Metamath Proof Explorer


Theorem mendvsca

Description: A specific scalar multiplication in the module endomorphism algebra. (Contributed by Stefan O'Rear, 3-Sep-2015)

Ref Expression
Hypotheses mendvscafval.a ⊢ 𝐴 = ( MEndo ‘ 𝑀 )
mendvscafval.v ⊢ · = ( ·𝑠 ‘ 𝑀 )
mendvscafval.b ⊢ 𝐵 = ( Base ‘ 𝐴 )
mendvscafval.s ⊢ 𝑆 = ( Scalar ‘ 𝑀 )
mendvscafval.k ⊢ 𝐾 = ( Base ‘ 𝑆 )
mendvscafval.e ⊢ 𝐸 = ( Base ‘ 𝑀 )
mendvsca.w ⊢ ∙ = ( ·𝑠 ‘ 𝐴 )
Assertion mendvsca ( ( 𝑋 ∈ 𝐾 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 ∙ 𝑌 ) = ( ( 𝐸 × { 𝑋 } ) ∘f · 𝑌 ) )

Proof

Step Hyp Ref Expression
1 mendvscafval.a ⊢ 𝐴 = ( MEndo ‘ 𝑀 )
2 mendvscafval.v ⊢ · = ( ·𝑠 ‘ 𝑀 )
3 mendvscafval.b ⊢ 𝐵 = ( Base ‘ 𝐴 )
4 mendvscafval.s ⊢ 𝑆 = ( Scalar ‘ 𝑀 )
5 mendvscafval.k ⊢ 𝐾 = ( Base ‘ 𝑆 )
6 mendvscafval.e ⊢ 𝐸 = ( Base ‘ 𝑀 )
7 mendvsca.w ⊢ ∙ = ( ·𝑠 ‘ 𝐴 )
8 sneq ⊢ ( 𝑥 = 𝑋 → { 𝑥 } = { 𝑋 } )
9 8 xpeq2d ⊢ ( 𝑥 = 𝑋 → ( 𝐸 × { 𝑥 } ) = ( 𝐸 × { 𝑋 } ) )
10 id ⊢ ( 𝑦 = 𝑌 → 𝑦 = 𝑌 )
11 9 10 oveqan12d ⊢ ( ( 𝑥 = 𝑋 ∧ 𝑦 = 𝑌 ) → ( ( 𝐸 × { 𝑥 } ) ∘f · 𝑦 ) = ( ( 𝐸 × { 𝑋 } ) ∘f · 𝑌 ) )
12 1 2 3 4 5 6 mendvscafval ⊢ ( ·𝑠 ‘ 𝐴 ) = ( 𝑥 ∈ 𝐾 , 𝑦 ∈ 𝐵 ↦ ( ( 𝐸 × { 𝑥 } ) ∘f · 𝑦 ) )
13 7 12 eqtri ⊢ ∙ = ( 𝑥 ∈ 𝐾 , 𝑦 ∈ 𝐵 ↦ ( ( 𝐸 × { 𝑥 } ) ∘f · 𝑦 ) )
14 ovex ⊢ ( ( 𝐸 × { 𝑋 } ) ∘f · 𝑌 ) ∈ V
15 11 13 14 ovmpoa ⊢ ( ( 𝑋 ∈ 𝐾 ∧ 𝑌 ∈ 𝐵 ) → ( 𝑋 ∙ 𝑌 ) = ( ( 𝐸 × { 𝑋 } ) ∘f · 𝑌 ) )