Metamath Proof Explorer


Theorem 2zrngnring

Description: R is not a unital ring. (Contributed by AV, 6-Jan-2020)

Ref Expression
Hypotheses 2zrng.e ⊢ E = z ∈ ℤ | ∃ x ∈ ℤ z = 2 ⁢ x
2zrngbas.r ⊢ R = ℂ fld ↾ 𝑠 E
2zrngmmgm.1 ⊢ M = mulGrp R
Assertion 2zrngnring ⊢ R ∉ Ring

Proof

Step Hyp Ref Expression
1 2zrng.e ⊢ E = z ∈ ℤ | ∃ x ∈ ℤ z = 2 ⁢ x
2 2zrngbas.r ⊢ R = ℂ fld ↾ 𝑠 E
3 2zrngmmgm.1 ⊢ M = mulGrp R
4 1 2 3 2zrngnmlid ⊢ ∀ b ∈ E ∃ a ∈ E b ⁢ a ≠ a
5 1 2 2zrngbas ⊢ E = Base R
6 3 5 mgpbas ⊢ E = Base M
7 1 2 2zrngmul ⊢ × = ⋅ R
8 3 7 mgpplusg ⊢ × = + M
9 6 8 isnmnd ⊢ ∀ b ∈ E ∃ a ∈ E b ⁢ a ≠ a → M ∉ Mnd
10 df-nel ⊢ M ∉ Mnd ↔ ¬ M ∈ Mnd
11 9 10 sylib ⊢ ∀ b ∈ E ∃ a ∈ E b ⁢ a ≠ a → ¬ M ∈ Mnd
12 4 11 ax-mp ⊢ ¬ M ∈ Mnd
13 12 3mix2i ⊢ ¬ R ∈ Grp ∨ ¬ M ∈ Mnd ∨ ¬ ∀ x ∈ Base R ∀ y ∈ Base R ∀ z ∈ Base R x ⁢ y + R z = x ⁢ y + R x ⁢ z ∧ x + R y ⁢ z = x ⁢ z + R y ⁢ z
14 3ianor ⊢ ¬ R ∈ Grp ∧ M ∈ Mnd ∧ ∀ x ∈ Base R ∀ y ∈ Base R ∀ z ∈ Base R x ⁢ y + R z = x ⁢ y + R x ⁢ z ∧ x + R y ⁢ z = x ⁢ z + R y ⁢ z ↔ ¬ R ∈ Grp ∨ ¬ M ∈ Mnd ∨ ¬ ∀ x ∈ Base R ∀ y ∈ Base R ∀ z ∈ Base R x ⁢ y + R z = x ⁢ y + R x ⁢ z ∧ x + R y ⁢ z = x ⁢ z + R y ⁢ z
15 13 14 mpbir ⊢ ¬ R ∈ Grp ∧ M ∈ Mnd ∧ ∀ x ∈ Base R ∀ y ∈ Base R ∀ z ∈ Base R x ⁢ y + R z = x ⁢ y + R x ⁢ z ∧ x + R y ⁢ z = x ⁢ z + R y ⁢ z
16 eqid ⊢ Base R = Base R
17 eqid ⊢ + R = + R
18 16 3 17 7 isring ⊢ R ∈ Ring ↔ R ∈ Grp ∧ M ∈ Mnd ∧ ∀ x ∈ Base R ∀ y ∈ Base R ∀ z ∈ Base R x ⁢ y + R z = x ⁢ y + R x ⁢ z ∧ x + R y ⁢ z = x ⁢ z + R y ⁢ z
19 15 18 mtbir ⊢ ¬ R ∈ Ring
20 19 nelir ⊢ R ∉ Ring