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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 ∈ ( ℤ≥ ‘ 𝑀 ) ↔ ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑀 ≤ 𝑁 ) )

Proof

Step Hyp Ref Expression
1 eluzel2 ⊢ ( 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) → 𝑀 ∈ ℤ )
2 simp1 ⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑀 ≤ 𝑁 ) → 𝑀 ∈ ℤ )
3 eluz1 ⊢ ( 𝑀 ∈ ℤ → ( 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ↔ ( 𝑁 ∈ ℤ ∧ 𝑀 ≤ 𝑁 ) ) )
4 ibar ⊢ ( 𝑀 ∈ ℤ → ( ( 𝑁 ∈ ℤ ∧ 𝑀 ≤ 𝑁 ) ↔ ( 𝑀 ∈ ℤ ∧ ( 𝑁 ∈ ℤ ∧ 𝑀 ≤ 𝑁 ) ) ) )
5 3 4 bitrd ⊢ ( 𝑀 ∈ ℤ → ( 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ↔ ( 𝑀 ∈ ℤ ∧ ( 𝑁 ∈ ℤ ∧ 𝑀 ≤ 𝑁 ) ) ) )
6 3anass ⊢ ( ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑀 ≤ 𝑁 ) ↔ ( 𝑀 ∈ ℤ ∧ ( 𝑁 ∈ ℤ ∧ 𝑀 ≤ 𝑁 ) ) )
7 5 6 bitr4di ⊢ ( 𝑀 ∈ ℤ → ( 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ↔ ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑀 ≤ 𝑁 ) ) )
8 1 2 7 pm5.21nii ⊢ ( 𝑁 ∈ ( ℤ≥ ‘ 𝑀 ) ↔ ( 𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝑀 ≤ 𝑁 ) )