HOL

Symmetry in HOL

Symmetry in HOL:

\[\cfrac{Γ ⊢ s=u}{Γ ⊢ u=s}\texttt{SYM}\]

Recall that:

1. Reflexivity of equality \(\cfrac{}{⊢ t=t} REFL\)

2. Transitivity of equality (could be derived from other rules, but added just for efficiency)

3. Congruence of application $s(u) = t(v)$ whenever $u=t$ ($MK_COMB$: make combinators)

4. Congruence of abstraction $λx. u = λx. v$ whenever $u=v$

5. $β$-reduction: $(λx. t)x = t$

6. Axiom rule $\lbrace p \rbrace ⊢ p$

7. Modus ponens $q$ whenever $p ⟺q$ and $p$

8. Building equivalence: \(\cfrac{Γ ⊢ p \quad Δ ⊢ q}{(Γ\backslash \lbrace q \rbrace) ∪ (Δ \backslash \lbrace p \rbrace) ⊢ p ⟺ q}\)

9. Instantiation of terms: \(\cfrac{Γ[x_1, ⋯, x_n] ⊢ p[x_1, ⋯, x_n]}{Γ[t_1, ⋯, t_n] ⊢ p[t_1, ⋯, t_n]}\)

10. Instantiation of type variables: \(\cfrac{Γ[α_1, ⋯, α_n] ⊢ p[α_1, ⋯, α_n]}{Γ[γ_1, ⋯, γ_n] ⊢ p[γ_1, ⋯, γ_n]}\)

And the $η$-rule can be added:

\[\cfrac{}{⊢ λx.(t x) = t} \texttt{ETA} \text{ if } x ∉ fv(t)\]

Let’s assume \(\cfrac{}{Γ ⊢ s=u}\)

We can write

\[(=): ∀α, α ⟶ α ⟶ Bool\]

So that:

\[\cfrac{}{⊢ (=) = (=)}REFL\]

and by $MK_COMB$:

\[\cfrac{}{Γ ⊢ (=)s = (=)u}\]

Then, by applying these functions to $s=s$ by $MK_COMB$ again:

\[\cfrac{}{Γ ⊢ \underbrace{(=) \, s \, s}_{bool} ⟺ (=) \, u \, s}\]

And finally, by modus ponens:

\[\cfrac{}{Γ ⊢ (=) \, u \, s}\]

Transitivity

Same thing for the transitivity (almost the same proof):

\[\cfrac{}{⊢ (=)s = (=)s}REFL\]

then, with $MK_COMB$ on our assumption $t=u$

\[\cfrac{}{⊢ (=) \, s\, t = (=)\, s\, u}REFL\]

then one concludes with the modus ponens on our assumption $s=t$

New beta rule

From

\[\cfrac{}{⊢ (λx. t)x = t} \texttt{BETA}\]

we want to derive

\[\cfrac{}{⊢ (λx. t)u = t[x ← u]} \texttt{BETA'}\]

Let’s begin with

\[\cfrac{}{⊢ (λx. t)x = t}\]

By instantiating $x ← u$:

\[\cfrac{}{⊢ ((λx. t)x = t)[x ← u]}\]

i.e.

\[\cfrac{}{⊢ (λx. t)u = t[x ← u]}\]

Implication

\[p ⟹ q ≝ p ∧ q ⟺ p\]

Why does it work?

Introduction

\[\cfrac{Γ, p ⊢ q}{Γ ⊢ p ⟹ q} \texttt{INTRO}\]

We want to prove

\[\cfrac{}{Γ ⊢ \begin{cases} p ⟹ q \\ p ∧ q ⟺ p \end{cases}}\] \[\cfrac{\cfrac{p ⊢ p \quad Γ, p ⊢ q}{Γ, p ⊢ p∧q} \texttt{∧-INTRO}\qquad \cfrac{p ∧ q ⊢ p ∧ q}{p ∧ q ⊢ p} \texttt{∧-Elim-left} }{Γ ⊢ \begin{cases} p ⟹ q \\ p ∧ q ⟺ p \end{cases}}\]

where

\[\cfrac{Γ ⊢ p \quad Δ ⊢ q}{Γ ∪ Δ ⊢ p ∧ q} \texttt{∧-INTRO}\]

Elimination

We use symmetry, then modus ponens with $Δ ⊢ p$ to get $Γ ∪ Δ ⊢ p ∧ q$, and then by elim:

\[Γ ∪ Δ ⊢ q\]

Universal quantification $∀$

\[\underbrace{∀x\underbrace{P(x)}_{α ⟶ bool}}_{bool} ≝ P = λx. ⊤\]

Impredicative encoding: we can quantify over all propositions.

Let’s show

\[\cfrac{}{Γ ⊢ \begin{cases} ∀x. P(x) \\ P = λx. ⊤ \end{cases}}\]

under the assumption

\[\cfrac{}{Γ ⊢ P(x)} \text{ when } x ∉ fv(Γ)\]

With $η$-expansion:

\[\cfrac{ \cfrac{ \cfrac{ \cfrac{}{ Γ ⊢ P(x) } \text{ when } x ∉ fv(Γ) }{Γ ⊢ P(x) ⟺ ⊤} }{Γ ⊢ λx. P(x) = λx. ⊤} }{Γ ⊢ \begin{cases} ∀x. P(x) \\ P = λx. ⊤ \end{cases}}\]

To avoid extensionality, we could have defined:

\[∀x. P(x) ≝ λx. P(x) = λx. ⊤\]

The other direction:

\[\cfrac{ Γ ⊢ \begin{cases} ∀x. P(x) \\ P = λx. ⊤ \end{cases} \qquad t=t} {\cfrac{ Γ ⊢ P(t) = \underbrace{(λx. ⊤)t}_{\overset{β}{=} ⊤}} { Γ ⊢ P(t) } }\]

False $⊥$

\[⊥ ≝ ∀ \underbrace{p}_{bool}. p\]

Then: $⊥$-Elim = $∀$-Elim:

\[\cfrac{Γ ⊢ ∀p.p}{Γ ⊢ p}\]

Existential quantification $∃$

\[∃x. P(x) ≝ ∀q. (∀x. P(x) ⟹ q) ⟹ q\]

Elimination:

\[\cfrac{Γ ⊢ ∃x. P(x) \qquad Δ, P(x) ⊢ C}{Γ, Δ ⊢ C} \text{ for } x ∉ fv(Δ, C)\]

Introduction:

\[\cfrac{ ⊢ P(t) ⟹ q \qquad ⊢ P(t) } { \cfrac{ ∀x. P(x) ⟹ q ⊢ q} {\cfrac{ ⊢ (∀x. P(x) ⟹ q) ⟹ q } { ⊢ ∃x. P(x) } }}\]

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In HOL, there is a (private) type thm (coming from LCF) ⟶ implemented in OCamL.