Metamath Proof Explorer


Theorem ismhmd

Description: Deduction version of ismhm . (Contributed by SN, 27-Jul-2024)

Ref Expression
Hypotheses ismhmd.b ⊢ B = Base S
ismhmd.c ⊢ C = Base T
ismhmd.p ⊢ + ˙ = + S
ismhmd.q ⊢ ⨣ ˙ = + T
ismhmd.0 ⊢ 0 ˙ = 0 S
ismhmd.z ⊢ Z = 0 T
ismhmd.s ⊢ φ → S ∈ Mnd
ismhmd.t ⊢ φ → T ∈ Mnd
ismhmd.f ⊢ φ → F : B ⟶ C
ismhmd.a ⊢ φ ∧ x ∈ B ∧ y ∈ B → F ⁡ x + ˙ y = F ⁡ x ⨣ ˙ F ⁡ y
ismhmd.h ⊢ φ → F ⁡ 0 ˙ = Z
Assertion ismhmd ⊢ φ → F ∈ S MndHom T

Proof

Step Hyp Ref Expression
1 ismhmd.b ⊢ B = Base S
2 ismhmd.c ⊢ C = Base T
3 ismhmd.p ⊢ + ˙ = + S
4 ismhmd.q ⊢ ⨣ ˙ = + T
5 ismhmd.0 ⊢ 0 ˙ = 0 S
6 ismhmd.z ⊢ Z = 0 T
7 ismhmd.s ⊢ φ → S ∈ Mnd
8 ismhmd.t ⊢ φ → T ∈ Mnd
9 ismhmd.f ⊢ φ → F : B ⟶ C
10 ismhmd.a ⊢ φ ∧ x ∈ B ∧ y ∈ B → F ⁡ x + ˙ y = F ⁡ x ⨣ ˙ F ⁡ y
11 ismhmd.h ⊢ φ → F ⁡ 0 ˙ = Z
12 10 ralrimivva ⊢ φ → ∀ x ∈ B ∀ y ∈ B F ⁡ x + ˙ y = F ⁡ x ⨣ ˙ F ⁡ y
13 9 12 11 3jca ⊢ φ → F : B ⟶ C ∧ ∀ x ∈ B ∀ y ∈ B F ⁡ x + ˙ y = F ⁡ x ⨣ ˙ F ⁡ y ∧ F ⁡ 0 ˙ = Z
14 1 2 3 4 5 6 ismhm ⊢ F ∈ S MndHom T ↔ S ∈ Mnd ∧ T ∈ Mnd ∧ F : B ⟶ C ∧ ∀ x ∈ B ∀ y ∈ B F ⁡ x + ˙ y = F ⁡ x ⨣ ˙ F ⁡ y ∧ F ⁡ 0 ˙ = Z
15 7 8 13 14 syl21anbrc ⊢ φ → F ∈ S MndHom T