Lecture 14: The $\bullet ; \bullet ⊢_o \bullet$ Proof System

$⊢_O = ⊢_I = ⊢_C \qquad hC ⊆ hH \qquad ⊢_O = ⊢_I$

Last time: $⊢_O$

• Standard R-H rules: $Σ: 𝒫 ⊢ G$
• New backchaining rules: $Σ: 𝒫 ⇓ D ⊢ \underbrace{A}_{\text{atomic}}$

e.g.

$\cfrac{𝒫 ⊢ G \qquad 𝒫 ⇓ D ⊢ A}{𝒫 ⇓ G ⇒ D ⊢ A}\supseteq L$ $\cfrac{}{𝒫 ⇓ A ⊢ A}$ $\text{(contraction) }\cfrac{𝒫 ⇓ D ⊢ A}{𝒫 ⊢ A} \qquad \text{Decide } D ∈ 𝒫$ $\cfrac{Σ: 𝒫, D ⊢ G}{Σ: 𝒫 ⊢ D ⇒ G}$

Dynamics of Proof Search: if $Σ’: 𝒫’ ⊢ A’$ appears inside the proof of $Σ: 𝒫 ⊢ A$

• If $𝒫$ is $hC$, then $Σ = Σ’, \; 𝒫 = 𝒫’$ (in other words, flat)
• If $𝒫$ is $hH$, then $Σ ⊆ Σ’, \; 𝒫 ⊆ 𝒫’$ (program and signature are “stack-based”)
$\cfrac{Σ, y: τ: 𝒫 ⊢ G(x)}{Σ: 𝒫 ⊢ ∀ x_τ. G}$

Program example:

kind nat type % natural
type z nat % zero
type s nat->nat % successor
kind list type
type :: nat->list->list
type nil list


We don’t have functions, but relations.

Let’s try to show

$𝒫 ⊢ max \; L \; \qquad \text{ iff } \qquad L \text{ is a list and } max \; [\,] \; = 0 \\ 𝒫 ⊢ leq \; N \; M \qquad \text{ iff } \qquad N ≤ M \\ 𝒫 ⊢ greater \; N \; M \qquad \text{ iff } \qquad N > M \\$
∀ N:nat. leq z N.
∀ N M:nat. leq (s N) (s M) :- leq N M.

greater N M :- leq (s M) N.

% M1
max L M :- maxx L z M. % accumulator
% M2
maxx (X :: L) A M :- leq X A, maxx L A M.
% M3
maxx (X :: L) A M :- greater X A, maxx L X M.
% M4
maxx nil A A.


NB: In Horn clause logic, it’s not possible to make a block (compacting it into one clause) out of $M_1, M_2, M_3$.

But in other logic, it is:

% M5
maxx L M :- (M2 /\ M3 /\ M4) -> maxx L z M.


and then

$\infer{\underbrace{𝒫}_{M_5} ⊢ max \; [2, 3, 1] \; 3}{ \infer{𝒫 ⊢ (M_2 ∧ M_3 ∧ M_3) ⇒ maxx \; [2, 3, 1] \; 0 \; 3}{ \infer{𝒫, M_2, M_3, M_3 ⊢ maxx \; [2, 3, 1] \; 0 \; 3}{ & } } }$

Another example:

$\infer{𝒫_0 ⊢ 𝒫_1 ⇒ \Big(((𝒫_2 ⇒ G_2) ∨ (𝒫_3 ⇒ G_3)) ∨ G_4\Big)}{ \infer{𝒫_0, 𝒫_1, 𝒫_2 ⊢ G_2 }{ & } & \infer{𝒫_0, 𝒫_1, 𝒫_3 ⊢ G_3 }{ & } & \infer{𝒫_0, 𝒫_1 ⊢ G_4 }{ & } }$

NB: note that there’s not the intended notion of scope anymore in classical logic, because $((𝒫_2 ⇒ G_2) ∨ (𝒫_3 ⇒ G_3)) \;≡\; ¬ 𝒫_2 ∨ G_2 ∨ ¬ 𝒫_3 ∨ G_3$

Monotonicity property

Let’s say we have an abstract “container” $𝒜$:

$𝒜, 𝒫 ⊢ maxa \; M \qquad \text{ if } 𝒜 = \lbrace a(n_1), …, a(n_m)\rbrace \text{ where } m ≥ 1 \text{ and } M = \max \lbrace n_1, …, n_m \rbrace$

E.g.:

it’s not possible to prove that, because of the monotonicity property:

$\text{ If } 𝒫 ⊆ 𝒫' \text{ and } 𝒫 ⊢ G \text{ then } 𝒫' ⊢ G$

So if

$a(1), a(2), 𝒫 ⊢ maxa \; 5\\ \text{ then } a(3), a(1), a(2), 𝒫 ⊢ maxa \; 5\\$

Pumping Lemma

adj ...
path ...
disconnected :- ...

$adj/G, 𝒫 ⊢ \texttt{disconnected} \qquad \text{ iff } \qquad G \text{ is disconnected}$

⟹ not possible to prove either, because of the monotonicity property (you can add edges until something disconnected becomes connected).

Back to Linear Logic

We will only care about the linear implication $⊸$

Recall that hH (hereditary Harrop formulas) make use of $⊤, \&, ∀, ⇒$ (no resort to $∨, ∃$).

$B ⇒ C \; ≝ \; (!B) ⊸ C \;≡\; ?B^⊥ ⅋ C$

When it comes to LL, we will only consider the connectives:

$⊥, ⅋, ⊤, \&, ∀, ⇒, ⊸$

This set of connectives is complete: all the others can be expressed in terms of these.

$B^⊤ = B ⊸ ⊥\\ B ⊗ C = (B^⊥ ⅋ C^⊥)^⊥\\ !B = (B ⇒ ⊥) ⇒ ⊥\\ ?B = (B ⊸ ⊥) ⇒ ⊥ = (!(B^⊥))^⊥$

But it is redundant, the par is not actually needed:

$B \,⅋\, C = (B^⊥) ⊸ C = (B ⊸ ⊥) ⊸ C$

Polarity of connectives

Positives vs. Negatives

NB: $B \;≡\; C$ means $⊢ (B ⊸ C) \& (C ⊸ B)$ (or equivalently: with the tensor, as the context is empty)

What does it mean for a connective to be positive:

Assume $B \,\;≡\;\, !B \text{ and } C \,\;≡\;\, !C \\ \text{ then } \\ \begin{cases} B ⊗ C \;≡\; !(B ⊗ C) \\ B ⊕ C \;≡\; !(B ⊕ C) \\ !B \;≡\; !!B\\ 1 \;≡\; !1\\ 0 \;≡\; !0\\ \end{cases}$

It is very important because of promotion:

$\cfrac{!Γ ⊢ B}{!Γ ⊢ !B} \quad\text{prom}$

If $!C ⊕ !B ∈ Γ$ is not stricto sensu promotable, it is in a way promotable, since

$!C ⊕ !D \;≡\; !(!C ⊕ !D)$

As for negatives:

Assume $B \,\;≡\;\, ?B \text{ and } C \,\;≡\;\, ?C \\ \text{ then } \\ \begin{cases} B ⅋ C \;≡\; ?(B ⊗ C) \\ B \& C \;≡\; ?(B ⊕ C) \\ ?B \;≡\; ??B\\ ⊤ \;≡\; ?⊤\\ ⊥ \;≡\; ?⊥\\ \end{cases}$

(cf. Fig. 7.5 (p76) in Dale Miller’s lecture notes)

$\underbrace{B_1, …, B_n}_{\text{unbounded/classical zone}} \quad ; \quad \underbrace{C_1, …, C_n}_{\text{bounded/linear zone}} ⊢ E$

where the intended meaning is

$!B_1, …, !B_n\; ; \; C_1, …, C_n ⊢ E$

but we rule the bang $!$ out from the syntax.

Here are the connectives we consider:

$𝒩_1 = \lbrace ⊤, \&, ⊸, ⇒, ∀ \rbrace$

How do the rules differ?

$\cfrac{Σ: Γ \quad ; \quad Δ, B ⊢ C}{Σ: Γ; Δ ⊢ B ⊸ C} ⊸R$ $\cfrac{Σ: Γ, B \quad ; \quad Δ ⊢ C}{Σ: Γ \quad ; \quad Δ ⊢ B ⇒ C} ⇒R$ $\cfrac{Γ \quad;\quad Δ_1 ⊢ B \qquad Γ \quad;\quad Δ_2, C ⊢ E}{Γ \quad;\quad Δ_1, Δ_2, B ⊸ C ⊢ E}$

⟹ the classical zone is treated additively, whereas the linear zone is treated multiplicatively

$\cfrac{Γ \quad;\quad ⊢ B \qquad Γ \quad;\quad Δ_1, Δ_2, C ⊢ E}{Γ \quad;\quad Δ_1, Δ_2, B ⇒ C ⊢ E}$

because we implicitely resort to promotion on $B$.

$\cfrac{Γ, B \quad;\quad Δ, B ⊢ C}{Γ, B \quad;\quad Δ ⊢ C} \text{ absorb (from top to bottom: } B \text{ is absorbed)}$ $\cfrac{}{Γ \quad;\quad A ⊢ A} \text{ init}$

NB: the classical initial rule can be derived:

$\infer{Γ, A \quad;\quad ⊢ A}{ \infer[\text{init}]{Γ, A \quad;\quad A ⊢ A}{\phantom{Γ, A \quad;\quad A ⊢ A}} }$

$\bullet ; \bullet ⊢_o \bullet$ proof system

Recall $⊢_o$ (uniform proof + backchaining): for $⊤, \&, ⇒, ∀$. How do we handle $⊸$ on top of that?

Negative connectives have invertible right-intro rules

$\cfrac{Γ \quad;\quad Δ ⊢ B \qquad Γ \quad;\quad Δ ⊢ C}{Γ \quad;\quad Δ ⊢ B \& C}$

$Γ \quad;\quad Δ ⊢ B \& C$ is provable iff $Γ \quad;\quad Δ ⊢ B$ and $Γ \quad;\quad Δ ⊢ C$ are.

Akin to

$𝒫 ⊢ A \text{ becoming } 𝒫 ⇓ D ⊢ A$ $Γ \quad;\quad Δ ⊢ A \text{ turns into } Γ \quad;\quad Δ ⇓ D ⊢ A$ $\cfrac{Γ \quad;\quad Δ ⇓ D ⊢ A}{Γ \quad;\quad Δ, D ⊢ A} \text{ decide}$ $\cfrac{Γ, D \quad;\quad Δ ⇓ D ⊢ A}{Γ, D \quad;\quad Δ ⊢ A} \text{ decide } !$ $\cfrac{}{Γ \quad;\quad \bullet ⇓ A ⊢ A} \text{ init }$ $\cfrac{Γ \quad;\quad Δ_1 ⊢ G \qquad Γ \quad;\quad Δ_2 ⇓ D ⊢ A}{Γ \quad;\quad Δ_1, Δ_2 ⇓ G ⊸ D ⊢ A}$ $\cfrac{Γ \quad;\quad \bullet ⊢ G \qquad Γ \quad;\quad Δ ⇓ D ⊢ A}{Γ \quad;\quad Δ ⇓ G ⇒ D ⊢ A}$ $\cfrac{Γ \quad;\quad Δ ⇓ D_i ⊢ A}{Γ \quad;\quad Δ ⇓ D_1 \& D_2 ⊢ A} \quad i= 1, 2$ $\cfrac{Γ \quad;\quad Δ ⇓ D[t/x] ⊢ A}{Γ \quad;\quad Δ ⇓ ∀x \, D ⊢ A}$

Embed C+I into LL

Girard’s Translation:

$A^° \quad = \quad A\\ ⊤^° \quad = \quad 1\\ (B_1 ∧ B_2)^° \quad = \quad (B_1)^° \& (B_2)^°\\ (B_1 ⇒ B_2)^° \quad = \quad !(B_1)^° ⊸ (B_2)^°\\$

But Girard’s translation doesn’t allow us to expand and properly integrate linear contexts into our sequents.

Instead: Positive/Negative mapping: to bang half as many formulas:

$(A)^+ = (A)^- = A\\ (B_1 ∧ B_2)^+ \quad = \quad (B_1)^+ \& (B_2)^+\\ (B_1 ∧ B_2)^- \quad = \quad (B_1)^- \& (B_2)^-\\ (B_1 ⇒ B_2)^+ \quad = \quad (B_1)^- ⇒ (B_2)^+\\ (B_1 ⇒ B_2)^- \quad = \quad (B_1)^+ ⊸ (B_2)^-\\$

Therefore:

$A_1 ∧ ⋯ ∧ A_n ⇒ A_0 \text{ is negatively translated into } A_1 ⊗ ⋯ ⊗ A_n ⊸ A_0$
A_0 :- A_1, …, A_n


Th:

$Γ ⊢_I B \quad\text{ iff }\quad Γ \quad;\quad \bullet ⊢_ℒ (B)^+$
prove (A -> B) :- (prove A ⇒ prove B)

% translated into
(prove A ⇒ prove B) ⊸ prove (A -> B)


In a way:

$\infer[⊸]{A → B}{ \infer{B}{ \infer{\vdots ⇒}{A} } }$

Switch example, or how to defeat monotonicity

toggle G :- on, off ⊸ G.
toggle G :- off, on ⊸ G.


where $on, off$ are constants of type $o$ (propositional symbols).

$D_2: \qquad ∀ G \; (on ⊗ (off ⊸ G) ⊸ toggle \; G)\\ D_1: \qquad ∀ G \; (off ⊗ (on ⊸ G) ⊸ toggle \; G)\\$ $\infer{Γ \quad;\quad Δ, on ⊢ toggle \; G}{ \infer{\vdots}{ \infer{Γ \quad;\quad Δ, off ⊢ G}{} } }$

NB: non-logical symbol at toplevel ⟶ seen as atomic

How to prove:

$\infer{Γ \quad;\quad Δ, on ⊢ toggle \; G}{ \infer{Γ \quad;\quad Δ, on ⇓ D_2 ⊢ toggle \; G}{ \infer{Γ \quad;\quad Δ, on ⇓ (on ⊗ (off ⊸ G)) ⊸ toggle \; G ⊢ toggle \; G}{ \infer{Γ \quad;\quad Δ, on ⊢ on ⊗ (off ⊸ G) }{ \infer{Γ \quad;\quad on ⊢ on}{ & } & \infer{Γ \quad;\quad Δ ⊢ off ⊸ G }{ \infer{Γ \quad;\quad Δ, off ⊢ G }{} } } & \infer[init]{Γ \quad;\quad \bullet ⇓ toggle \; G ⊢ toggle \; G}{\phantom{Γ \quad;\quad \bullet ⇓ toggle \; G ⊢ toggle \; G}} } } }$

Register

Other example:

$\infer{Γ \quad;\quad Δ, reg \; n ⊢ inc \; G}{ \infer{\vdots}{ \infer{Γ \quad;\quad Δ, reg \; (n+1) ⊢ G}{} } }$ $inc \; G ⊢ \underbrace{reg \; N}_{\text{read + delete}} ⊗ (\underbrace{reg \; (N+1)}_{\text{write}} ⊸ G)$

Multi-set rewriting

Or even:

p :- a ⊗ a ⊗ b ⊗ (c ⊗ a ⊸ q)

$\infer{Γ \quad;\quad Δ, c, a ⊢ q}{ \infer{\vdots}{ \infer{Γ \quad;\quad Δ, a, a, b ⊢ p}{} } }$

Computing permutations

p :- perm L K

load nil K :- unload K.
load (X::L) K :- item X ⊸ load L K.
unload (X::L) :- item X, unload L.
perm L K :- load L K.

$\infer{Γ \quad;\quad \underbrace{\bullet}_{\text{the floor is empty}} ⊢ perm \; [1, 2, 3] \; [3, 1, 2]}{ \infer{Γ \quad;\quad \bullet ⊢ load \; [1, 2, 3] \; [3, 1, 2]}{ \infer{Γ \quad;\quad \bullet ⊢ item \; 1 ⊸ load \; [2, 3] \; [3, 1, 2]}{ \infer{Γ \quad;\quad item \; 1 ⊢ load \; [2, 3] \; [3, 1, 2]}{ \infer{\vdots}{ \infer{Γ \quad;\quad item \; 1, item \; 2, item \; 3 ⊢ load \; [] \; [3, 1, 2]}{ \infer{Γ \quad;\quad item \; 1, item \; 2, item \; 3 ⊢ unload \; [3, 1, 2]}{ \infer{Γ \quad;\quad item \; 3 ⊢ item \; 3}{} & \infer{Γ \quad;\quad item \; 1, item \; 2 ⊢ unload \; [1, 2]}{ \infer{\vdots}{ \infer{Γ \quad;\quad \bullet ⊢ unload \; []}{} } } } } } } } } }$

Problem: if you have already something in the context at the beginning, it fails at the end:

$\infer{Γ \quad;\quad \underbrace{sth}_{\text{the floor is NOT empty}} ⊢ perm \; [1, 2, 3] \; [3, 1, 2]}{ \infer{Γ \quad;\quad sth ⊢ load \; [1, 2, 3] \; [3, 1, 2]}{ \infer{Γ \quad;\quad sth ⊢ item \; 1 ⊸ load \; [2, 3] \; [3, 1, 2]}{ \infer{Γ \quad;\quad sth, item \; 1 ⊢ load \; [2, 3] \; [3, 1, 2]}{ \infer{\vdots}{ \infer{Γ \quad;\quad sth, item \; 1, item \; 2, item \; 3 ⊢ load \; [] \; [3, 1, 2]}{ \infer{Γ \quad;\quad sth, item \; 1, item \; 2, item \; 3 ⊢ unload \; [3, 1, 2]}{ \infer{Γ \quad;\quad item \; 3 ⊢ item \; 3}{} & \infer{Γ \quad;\quad sth, item \; 1, item \; 2 ⊢ unload \; [1, 2]}{ \infer{\vdots}{ \infer{Γ \quad;\quad sth ⊢ unload \; []}{\text{doesn't work!}} } } } } } } } } }$

Workaround:

$\cfrac{Γ \quad;\quad \bullet ⊢ load \; L \; K \qquad Γ \quad;\quad Δ ⊢ goal(K)}{Γ \quad;\quad Δ ⊢ perm \; L \; K ⊗ goal(K)}$

whence the critical use of the tensor!

Reverse lists

Idea:

⊢ rv [] [3, 2, 1]
⊢ rv [3] [2, 1]
⊢ rv [2, 3] [1]
⊢ rv [1, 2, 3] []


In Prolog, scheme:

$reverse ⊢ rv ⟹ (rv ⊢ rv) ⟹ rv$
reverse L K :- ∀ rv, (rv [] K ⇒ (∀ X, L, K. (rv (X::L) K :- rv L (X::K))) ⇒ rv L [])

$\infer{⊢ reverse \; [1, 2, 3] \; [3, 2, 1]}{ \infer{rv \; [] \; [3, 2, 1], ∀(rv ⊢ rv) ⊢ rv \; [1, 2, 3] \; []}{ \infer{⊢ rv \; [1, 2, 3] \; []}{ \vdots} } }$

NB: We actually have

$reverse \overbrace{⊢}^{⊸} rv \overbrace{⊸}^{⟹} (rv \underbrace{⊢}_{⊸} rv) \overbrace{⊸}^{⟹} rv$

because the base case (contrary to the recursive case) is used exactly once.

Property we want to ensure:

$𝒫 ⊢ reverse \; L \; K ⟹ 𝒫 ⊢ reverse \; K \; L$ $p ⊸ q \; ≡ \; q^⊥ ⊸ p^⊥$

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