Package gfp-div-exp-thm: Correctness of a GF(p) exponentiation algorithm based on division

Information

namegfp-div-exp-thm
version1.43
descriptionCorrectness of a GF(p) exponentiation algorithm based on division
authorJoe Leslie-Hurd <joe@gilith.com>
licenseMIT
provenanceHOL Light theory extracted on 2014-11-01
checksum1d8babf0bfe15cccb75328f209d69ad07e057567
requiresbase
gfp-def
gfp-div-def
gfp-div-exp-def
gfp-div-thm
gfp-thm
natural-fibonacci
showData.Bool
Data.List
Number.GF(p)
Number.Natural
Number.Natural.Fibonacci

Files

Theorems

x n.
    x n =
    if n = 0 then 1 else if x = 0 then 0 else expDiv 1 1 x 1 (encode n)

x n d f p l.
    ¬(x = 0) ¬(n = 0) ¬(d = 0)
    expDiv n d (x f) (inv (x p)) l =
    (n / d) * x decode.dest f p l
    expDiv n d (inv (x f)) (x p) l = (d / n) * x decode.dest f p l

External Type Operators

External Constants

Assumptions

¬

¬

p. p

(¬) = λp. p

() = λp. p = λx.

t. t

t. t t

t. t t t

t. t t

n. decode (encode n) = n

¬(1 = 0)

x. x 1 = x

() = λp q. p q p

inv 1 = 1

t. (t ) (t )

x. x 0 = 1

x. x * 1 = x

x. x / 1 = x

x. 1 * x = x

t1 t2. (if then t1 else t2) = t2

t1 t2. (if then t1 else t2) = t1

f p. decode.dest f p [] = 0

l. decode l = decode.dest 1 0 l

x y. x = y y = x

x y. x * y = y * x

m n. m + n = n + m

() = λp q. (λf. f p q) = λf. f

() = λp. q. (x. p x q) q

x. ¬(x = 0) inv (inv x) = x

x. ¬(x = 0) ¬(inv x = 0)

() = λp q. r. (p r) (q r) r

x y z. x = y y = z x = z

p q r. p q r p q r

x y z. x * y * z = x * (y * z)

x. ¬(x = 0) inv x * x = 1

x y. ¬(x = 0) x * (y / x) = y

x y. ¬(x = 0) (y / x) * x = y

x y. ¬(x = 0) y * x / x = y

x y. ¬(x = 0) y / x = y * inv x

x m n. x m * x n = x (m + n)

x y. x * y = 0 x = 0 y = 0

x n. x n = 0 x = 0 ¬(n = 0)

p. p [] (h t. p t p (h :: t)) l. p l

x y z. x * y = x * z x = 0 y = z

x y z. y * x = z * x x = 0 y = z

x y z. ¬(x = 0) x * y = x * z y = z

x y z. ¬(x = 0) y * x = z * x y = z

x y. ¬(x = 0) ¬(y = 0) ¬(y / x = 0)

p c x y. p (if c then x else y) (c p x) (¬c p y)

b n d f p. expDiv b n d f p [] = if b then n / d else d / n

f p h t.
    decode.dest f p (h :: t) =
    let s f + p in let n decode.dest s f t in if h then s + n else n

b n d f p h t.
    expDiv b n d f p (h :: t) =
    let s p / f in expDiv (¬b) d (if h then n / s else n) s f t