Metamath Proof Explorer


Theorem lmod1lem5

Description: Lemma 5 for lmod1 . (Contributed by AV, 28-Apr-2019)

Ref Expression
Hypothesis lmod1.m ⊢ M = Base ndx I + ndx I I I Scalar ⁡ ndx R ∪ ⋅ ndx x ∈ Base R , y ∈ I ⟼ y
Assertion lmod1lem5 ⊢ I ∈ V ∧ R ∈ Ring → 1 Scalar ⁡ M ⋅ M I = I

Proof

Step Hyp Ref Expression
1 lmod1.m ⊢ M = Base ndx I + ndx I I I Scalar ⁡ ndx R ∪ ⋅ ndx x ∈ Base R , y ∈ I ⟼ y
2 fvex ⊢ Base R ∈ V
3 snex ⊢ I ∈ V
4 2 3 pm3.2i ⊢ Base R ∈ V ∧ I ∈ V
5 4 a1i ⊢ I ∈ V ∧ R ∈ Ring → Base R ∈ V ∧ I ∈ V
6 mpoexga ⊢ Base R ∈ V ∧ I ∈ V → x ∈ Base R , y ∈ I ⟼ y ∈ V
7 1 lmodvsca ⊢ x ∈ Base R , y ∈ I ⟼ y ∈ V → x ∈ Base R , y ∈ I ⟼ y = ⋅ M
8 5 6 7 3syl ⊢ I ∈ V ∧ R ∈ Ring → x ∈ Base R , y ∈ I ⟼ y = ⋅ M
9 8 eqcomd ⊢ I ∈ V ∧ R ∈ Ring → ⋅ M = x ∈ Base R , y ∈ I ⟼ y
10 simprr ⊢ I ∈ V ∧ R ∈ Ring ∧ x = 1 Scalar ⁡ M ∧ y = I → y = I
11 1 lmodsca ⊢ R ∈ Ring → R = Scalar ⁡ M
12 11 adantl ⊢ I ∈ V ∧ R ∈ Ring → R = Scalar ⁡ M
13 12 eqcomd ⊢ I ∈ V ∧ R ∈ Ring → Scalar ⁡ M = R
14 13 fveq2d ⊢ I ∈ V ∧ R ∈ Ring → 1 Scalar ⁡ M = 1 R
15 eqid ⊢ Base R = Base R
16 eqid ⊢ 1 R = 1 R
17 15 16 ringidcl ⊢ R ∈ Ring → 1 R ∈ Base R
18 17 adantl ⊢ I ∈ V ∧ R ∈ Ring → 1 R ∈ Base R
19 14 18 eqeltrd ⊢ I ∈ V ∧ R ∈ Ring → 1 Scalar ⁡ M ∈ Base R
20 snidg ⊢ I ∈ V → I ∈ I
21 20 adantr ⊢ I ∈ V ∧ R ∈ Ring → I ∈ I
22 9 10 19 21 21 ovmpod ⊢ I ∈ V ∧ R ∈ Ring → 1 Scalar ⁡ M ⋅ M I = I