correzioni Gabriel piu` apici

tesi/tesi_unicode.org

@@ -1,4 +1,5 @@
 \begin{comment}
+TODO: not all todos are explicit. Check every comment section
 TODO: chiedi a Gabriel se T e S vanno bene, ma prima controlla che siano coerenti
 * TODO Scaletta [1/6]
   - [X] Introduction
@@ -1251,10 +1252,10 @@ $(\EquivTEX  S {C_S} {C_T} G)$, defined below.
 * Correctness of the algorithm
 Running a program tₛ or its translation 〚tₛ〛 against an input vₛ
 produces as a result /r/ in the following way:
-| ( 〚tₛ〛ₛ(vₛ) = Cₛ(vₛ) ) → r
+| ( 〚tₛ〛ₛ(vₛ) ≡ Cₛ(vₛ) ) → r
 | tₛ(vₛ) → r
 Likewise
-| ( 〚tₜ〛ₜ(vₜ) = Cₜ(vₜ) ) → r
+| ( 〚tₜ〛ₜ(vₜ) ≡ Cₜ(vₜ) ) → r
 | tₜ(vₜ) → r
 where result r ::= guard list * (Match blackbox | NoMatch | Absurd)
 and guard ::= blackbox.
@@ -1268,10 +1269,10 @@ of decision trees
 | tₛ ≃ tₜ := ∀vₛ≃vₜ, tₛ(vₛ) = tₜ(vₜ)
 | Cₛ ≃ Cₜ := ∀vₛ≃vₜ, Cₛ(vₛ) = Cₜ(vₜ)
 
-The proposed equivalence algorithm that works on a couple of
+The result of the proposed equivalence algorithm is /Yes/ or /No(vₛ,
-decision trees is returns either /Yes/ or /No(vₛ, vₜ)/ where vₛ and
+vₜ)/. In particular, in the negative case, vₛ and vₜ are a couple of
-vₜ are a couple of possible counter examples for which the constraint
+possible counter examples for which the decision trees produce a
-trees produce a different result.
+different result.
 
 ** Statements
 Theorem. We say that a translation of a source program to a decision tree
@@ -1312,12 +1313,12 @@ target program we say:
 
 *** Proof of the correctness of the translation from source programs to source decision trees
 
-We define a source term tₛ as a collection of patterns pointing to blackboxes
+We define the source term tₛ as a collection of patterns pointing to blackboxes
 | tₛ ::= (p → bb)^{i∈I}
 
 A pattern is defined as either a constructor pattern, an or pattern or
 a constant pattern
-| p ::= | K(pᵢ)ⁱ, i ∈ I | (p|q) | n ∈ ℕ
+| p ::= \vert K(pᵢ)ⁱ, i ∈ I \vert (p\vert{}q) \vert n ∈ ℕ
 
 A decision tree is defined as either a Leaf, a Failure terminal or
 an intermediate node with different children sharing the same accessor /a/
@@ -1327,9 +1328,9 @@ possible input values /S/. The fallback is not needed when the user
 doesn't use a wildcard pattern.
 %%% Give example of thing
 
-| Cₛ ::= Leaf bb | Switch(a, (Kᵢ → Cᵢ)^{i∈S} , C?)
+| Cₛ ::= Leaf bb \vert Switch(a, (Kᵢ → Cᵢ)^{i∈S} , C?)
-| a ::= Here | n.a
+| a ::= Here \vert n.a
-| vₛ ::= K(vᵢ)^{i∈I} | n ∈ ℕ
+| vₛ ::= K(vᵢ)^{i∈I} \vert n ∈ ℕ
 
 \begin{comment}
 Are K and Here clear here?
@@ -1361,12 +1362,12 @@ We proceed to show the correctness of the invariant by a case
 analysys.
 
 Base cases:
-1. [| ∅, (∅ → bbᵢ)ⁱ |] := Leaf bbᵢ where i := min(I), that is a
+1. [| ∅, (∅ → bbᵢ)ⁱ |] ≡ Leaf bbᵢ where i := min(I), that is a
    decision tree [|ms|] defined by an empty accessor and empty
    patterns pointing to blackboxes /bbᵢ/. This respects the invariant
    because a decomposition matrix in the case of empty rows returns
-   the first expression and we known that (Leaf bb)(v) := Match bb
+   the first expression and (Leaf bb)(v) := Match bb
-2. [| (aⱼ)ʲ, ∅ |] := Failure 
+2. [| (aⱼ)ʲ, ∅ |] ≡ Failure 
 
 Regarding non base cases:
 Let's first define
@@ -1397,9 +1398,9 @@ 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
+sense: if the first pattern of an input vector (vᵢ)ⁱ starts with one
-of the head constructors k, then running (v_i)^i against m is the same
+of the head constructors k, then running (vᵢ)ⁱ against m is the same
-as running it against the submatrix m_k; otherwise (its head
+as running it against the submatrix mₖ; otherwise (its head
 constructor is none of the k) it is equivalent to running it against
 the wildcard submatrix.
 
@@ -1466,21 +1467,23 @@ TODO: put ^i∈I where needed
 \end{comment}
 \subsubsection{The trimming lemma}
 The trimming lemma allows to reduce the size of a decision tree given
-an accessor → π relation (TODO: expand)
+an accessor /a/ → π relation (TODO: expand)
-| ∀vₜ ∈ (a→π), Cₜ(vₜ) = C_{t/a→π(kᵢ)}(vₜ)
+| ∀vₜ ∈ (a→π), Cₜ(vₜ) = C_{t/a→π}(vₜ)
 We prove this by induction on Cₜ:
-    a. Cₜ = Leaf_{bb}: when the decision tree is a leaf terminal, we
-    know that 
-    | Leaf_{bb/a→π}(v) = Leaf_{bb}(v)
-    That means that the result of trimming on a Leaf is the Leaf itself
-    b. The same applies to Failure terminal
-    | Failure_{/a→π}(v) = Failure(v)
-    c. When Cₜ = Switch(b, (π→Cᵢ)ⁱ)_{/a→π}  then
-    we look at the accessor /a/ of the subtree Cᵢ and
-    we define πᵢ' =  πᵢ if a≠b else πᵢ∩π 
-    Trimming a switch node yields the following result:
-    | Switch(b, (π→Cᵢ)ⁱ)_{/a→π} :=  Switch(b, (π'ᵢ→C_{i/a→π})ⁱ)
 
+- Cₜ = Leaf_{bb}: when the decision tree is a leaf terminal, the result of trimming on a Leaf is the Leaf itself
+    | Leaf_{bb/a→π}(v) = Leaf_{bb}(v)
+- The same applies to Failure terminal
+    | Failure_{/a→π}(v) = Failure(v)
+- When Cₜ = Switch(b, (π→Cᵢ)ⁱ)_{/a→π} then we look at the accessor
+    /a/ of the subtree Cᵢ and we define πᵢ' = πᵢ if a≠b else πᵢ∩π Trimming
+    a switch node yields the following result:
+    | Switch(b, (π→Cᵢ)^{i∈I})_{/a→π} :=  Switch(b, (π'ᵢ→C_{i/a→π})^{i∈I})
+
+\begin{comment}
+TODO: understand how to properly align lists
+check that every list is aligned
+\end{comment}
 \begin{comment}
 Actually in the proof.org file I transcribed:
     e. Unreachabe → ⊥
@@ -1488,14 +1491,14 @@ This is not correct because you don't have Unreachable nodes in target decision
 \end{comment}
 
 For the trimming lemma we have to prove that running the value vₜ against
-the decistion tree Cₜ is the same as running vₜ against the tree
+the decision tree Cₜ is the same as running vₜ against the tree
 C_{trim} that is the result of the trimming operation on Cₜ
-| Cₜ(vₜ) = C_{trim}(vₜ) = Switch(b, (πᵢ'→C_{i/a→π})ⁱ)(vₜ)
+| Cₜ(vₜ) = C_{trim}(vₜ) = Switch(b, (πᵢ'→C_{i/a→π})^{i∈I})(vₜ)
 We can reason by first noting that when vₜ∉(b→πᵢ)ⁱ the node must be a Failure node.
-In the case where ∃k| vₜ∈(b→πₖ) then we can prove that
+In the case where ∃k \vert{} vₜ∈(b→πₖ) then we can prove that
-| C_{k/a→π}(vₜ) = Switch(b, (πᵢ'→C_{i/a→π})ⁱ)(vₜ)
+| C_{k/a→π}(vₜ) = Switch(b, (πᵢ'→C_{i/a→π})^{i∈I})(vₜ)
 because when a ≠ b then πₖ'= πₖ and this means that vₜ∈πₖ'
-while when a = b  then πₖ'=(πₖ∩π) and vt∈πₖ' because:
+while when a = b  then πₖ'=(πₖ∩π) and vₜ∈πₖ' because:
 - by the hypothesis, vₜ∈π
 - we are in the case where vₜ∈πₖ
 So vₜ ∈ πₖ' and by induction
@@ -1517,7 +1520,7 @@ Covering lemma:
 \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)
+| equiv(S, Cₛ, Cₜ) → Yes \vert{} No(vₛ, vₜ)
 
 When the algorithm returns Yes and the input space is covered by /Cₛ/
 we can say that the couple of decision trees are the same for
@@ -1563,8 +1566,8 @@ I think the unreachable case should go at the end.
    |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, Switch(a, (πᵢ → Cₜᵢ)ⁱ)) :=  Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)ⁱ)
+   | equiv(S, Failure as Cₛ, Switch(a, (πᵢ → Cₜᵢ)^{i∈I})) :=  Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)^{i∈I})
-   | equiv(S, Leaf bbₛ as Cₛ, Switch(a, (πᵢ → Cₜᵢ)ⁱ)) :=  Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)ⁱ)
+   | equiv(S, Leaf bbₛ as Cₛ, Switch(a, (πᵢ → Cₜᵢ)^{i∈I})) :=  Forall(equiv( S∩a→π(kᵢ)), Cₛ, Cₜᵢ)^{i∈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
@@ -1577,13 +1580,13 @@ I think the unreachable case should go at the end.
    | equiv(Sᵢ, Cₛ, Cₜᵢ) = No(vₛ, vₜ) for some minimal k∈I
 4. When we have a Switch on the right we define πₙ as the domain of
    values not covered but the union of the constructors kᵢ
-   | πₙ = ¬(⋃π(kᵢ)ⁱ)
+   | πₙ = ¬(⋃π(kᵢ)^{i∈I})
    The algorithm proceeds by trimming
-   | equiv(S, Switch(a, (kᵢ → Cₛᵢ)ⁱ, C_{sf}), Cₜ) := 
+   | equiv(S, Switch(a, (kᵢ → Cₛᵢ)^{i∈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}))
+   | Forall(equiv( S∩(a→π(kᵢ)^{i∈I}), Cₛᵢ, C_{t/a→π(kᵢ)})^{i∈I} +++   equiv(S∩(a→πₙ), 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
+    | Forall(equiv( S∩(a→π(kᵢ)^{i∈I}), Cₛᵢ, C_{t/a→π(kᵢ)})^{i∈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