Metamath Proof Explorer


Theorem 2nodd

Description: 2 is not an odd integer. (Contributed by AV, 3-Feb-2020)

Ref Expression
Hypothesis oddinmgm.e ⊢ 𝑂 = { 𝑧 ∈ ℤ ∣ ∃ 𝑥 ∈ ℤ 𝑧 = ( ( 2 · 𝑥 ) + 1 ) }
Assertion 2nodd 2 ∉ 𝑂

Proof

Step Hyp Ref Expression
1 oddinmgm.e ⊢ 𝑂 = { 𝑧 ∈ ℤ ∣ ∃ 𝑥 ∈ ℤ 𝑧 = ( ( 2 · 𝑥 ) + 1 ) }
2 halfnz ⊢ ¬ ( 1 / 2 ) ∈ ℤ
3 eleq1 ⊢ ( ( 1 / 2 ) = 𝑥 → ( ( 1 / 2 ) ∈ ℤ ↔ 𝑥 ∈ ℤ ) )
4 2 3 mtbii ⊢ ( ( 1 / 2 ) = 𝑥 → ¬ 𝑥 ∈ ℤ )
5 4 con2i ⊢ ( 𝑥 ∈ ℤ → ¬ ( 1 / 2 ) = 𝑥 )
6 1cnd ⊢ ( 𝑥 ∈ ℤ → 1 ∈ ℂ )
7 zcn ⊢ ( 𝑥 ∈ ℤ → 𝑥 ∈ ℂ )
8 2cnd ⊢ ( 𝑥 ∈ ℤ → 2 ∈ ℂ )
9 2ne0 ⊢ 2 ≠ 0
10 9 a1i ⊢ ( 𝑥 ∈ ℤ → 2 ≠ 0 )
11 6 7 8 10 divmul2d ⊢ ( 𝑥 ∈ ℤ → ( ( 1 / 2 ) = 𝑥 ↔ 1 = ( 2 · 𝑥 ) ) )
12 5 11 mtbid ⊢ ( 𝑥 ∈ ℤ → ¬ 1 = ( 2 · 𝑥 ) )
13 eqcom ⊢ ( 2 = ( ( 2 · 𝑥 ) + 1 ) ↔ ( ( 2 · 𝑥 ) + 1 ) = 2 )
14 13 a1i ⊢ ( 𝑥 ∈ ℤ → ( 2 = ( ( 2 · 𝑥 ) + 1 ) ↔ ( ( 2 · 𝑥 ) + 1 ) = 2 ) )
15 8 7 mulcld ⊢ ( 𝑥 ∈ ℤ → ( 2 · 𝑥 ) ∈ ℂ )
16 subadd2 ⊢ ( ( 2 ∈ ℂ ∧ 1 ∈ ℂ ∧ ( 2 · 𝑥 ) ∈ ℂ ) → ( ( 2 − 1 ) = ( 2 · 𝑥 ) ↔ ( ( 2 · 𝑥 ) + 1 ) = 2 ) )
17 16 bicomd ⊢ ( ( 2 ∈ ℂ ∧ 1 ∈ ℂ ∧ ( 2 · 𝑥 ) ∈ ℂ ) → ( ( ( 2 · 𝑥 ) + 1 ) = 2 ↔ ( 2 − 1 ) = ( 2 · 𝑥 ) ) )
18 8 6 15 17 syl3anc ⊢ ( 𝑥 ∈ ℤ → ( ( ( 2 · 𝑥 ) + 1 ) = 2 ↔ ( 2 − 1 ) = ( 2 · 𝑥 ) ) )
19 2m1e1 ⊢ ( 2 − 1 ) = 1
20 19 a1i ⊢ ( 𝑥 ∈ ℤ → ( 2 − 1 ) = 1 )
21 20 eqeq1d ⊢ ( 𝑥 ∈ ℤ → ( ( 2 − 1 ) = ( 2 · 𝑥 ) ↔ 1 = ( 2 · 𝑥 ) ) )
22 14 18 21 3bitrd ⊢ ( 𝑥 ∈ ℤ → ( 2 = ( ( 2 · 𝑥 ) + 1 ) ↔ 1 = ( 2 · 𝑥 ) ) )
23 12 22 mtbird ⊢ ( 𝑥 ∈ ℤ → ¬ 2 = ( ( 2 · 𝑥 ) + 1 ) )
24 23 nrex ⊢ ¬ ∃ 𝑥 ∈ ℤ 2 = ( ( 2 · 𝑥 ) + 1 )
25 24 intnan ⊢ ¬ ( 2 ∈ ℤ ∧ ∃ 𝑥 ∈ ℤ 2 = ( ( 2 · 𝑥 ) + 1 ) )
26 eqeq1 ⊢ ( 𝑧 = 2 → ( 𝑧 = ( ( 2 · 𝑥 ) + 1 ) ↔ 2 = ( ( 2 · 𝑥 ) + 1 ) ) )
27 26 rexbidv ⊢ ( 𝑧 = 2 → ( ∃ 𝑥 ∈ ℤ 𝑧 = ( ( 2 · 𝑥 ) + 1 ) ↔ ∃ 𝑥 ∈ ℤ 2 = ( ( 2 · 𝑥 ) + 1 ) ) )
28 27 1 elrab2 ⊢ ( 2 ∈ 𝑂 ↔ ( 2 ∈ ℤ ∧ ∃ 𝑥 ∈ ℤ 2 = ( ( 2 · 𝑥 ) + 1 ) ) )
29 25 28 mtbir ⊢ ¬ 2 ∈ 𝑂
30 29 nelir ⊢ 2 ∉ 𝑂