# 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 ) )`