | equiv(S, (Leaf bbₛ|Failure) as Cₛ, Node(a, (πᵢ → Cₜᵢ)ⁱ)) :=  Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)ⁱ)
-   The statement holds because defined let Sᵢ := S∩(a→πᵢ)
-   either the algorithm is true for every sub-input space Sᵢ and
+2. When there are /Failure/ nodes at both sides the result is /Yes/:
+   |equiv(S, Failure, Failure) := Yes
+   Given that ∀v, Failure(v) = Failure, the statement holds.
+3. When we have a Leaf or a Failure at the left side:
+   | equiv(S, Failure as Cₛ, Node(a, (πᵢ → Cₜᵢ)ⁱ)) :=  Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)ⁱ)
+   | equiv(S, Leaf bbₛ as Cₛ, Node(a, (πᵢ → Cₜᵢ)ⁱ)) :=  Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)ⁱ)
+   The algorithm either returns Yes for every sub-input space Sᵢ := S∩(a→π(kᵢ)) and
    subtree Cₜᵢ
    | equiv(Sᵢ, Cₛ, Cₜᵢ) = Yes ∀i
    or we have a counter example vₛ, vₜ for which
    | vₛ≃vₜ∈Sₖ ∧ cₛ(vₛ) ≠ Cₜₖ(vₜ)
    then because 
-   | vₜ∈(a→πₖ) ⇒ Cₜ(vₜ) = Cₜₖ(vₜ)
-   then
+   | vₜ∈(a→πₖ) → Cₜ(vₜ) = Cₜₖ(vₜ) ,
    | vₛ≃vₜ∈S ∧ Cₛ(vₛ)≠Cₜ(vₜ) 
-   and the result of the algorithm is
+   we can say that
    | equiv(Sᵢ, Cₛ, Cₜᵢ) = No(vₛ, vₜ) for some minimal k∈I
-4. equiv(S, Node(a, (kᵢ → Cₛᵢ)ⁱ, C_{sf}), Cₜ) := 
-    let π' = ⋃π(kᵢ) ∀i in
-    Forall(equiv( S∩(a→π(kᵢ)ⁱ), Cₛᵢ, C_{t/a→π(kᵢ)})ⁱ +++   equiv(S∩(a→π(kᵢ)), Cₛ, C_{/a̸¬̸π'}))
-    The statement  holds because:
-    a. Forall(equiv( S∩(a→π(kᵢ)ⁱ), Cₛᵢ, C_{t/a→π(kᵢ)})ⁱ = Yes
-       In the yes case let's reason by case analysis:
-       i. When k∈(kᵢ)ⁱ
-            there is a k=kₖ for some k and this means that Cₛ(vₛ) = Cₛᵢ(vₛ)
-            By induction we know that  Cₛᵢ(vₛ) = c_{t/a→πᵢ}(vₜ)
-            and because of the trimming lemma:
-            Cₜ(vₜ) = C_{t/a→πᵢ}(vₜ)
-            Putting all together:
-            Cₛ(vₛ) = Cₛᵢ(vₛ) = C_{t/a→πᵢ}(vₜ) = Cₜ(vₜ)
+4. When we have a Node on the right we define πₙ as the domain of
+   values not covered but the union of the constructors kᵢ
+   | πₙ = ¬(⋃π(kᵢ)ⁱ)
+   The algorithm proceeds by trimming
+   | equiv(S, Node(a, (kᵢ → Cₛᵢ)ⁱ, C_{sf}), Cₜ) := 
+   | Forall(equiv( S∩(a→π(kᵢ)ⁱ), Cₛᵢ, C_{t/a→π(kᵢ)})ⁱ +++   equiv(S∩(a→π(kᵢ)), Cₛ, C_{a→πₙ}))
+    The statement still holds and we show this by first analyzing the
+   /Yes/ case:
+    | Forall(equiv( S∩(a→π(kᵢ)ⁱ), Cₛᵢ, C_{t/a→π(kᵢ)})ⁱ = Yes
+    The constructor k is either included in the set of constructors kᵢ:
+    | k \vert k∈(kᵢ)ⁱ ∧ Cₛ(vₛ) = Cₛᵢ(vₛ)
+    We also know that 
+    | (1) Cₛᵢ(vₛ) = C_{t/a→πᵢ}(vₜ)      
+    | (2) C_{T/a→πᵢ}(vₜ) = Cₜ(vₜ)      
+    (1) is true by induction and (2) is a consequence of the trimming lemma.
+    Putting everything together:
+    | Cₛ(vₛ) = Cₛᵢ(vₛ) = C_{T/a→πᵢ}(vₜ) = Cₜ(vₜ)
             
-       ii. when k∉(kᵢ)ⁱ ???
+    When the k∉(kᵢ)ⁱ [TODO]
 
-    b. Forall(...) = No(vₛ, vₜ)
-        for a minimum k, equiv(Sₖ, Cₛₖ, C_{t/a→πₖ} = No(vₛ, vₜ)
-           then Cₛₖ(vₛ) ≠ C_{t/a→πₖ}(vₜ)  and C_{t/a→πₖ}(vₜ) = Cₜ(vt) 
-             => (Cₛₖ(vₛ) = Cₛ(vₛ)) ≠ Cₜ(vₜ)
-        # Same for fallback?
+    The auxiliary Forall function returns /No(vₛ, vₜ)/ when, for a minimum  k,
+    | equiv(Sₖ, Cₛₖ, C_{T/a→πₖ} = No(vₛ, vₜ)
+    Then we can say that 
+    | Cₛₖ(vₛ) ≠ C_{t/a→πₖ}(vₜ)
+    that is enough for proving that 
+    | Cₛₖ(vₛ) ≠ (C_{t/a→πₖ}(vₜ) = Cₜ(vₜ))