Metamath Proof Explorer


Theorem funcringcsetcALTV2lem6

Description: Lemma 6 for funcringcsetcALTV2 . (Contributed by AV, 15-Feb-2020) (New usage is discouraged.)

Ref Expression
Hypotheses funcringcsetcALTV2.r ⊢ R = RingCat ⁡ U
funcringcsetcALTV2.s ⊢ S = SetCat ⁡ U
funcringcsetcALTV2.b ⊢ B = Base R
funcringcsetcALTV2.c ⊢ C = Base S
funcringcsetcALTV2.u ⊢ φ → U ∈ WUni
funcringcsetcALTV2.f ⊢ φ → F = x ∈ B ⟼ Base x
funcringcsetcALTV2.g ⊢ φ → G = x ∈ B , y ∈ B ⟼ I ↾ x RingHom y
Assertion funcringcsetcALTV2lem6 ⊢ φ ∧ X ∈ B ∧ Y ∈ B ∧ H ∈ X RingHom Y → X G Y ⁡ H = H

Proof

Step Hyp Ref Expression
1 funcringcsetcALTV2.r ⊢ R = RingCat ⁡ U
2 funcringcsetcALTV2.s ⊢ S = SetCat ⁡ U
3 funcringcsetcALTV2.b ⊢ B = Base R
4 funcringcsetcALTV2.c ⊢ C = Base S
5 funcringcsetcALTV2.u ⊢ φ → U ∈ WUni
6 funcringcsetcALTV2.f ⊢ φ → F = x ∈ B ⟼ Base x
7 funcringcsetcALTV2.g ⊢ φ → G = x ∈ B , y ∈ B ⟼ I ↾ x RingHom y
8 1 2 3 4 5 6 7 funcringcsetcALTV2lem5 ⊢ 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