Entry Value
Name SURJ
Conclusion !s t. SURJ s t = {f | (!x. x IN s ==> f x IN t) /\ (!x. x IN t ==> (?y. y IN s /\ f y = x))}
Constructive Proof Yes
Axiom 
N|A
Classical Lemmas N|A
Constructive Lemmas
  • T
  • T <=> (\p. p) = (\p. p)
  • (/\) = (\p q. (\f. f p q) = (\f. f T T))
  • (==>) = (\p q. p /\ q <=> p)
  • (!) = (\p. p = (\x. T))
    		•
    Contained Package set-def
    Comment Standard HOL library retrieved from OpenTheory
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