Metamath Proof Explorer


Theorem dvdsr01

Description: In a ring, zero is divisible by all elements. ("Zero divisor" as a term has a somewhat different meaning, see df-rlreg .) (Contributed by Stefan O'Rear, 29-Mar-2015)

Ref Expression
Hypotheses dvdsr0.b ⊢ B = Base R
dvdsr0.d ⊢ ∥ ˙ = ∥ r ⁡ R
dvdsr0.z ⊢ 0 ˙ = 0 R
Assertion dvdsr01 ⊢ R ∈ Ring ∧ X ∈ B → X ∥ ˙ 0 ˙

Proof

Step Hyp Ref Expression
1 dvdsr0.b ⊢ B = Base R
2 dvdsr0.d ⊢ ∥ ˙ = ∥ r ⁡ R
3 dvdsr0.z ⊢ 0 ˙ = 0 R
4 1 3 ring0cl ⊢ R ∈ Ring → 0 ˙ ∈ B
5 eqid ⊢ ⋅ R = ⋅ R
6 1 5 3 ringlz ⊢ R ∈ Ring ∧ X ∈ B → 0 ˙ ⋅ R X = 0 ˙
7 oveq1 ⊢ x = 0 ˙ → x ⋅ R X = 0 ˙ ⋅ R X
8 7 eqeq1d ⊢ x = 0 ˙ → x ⋅ R X = 0 ˙ ↔ 0 ˙ ⋅ R X = 0 ˙
9 8 rspcev ⊢ 0 ˙ ∈ B ∧ 0 ˙ ⋅ R X = 0 ˙ → ∃ x ∈ B x ⋅ R X = 0 ˙
10 4 6 9 syl2an2r ⊢ R ∈ Ring ∧ X ∈ B → ∃ x ∈ B x ⋅ R X = 0 ˙
11 1 2 5 dvdsr2 ⊢ X ∈ B → X ∥ ˙ 0 ˙ ↔ ∃ x ∈ B x ⋅ R X = 0 ˙
12 11 adantl ⊢ R ∈ Ring ∧ X ∈ B → X ∥ ˙ 0 ˙ ↔ ∃ x ∈ B x ⋅ R X = 0 ˙
13 10 12 mpbird ⊢ R ∈ Ring ∧ X ∈ B → X ∥ ˙ 0 ˙