[英]What is a correct way to prove the next propositional logic statement using Curry–Howard correspondence?
我正在研究庫里-霍華德函授。
給定命題邏輯語句: (¬p -> q) -> ((¬p -> ¬q) -> p)
。
我需要在 OCaml 中定義一個類型(作為命題)和一個 function(作為證明)。
我想出了下一個代碼並卡住了:
type empty = | ;;
let ex58: (('p->empty) -> 'q) -> (('p->empty) -> ('q->empty)) -> 'p = fun f g -> g(f)
錯誤:
This expression has type ('p -> empty) -> 'q but an expression was expected of type 'p -> empty.
在進行此練習時,從為not
引入類型構造函數開始可能會更容易:
type empty = |
type 'a not = 'a -> empty
然后使用明確的全稱量化來重寫練習:
let proof_by_contradiction: type p q. (p not -> q) -> (p not -> q not) -> p =
...
這應該會稍微改善錯誤消息
錯誤:此表達式的類型為 p not -> q 但預期的表達式類型為 p not = p -> empty
在開始這個練習之前,嘗試一下可能會很有用
let proof_by_negation: type p q. (p -> q) -> (p -> q not) -> p not =
...
第一的。
我很確定這不是建設性的證明。
首先,請注意
¬¬p -> (¬p -> a)
成立(從¬¬p
和¬p
您首先獲得虛假證明,然后通過 ex falso quodlibet 您獲得任何a
)。
特別是,對於任何q
,
¬¬p -> ((¬p -> q) /\ (¬p -> ¬q)) // ("lemma")
成立(將先前的語句應用於a = q
和a = ¬q
)。
現在,如果你的原始陳述((¬p -> q) /\ (¬p -> ¬q)) -> p
是真的,那么你可以預先組合¬¬p -> ((¬p -> q) /\ (¬p -> ¬q))
,因此獲得¬¬p -> p
。 但這是雙重否定消除,眾所周知,這是不可建設性地證明的。
這是 Scala 3 中的完整結構(與 OCaml 有點密切相關;這里使用的語言子集應該很容易翻譯成 OCaml):
type ¬[A] = A => Nothing // negation
type /\[A, B] = (A, B) // conjunction / product
type Claim[P, Q] = (¬[P] => Q) => (¬[P] => ¬[Q]) => P // your claim
type DoubleNegationElimination[P] = ¬[¬[P]] => P
/** Ex falso quodlibet. */
def efq[X]: Nothing => X = f => f
/** Lemma, as explained above. */
def lemma[P, Q](a: ¬[¬[P]]): (¬[P] => Q) /\ (¬[P] => ¬[Q]) =
val left: ¬[P] => Q = notP => efq(a(notP))
val right: ¬[P] => ¬[Q] = notP => efq(a(notP))
(left, right)
/** This shows that if you could prove your claim for any `P`, `Q`,
* then you would also be able to prove double negation elimination
* for `P`.
*/
def claimImpliesDoubleNegationElimination[P, Q](
c: Claim[P, Q]
): DoubleNegationElimination[P] =
notNotP => {
val (left, right) = lemma[P, Q](notNotP)
c(left)(right)
}
/** This is an (incomplete, because impossible) proof of the double
* negation elimination for any `P`. It is incomplete, because it
* relies on the validity of your original claim.
*/
def doubleNegationElimination[P]: DoubleNegationElimination[P] =
claimImpliesDoubleNegationElimination(claim[P, Unit])
/** There cannot be a constructive proof of this, because otherwise
* we would obtain a constructive proof of `doubleNegationElimination`.
*/
def claim[P, Q]: Claim[P, Q] = ???
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