Package natural-add-thm: Properties of natural number addition

Information

namenatural-add-thm
version1.55
descriptionProperties of natural number addition
authorJoe Leslie-Hurd <joe@gilith.com>
licenseHOLLight
provenanceHOL Light theory extracted on 2014-11-17
checksum46a6460c94750b81d01162a3e1bf270403e773fa
requiresbool
natural-add-def
natural-def
natural-numeral
natural-order
showData.Bool
Number.Natural

Files

Theorems

m. m + 0 = m

m n. m m + n

m n. n m + n

m. suc m = m + 1

m n. ¬(m + n < m)

m n. ¬(n + m < m)

m n. m + n = n + m

m n. m + suc n = suc (m + n)

m n. m < m + n 0 < n

m n. n < m + n 0 < m

m n. m + n = m n = 0

m n. m + n = n m = 0

m n. m + n m n = 0

m n. n + m m n = 0

m n. m n d. n = m + d

m n. m < n d. n = m + suc d

m n p. m + (n + p) = m + n + p

m n p. m + n = m + p n = p

p m n. m + p = n + p m = n

m n p. m + n < m + p n < p

m n p. n + m < p + m n < p

m n p. m + n m + p n p

m n p. n + m p + m n p

m n. m + n = 0 m = 0 n = 0

m n p q. m < p n < q m + n < p + q

m n p q. m < p n q m + n < p + q

m n p q. m p n < q m + n < p + q

m n p q. m p n q m + n p + q

p q. (b. i. p i q i + b) b n. i. n i p i q i + b

External Type Operators

External Constants

Assumptions

¬

¬

bit0 0 = 0

n. 0 n

n. n n

p. p

t. t ¬t

m. ¬(m < 0)

(¬) = λp. p

a. x. x = a

t. (x. t) t

t. (x. t) t

t. (λx. t x) = t

() = λp. p = λx.

t. ¬¬t t

t. ( t) t

t. t

t. t t

t. t t

t. t t t

t. t

t. t t

t. t

t. t t

t. t

t. t t

t. t

n. ¬(suc n = 0)

n. 0 + n = n

t. ( t) ¬t

t. t ¬t

n. bit1 n = suc (bit0 n)

() = λp q. p q p

t. (t ) (t )

m. m 0 m = 0

n. bit0 (suc n) = suc (suc (bit0 n))

t1 t2. t1 t2 t2 t1

t1 t2. t1 t2 t2 t1

m n. m < n m n

m n. ¬(m n) n < m

() = λp q. (λf. f p q) = λf. f

p. ¬(x. p x) x. ¬p x

() = λp. q. (x. p x q) q

m n. suc m + n = suc (m + n)

m n. suc m = suc n m = n

m n. suc m < suc n m < n

() = λp q. r. (p r) (q r) r

p q. (x. p q x) p x. q x

p q. p (x. q x) x. p q x

m n. m < suc n m = n m < n

p q. (x. p x) q x. p x q

t1 t2 t3. (t1 t2) t3 t1 t2 t3

m n p. m n n p m p

m n. m suc n m = suc n m n

p. p 0 (n. p n p (suc n)) n. p n

p q. (x. p x q x) (x. p x) x. q x

p q. (x. p x) (x. q x) x. p x q x