Metamath Proof Explorer


Theorem lincval0

Description: The value of an empty linear combination. (Contributed by AV, 12-Apr-2019)

Ref Expression
Assertion lincval0 ( 𝑀𝑋 → ( ∅ ( linC ‘ 𝑀 ) ∅ ) = ( 0g𝑀 ) )

Proof

Step Hyp Ref Expression
1 0ex ∅ ∈ V
2 1 snid ∅ ∈ { ∅ }
3 fvex ( Base ‘ ( Scalar ‘ 𝑀 ) ) ∈ V
4 map0e ( ( Base ‘ ( Scalar ‘ 𝑀 ) ) ∈ V → ( ( Base ‘ ( Scalar ‘ 𝑀 ) ) ↑m ∅ ) = 1o )
5 3 4 mp1i ( 𝑀𝑋 → ( ( Base ‘ ( Scalar ‘ 𝑀 ) ) ↑m ∅ ) = 1o )
6 df1o2 1o = { ∅ }
7 5 6 eqtrdi ( 𝑀𝑋 → ( ( Base ‘ ( Scalar ‘ 𝑀 ) ) ↑m ∅ ) = { ∅ } )
8 2 7 eleqtrrid ( 𝑀𝑋 → ∅ ∈ ( ( Base ‘ ( Scalar ‘ 𝑀 ) ) ↑m ∅ ) )
9 0elpw ∅ ∈ 𝒫 ( Base ‘ 𝑀 )
10 9 a1i ( 𝑀𝑋 → ∅ ∈ 𝒫 ( Base ‘ 𝑀 ) )
11 lincval ( ( 𝑀𝑋 ∧ ∅ ∈ ( ( Base ‘ ( Scalar ‘ 𝑀 ) ) ↑m ∅ ) ∧ ∅ ∈ 𝒫 ( Base ‘ 𝑀 ) ) → ( ∅ ( linC ‘ 𝑀 ) ∅ ) = ( 𝑀 Σg ( 𝑣 ∈ ∅ ↦ ( ( ∅ ‘ 𝑣 ) ( ·𝑠𝑀 ) 𝑣 ) ) ) )
12 8 10 11 mpd3an23 ( 𝑀𝑋 → ( ∅ ( linC ‘ 𝑀 ) ∅ ) = ( 𝑀 Σg ( 𝑣 ∈ ∅ ↦ ( ( ∅ ‘ 𝑣 ) ( ·𝑠𝑀 ) 𝑣 ) ) ) )
13 mpt0 ( 𝑣 ∈ ∅ ↦ ( ( ∅ ‘ 𝑣 ) ( ·𝑠𝑀 ) 𝑣 ) ) = ∅
14 13 a1i ( 𝑀𝑋 → ( 𝑣 ∈ ∅ ↦ ( ( ∅ ‘ 𝑣 ) ( ·𝑠𝑀 ) 𝑣 ) ) = ∅ )
15 14 oveq2d ( 𝑀𝑋 → ( 𝑀 Σg ( 𝑣 ∈ ∅ ↦ ( ( ∅ ‘ 𝑣 ) ( ·𝑠𝑀 ) 𝑣 ) ) ) = ( 𝑀 Σg ∅ ) )
16 eqid ( 0g𝑀 ) = ( 0g𝑀 )
17 16 gsum0 ( 𝑀 Σg ∅ ) = ( 0g𝑀 )
18 15 17 eqtrdi ( 𝑀𝑋 → ( 𝑀 Σg ( 𝑣 ∈ ∅ ↦ ( ( ∅ ‘ 𝑣 ) ( ·𝑠𝑀 ) 𝑣 ) ) ) = ( 0g𝑀 ) )
19 12 18 eqtrd ( 𝑀𝑋 → ( ∅ ( linC ‘ 𝑀 ) ∅ ) = ( 0g𝑀 ) )