Metamath Proof Explorer


Theorem funcringcsetclem5ALTV

Description: Lemma 5 for funcringcsetcALTV . (Contributed by AV, 15-Feb-2020) (New usage is discouraged.)

Ref Expression
Hypotheses funcringcsetcALTV.r ⊢ R = RingCatALTV ⁡ U
funcringcsetcALTV.s ⊢ S = SetCat ⁡ U
funcringcsetcALTV.b ⊢ B = Base R
funcringcsetcALTV.c ⊢ C = Base S
funcringcsetcALTV.u ⊢ φ → U ∈ WUni
funcringcsetcALTV.f ⊢ φ → F = x ∈ B ⟼ Base x
funcringcsetcALTV.g ⊢ φ → G = x ∈ B , y ∈ B ⟼ I ↾ x RingHom y
Assertion funcringcsetclem5ALTV ⊢ φ ∧ X ∈ B ∧ Y ∈ B → X G Y = I ↾ X RingHom Y

Proof

Step Hyp Ref Expression
1 funcringcsetcALTV.r ⊢ R = RingCatALTV ⁡ U
2 funcringcsetcALTV.s ⊢ S = SetCat ⁡ U
3 funcringcsetcALTV.b ⊢ B = Base R
4 funcringcsetcALTV.c ⊢ C = Base S
5 funcringcsetcALTV.u ⊢ φ → U ∈ WUni
6 funcringcsetcALTV.f ⊢ φ → F = x ∈ B ⟼ Base x
7 funcringcsetcALTV.g ⊢ φ → G = x ∈ B , y ∈ B ⟼ I ↾ x RingHom y
8 7 adantr ⊢ φ ∧ X ∈ B ∧ Y ∈ B → G = x ∈ B , y ∈ B ⟼ I ↾ x RingHom y
9 oveq12 ⊢ x = X ∧ y = Y → x RingHom y = X RingHom Y
10 9 adantl ⊢ φ ∧ X ∈ B ∧ Y ∈ B ∧ x = X ∧ y = Y → x RingHom y = X RingHom Y
11 10 reseq2d ⊢ φ ∧ X ∈ B ∧ Y ∈ B ∧ x = X ∧ y = Y → I ↾ x RingHom y = I ↾ X RingHom Y
12 simprl ⊢ φ ∧ X ∈ B ∧ Y ∈ B → X ∈ B
13 simprr ⊢ φ ∧ X ∈ B ∧ Y ∈ B → Y ∈ B
14 ovexd ⊢ φ ∧ X ∈ B ∧ Y ∈ B → X RingHom Y ∈ V
15 14 resiexd ⊢ φ ∧ X ∈ B ∧ Y ∈ B → I ↾ X RingHom Y ∈ V
16 8 11 12 13 15 ovmpod ⊢ φ ∧ X ∈ B ∧ Y ∈ B → X G Y = I ↾ X RingHom Y