Metamath Proof Explorer


Theorem eluz2

Description: Membership in an upper set of integers. We use the fact that a function's value (under our function value definition) is empty outside of its domain to show M e. ZZ . (Contributed by NM, 5-Sep-2005) (Revised by Mario Carneiro, 3-Nov-2013)

Ref Expression
Assertion eluz2
|- ( N e. ( ZZ>= ` M ) <-> ( M e. ZZ /\ N e. ZZ /\ M <_ N ) )

Proof

Step Hyp Ref Expression
1 eluzel2
 |-  ( N e. ( ZZ>= ` M ) -> M e. ZZ )
2 simp1
 |-  ( ( M e. ZZ /\ N e. ZZ /\ M <_ N ) -> M e. ZZ )
3 eluz1
 |-  ( M e. ZZ -> ( N e. ( ZZ>= ` M ) <-> ( N e. ZZ /\ M <_ N ) ) )
4 ibar
 |-  ( M e. ZZ -> ( ( N e. ZZ /\ M <_ N ) <-> ( M e. ZZ /\ ( N e. ZZ /\ M <_ N ) ) ) )
5 3 4 bitrd
 |-  ( M e. ZZ -> ( N e. ( ZZ>= ` M ) <-> ( M e. ZZ /\ ( N e. ZZ /\ M <_ N ) ) ) )
6 3anass
 |-  ( ( M e. ZZ /\ N e. ZZ /\ M <_ N ) <-> ( M e. ZZ /\ ( N e. ZZ /\ M <_ N ) ) )
7 5 6 syl6bbr
 |-  ( M e. ZZ -> ( N e. ( ZZ>= ` M ) <-> ( M e. ZZ /\ N e. ZZ /\ M <_ N ) ) )
8 1 2 7 pm5.21nii
 |-  ( N e. ( ZZ>= ` M ) <-> ( M e. ZZ /\ N e. ZZ /\ M <_ N ) )