Metamath Proof Explorer


Theorem lfl0sc

Description: The (right vector space) scalar product of a functional with zero is the zero functional. Note that the first occurrence of ( V X. { .0. } ) represents the zero scalar, and the second is the zero functional. (Contributed by NM, 7-Oct-2014)

Ref Expression
Hypotheses lfl0sc.v 𝑉 = ( Base ‘ 𝑊 )
lfl0sc.d 𝐷 = ( Scalar ‘ 𝑊 )
lfl0sc.f 𝐹 = ( LFnl ‘ 𝑊 )
lfl0sc.k 𝐾 = ( Base ‘ 𝐷 )
lfl0sc.t · = ( .r𝐷 )
lfl0sc.o 0 = ( 0g𝐷 )
lfl0sc.w ( 𝜑𝑊 ∈ LMod )
lfl0sc.g ( 𝜑𝐺𝐹 )
Assertion lfl0sc ( 𝜑 → ( 𝐺f · ( 𝑉 × { 0 } ) ) = ( 𝑉 × { 0 } ) )

Proof

Step Hyp Ref Expression
1 lfl0sc.v 𝑉 = ( Base ‘ 𝑊 )
2 lfl0sc.d 𝐷 = ( Scalar ‘ 𝑊 )
3 lfl0sc.f 𝐹 = ( LFnl ‘ 𝑊 )
4 lfl0sc.k 𝐾 = ( Base ‘ 𝐷 )
5 lfl0sc.t · = ( .r𝐷 )
6 lfl0sc.o 0 = ( 0g𝐷 )
7 lfl0sc.w ( 𝜑𝑊 ∈ LMod )
8 lfl0sc.g ( 𝜑𝐺𝐹 )
9 1 fvexi 𝑉 ∈ V
10 9 a1i ( 𝜑𝑉 ∈ V )
11 2 4 1 3 lflf ( ( 𝑊 ∈ LMod ∧ 𝐺𝐹 ) → 𝐺 : 𝑉𝐾 )
12 7 8 11 syl2anc ( 𝜑𝐺 : 𝑉𝐾 )
13 2 lmodring ( 𝑊 ∈ LMod → 𝐷 ∈ Ring )
14 7 13 syl ( 𝜑𝐷 ∈ Ring )
15 4 6 ring0cl ( 𝐷 ∈ Ring → 0𝐾 )
16 14 15 syl ( 𝜑0𝐾 )
17 4 5 6 ringrz ( ( 𝐷 ∈ Ring ∧ 𝑘𝐾 ) → ( 𝑘 · 0 ) = 0 )
18 14 17 sylan ( ( 𝜑𝑘𝐾 ) → ( 𝑘 · 0 ) = 0 )
19 10 12 16 16 18 caofid1 ( 𝜑 → ( 𝐺f · ( 𝑉 × { 0 } ) ) = ( 𝑉 × { 0 } ) )