Metamath Proof Explorer


Theorem funcringcsetclem6ALTV

Description: Lemma 6 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 funcringcsetclem6ALTV ⊢ φ ∧ X ∈ B ∧ Y ∈ B ∧ H ∈ X RingHom Y → X G Y ⁡ H = H

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 1 2 3 4 5 6 7 funcringcsetclem5ALTV ⊢ φ ∧ X ∈ B ∧ Y ∈ B → X G Y = I ↾ X RingHom Y
9 8 3adant3 ⊢ φ ∧ X ∈ B ∧ Y ∈ B ∧ H ∈ X RingHom Y → X G Y = I ↾ X RingHom Y
10 9 fveq1d ⊢ φ ∧ X ∈ B ∧ Y ∈ B ∧ H ∈ X RingHom Y → X G Y ⁡ H = I ↾ X RingHom Y ⁡ H
11 fvresi ⊢ H ∈ X RingHom Y → I ↾ X RingHom Y ⁡ H = H
12 11 3ad2ant3 ⊢ φ ∧ X ∈ B ∧ Y ∈ B ∧ H ∈ X RingHom Y → I ↾ X RingHom Y ⁡ H = H
13 10 12 eqtrd ⊢ φ ∧ X ∈ B ∧ Y ∈ B ∧ H ∈ X RingHom Y → X G Y ⁡ H = H