correzioni gabriel #2
This commit is contained in:
Francesco Mecca
parent
14cf53ee8d
commit
97360b7f1d
4 changed files with 237 additions and 70 deletions
|
|
@ -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}
|
||||
|
|
|
|||
139
tesi/drafts/partial-decomposition-proof.org
Normal file
139
tesi/drafts/partial-decomposition-proof.org
Normal file
|
|
@ -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
|
||||
BIN
tesi/tesi.pdf
BIN
tesi/tesi.pdf
Binary file not shown.
|
|
@ -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 Idx(k) := [0; arity(k)[
|
||||
| let First(∅) := ⊥
|
||||
| let First((aⱼ)ʲ) := a_{min(j∈J≠∅)}
|
||||
\[
|
||||
m := ((a_i)^i ((p_{ij})^i \to e_j)^{ij})
|
||||
\]
|
||||
\[
|
||||
(k_k)^k := headconstructor(p_{i0})^i
|
||||
\]
|
||||
| Groups(m) := (k_{k} \to ((a)_{0.l})^{l \in Idx(k_{k})} +++ (a_{i})^{i\in{}I\DZ}),
|
||||
|(
|
||||
| \quad if p_{0j} is k(q_{l}) then
|
||||
| \quad \quad \quad (qₗ)^{l \in Idx(k_k)} +++ (p_{ij})^{i\in{}I\DZ} \to e_{j}
|
||||
| \quad if p_{0j} is "_" then
|
||||
| \quad \quad \quad ("_")^{l \in Idx(k_{k})} +++ (p_{ij})^{i\in{}I\DZ} \to e_{j}
|
||||
| \quad else \bot
|
||||
|)^{j ∈ J},
|
||||
| ((a_{i})^{i\in{}I\DZ}, ((p_{ij})^{i\in{}I\DZ} \to eⱼ if p_{0j} is \_ else \bot)^{j\in{}J})
|
||||
Groups(m) is an auxiliary function that source a matrix m into
|
||||
Let's first define some auxiliary functions
|
||||
- The 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) := ⊥ \quad \quad \quad if ∀i, xᵢ = ⊥
|
||||
| First((xᵢ)ⁱ) := x_min{i \vert{} xᵢ ≠ ⊥} \quad if ∃i, xᵢ ≠ ⊥
|
||||
- definition of group decomposition:
|
||||
| let constrs((pᵢ)^{i ∈ I}) = { K \vert{} K = constr(pᵢ), i ∈ I }
|
||||
| let Groups(m) where m = ((aᵢ)ⁱ ((pᵢⱼ)ⁱ → eⱼ)ⁱʲ) =
|
||||
| \quad let (Kₖ)ᵏ = constrs(pᵢ₀)ⁱ in
|
||||
| \quad ( Kₖ →
|
||||
| \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
|
||||
of the head constructors k, then running (vᵢ)ⁱ against m is the same
|
||||
as running it against the submatrix mₖ; otherwise (its head
|
||||
constructor is none of the k) it is equivalent to running it against
|
||||
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 \[Groups(m) = (k_r \to m_r)^k, m_{wild}\].
|
||||
We formalize this intuition as follows
|
||||
*** Lemma (Groups):
|
||||
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 $(v_i)^i$ be an input matrix with $v_0 = k(v'_l)^l$ for some k.
|
||||
We proceed by case analysis:
|
||||
- either k is one of the kₖ for some k
|
||||
- or k is none of the (kₖ)ᵏ
|
||||
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 such that
|
||||
the first row does not contain an or-pattern or a binding to a
|
||||
variable.
|
||||
|
||||
Both m(vᵢ)ⁱ and mₖ(vₖ)ᵏ are defined as the first matching result of
|
||||
a family over each row rⱼ of a matrix
|
||||
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}.
|
||||
|
||||
We know, from the definition of
|
||||
Groups(m), that mₖ is
|
||||
| ((a){0.l})^{l∈Idx(kₖ)} +++ (aᵢ)^{i∈I\DZ}),
|
||||
| (
|
||||
| \quad if p_{0j} is k(qₗ) then
|
||||
| \quad \quad (qₗ)ˡ +++ (p_{ij})^{i∈I\DZ } → eⱼ
|
||||
| \quad if p_{0j} is _ then
|
||||
| \quad \quad (_)ˡ +++ (p_{ij})^{i∈I\DZ} → eⱼ
|
||||
| \quad else ⊥
|
||||
| )^{j∈J}
|
||||
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
|
||||
| \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
|
||||
| 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})}
|
||||
We prove this first case by a second case analysis on p₀ⱼ.
|
||||
|
||||
For this reason, if we can prove that
|
||||
| ∀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.
|
||||
TODO
|
||||
|
||||
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
|
||||
|
||||
|
||||
|
|
|
|||
Loading…
Reference in a new issue