Metamath Proof Explorer


Theorem lsmsnpridl

Description: The product of the ring with a single element is equal to the principal ideal generated by that element. (Contributed by Thierry Arnoux, 21-Jan-2024)

Ref Expression
Hypotheses lsmsnpridl.1 ⊢ B = Base R
lsmsnpridl.2 ⊢ G = mulGrp R
lsmsnpridl.3 ⊢ × ˙ = LSSum ⁡ G
lsmsnpridl.4 ⊢ K = RSpan ⁡ R
lsmsnpridl.5 ⊢ φ → R ∈ Ring
lsmsnpridl.6 ⊢ φ → X ∈ B
Assertion lsmsnpridl ⊢ φ → B × ˙ X = K ⁡ X

Proof

Step Hyp Ref Expression
1 lsmsnpridl.1 ⊢ B = Base R
2 lsmsnpridl.2 ⊢ G = mulGrp R
3 lsmsnpridl.3 ⊢ × ˙ = LSSum ⁡ G
4 lsmsnpridl.4 ⊢ K = RSpan ⁡ R
5 lsmsnpridl.5 ⊢ φ → R ∈ Ring
6 lsmsnpridl.6 ⊢ φ → X ∈ B
7 2 1 mgpbas ⊢ B = Base G
8 eqid ⊢ ⋅ R = ⋅ R
9 2 8 mgpplusg ⊢ ⋅ R = + G
10 2 fvexi ⊢ G ∈ V
11 10 a1i ⊢ φ → G ∈ V
12 ssidd ⊢ φ → B ⊆ B
13 7 9 3 11 12 6 elgrplsmsn ⊢ φ → x ∈ B × ˙ X ↔ ∃ y ∈ B x = y ⋅ R X
14 1 8 4 elrspsn ⊢ R ∈ Ring ∧ X ∈ B → x ∈ K ⁡ X ↔ ∃ y ∈ B x = y ⋅ R X
15 5 6 14 syl2anc ⊢ φ → x ∈ K ⁡ X ↔ ∃ y ∈ B x = y ⋅ R X
16 13 15 bitr4d ⊢ φ → x ∈ B × ˙ X ↔ x ∈ K ⁡ X
17 16 eqrdv ⊢ φ → B × ˙ X = K ⁡ X