Entry Value
Name LT_0
Conclusion !n. 0 < SUC n
Constructive Proof Yes
Axiom
N|A
Classical Lemmas N|A
Constructive Lemmas
  • T
  • !x. x = x
  • !p1 p2 q1 q2. (p1 ==> p2) /\ (q1 ==> q2) ==> p1 /\ q1 ==> p2 /\ q2
  • !p1 p2 q1 q2. (p1 ==> p2) /\ (q1 ==> q2) ==> p1 \/ q1 ==> p2 \/ q2
  • !t. (!x. t) <=> t
  • !t. F /\ t <=> F
  • !t. T /\ t <=> t
  • !t. t /\ F <=> F
  • !t. t /\ T <=> t
  • !t. t /\ t <=> t
  • !t. (F <=> t) <=> ~t
  • !t. (T <=> t) <=> t
  • !t. (t <=> F) <=> ~t
  • !t. (t <=> T) <=> t
  • !t. F ==> t <=> T
  • !t. T ==> t <=> t
  • !t. t ==> F <=> ~t
  • !t. t ==> T <=> T
  • !t. F \/ t <=> t
  • !t. T \/ t <=> T
  • !t. t \/ F <=> t
  • !t. t \/ T <=> T
  • !t. t \/ t <=> t
  • !t. t ==> t
  • !f y. (\x. f x) y = f y
  • !p q. (!x. p x ==> q x) ==> (?x. p x) ==> (?x. q x)
  • !p. p _0 /\ (!n. p n ==> p (SUC n)) ==> (!n. p n)
  • !m n. m < SUC n <=> m <= n
  • !m n. m < SUC n <=> m = n \/ m < n
  • !m n. m <= SUC n <=> m = SUC n \/ m <= n
  • !m. ~(m < 0)
  • !n. 0 <= n
  • !m. m <= 0 <=> m = 0
  • F <=> (!p. p)
  • T <=> (\p. p) = (\p. p)
  • (~) = (\p. p ==> F)
  • (/\) = (\p q. (\f. f p q) = (\f. f T T))
  • (==>) = (\p q. p /\ q <=> p)
  • (\/) = (\p q. !r. (p ==> r) ==> (q ==> r) ==> r)
  • (!) = (\p. p = (\x. T))
  • (?) = (\p. !q. (!x. p x ==> q) ==> q)
  • NUMERAL = (\n. n)
  • Contained Package natural-order-thm
    Comment Standard HOL library retrieved from OpenTheory
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