Metamath Proof Explorer


Theorem 0even

Description: 0 is an even integer. (Contributed by AV, 11-Feb-2020)

Ref Expression
Hypothesis 2zrng.e 𝐸 = { 𝑧 ∈ ℤ ∣ ∃ 𝑥 ∈ ℤ 𝑧 = ( 2 · 𝑥 ) }
Assertion 0even 0 ∈ 𝐸

Proof

Step Hyp Ref Expression
1 2zrng.e 𝐸 = { 𝑧 ∈ ℤ ∣ ∃ 𝑥 ∈ ℤ 𝑧 = ( 2 · 𝑥 ) }
2 0z 0 ∈ ℤ
3 2cn 2 ∈ ℂ
4 0zd ( 2 ∈ ℂ → 0 ∈ ℤ )
5 oveq2 ( 𝑥 = 0 → ( 2 · 𝑥 ) = ( 2 · 0 ) )
6 5 eqeq2d ( 𝑥 = 0 → ( 0 = ( 2 · 𝑥 ) ↔ 0 = ( 2 · 0 ) ) )
7 6 adantl ( ( 2 ∈ ℂ ∧ 𝑥 = 0 ) → ( 0 = ( 2 · 𝑥 ) ↔ 0 = ( 2 · 0 ) ) )
8 mul01 ( 2 ∈ ℂ → ( 2 · 0 ) = 0 )
9 8 eqcomd ( 2 ∈ ℂ → 0 = ( 2 · 0 ) )
10 4 7 9 rspcedvd ( 2 ∈ ℂ → ∃ 𝑥 ∈ ℤ 0 = ( 2 · 𝑥 ) )
11 3 10 ax-mp 𝑥 ∈ ℤ 0 = ( 2 · 𝑥 )
12 eqeq1 ( 𝑧 = 0 → ( 𝑧 = ( 2 · 𝑥 ) ↔ 0 = ( 2 · 𝑥 ) ) )
13 12 rexbidv ( 𝑧 = 0 → ( ∃ 𝑥 ∈ ℤ 𝑧 = ( 2 · 𝑥 ) ↔ ∃ 𝑥 ∈ ℤ 0 = ( 2 · 𝑥 ) ) )
14 13 elrab ( 0 ∈ { 𝑧 ∈ ℤ ∣ ∃ 𝑥 ∈ ℤ 𝑧 = ( 2 · 𝑥 ) } ↔ ( 0 ∈ ℤ ∧ ∃ 𝑥 ∈ ℤ 0 = ( 2 · 𝑥 ) ) )
15 2 11 14 mpbir2an 0 ∈ { 𝑧 ∈ ℤ ∣ ∃ 𝑥 ∈ ℤ 𝑧 = ( 2 · 𝑥 ) }
16 15 1 eleqtrri 0 ∈ 𝐸