Metamath Proof Explorer


Theorem rhmsubclem2

Description: Lemma 2 for rhmsubc . (Contributed by AV, 2-Mar-2020)

Ref Expression
Hypotheses rngcrescrhm.u ⊢ φ → U ∈ V
rngcrescrhm.c ⊢ C = RngCat ⁡ U
rngcrescrhm.r ⊢ φ → R = Ring ∩ U
rngcrescrhm.h ⊢ H = RingHom ↾ R × R
Assertion rhmsubclem2 ⊢ φ ∧ X ∈ R ∧ Y ∈ R → X H Y = X RingHom Y

Proof

Step Hyp Ref Expression
1 rngcrescrhm.u ⊢ φ → U ∈ V
2 rngcrescrhm.c ⊢ C = RngCat ⁡ U
3 rngcrescrhm.r ⊢ φ → R = Ring ∩ U
4 rngcrescrhm.h ⊢ H = RingHom ↾ R × R
5 opelxpi ⊢ X ∈ R ∧ Y ∈ R → X Y ∈ R × R
6 5 3adant1 ⊢ φ ∧ X ∈ R ∧ Y ∈ R → X Y ∈ R × R
7 6 fvresd ⊢ φ ∧ X ∈ R ∧ Y ∈ R → RingHom ↾ R × R ⁡ X Y = RingHom ⁡ X Y
8 df-ov ⊢ X H Y = H ⁡ X Y
9 4 fveq1i ⊢ H ⁡ X Y = RingHom ↾ R × R ⁡ X Y
10 8 9 eqtri ⊢ X H Y = RingHom ↾ R × R ⁡ X Y
11 df-ov ⊢ X RingHom Y = RingHom ⁡ X Y
12 7 10 11 3eqtr4g ⊢ φ ∧ X ∈ R ∧ Y ∈ R → X H Y = X RingHom Y