Metamath Proof Explorer


Theorem nvsz

Description: Anything times the zero vector is the zero vector. (Contributed by NM, 28-Nov-2007) (Revised by Mario Carneiro, 21-Dec-2013) (New usage is discouraged.)

Ref Expression
Hypotheses nvsz.4 ⊢ 𝑆 = ( ·𝑠OLD ‘ 𝑈 )
nvsz.6 ⊢ 𝑍 = ( 0vec ‘ 𝑈 )
Assertion nvsz ( ( 𝑈 ∈ NrmCVec ∧ 𝐴 ∈ ℂ ) → ( 𝐴 𝑆 𝑍 ) = 𝑍 )

Proof

Step Hyp Ref Expression
1 nvsz.4 ⊢ 𝑆 = ( ·𝑠OLD ‘ 𝑈 )
2 nvsz.6 ⊢ 𝑍 = ( 0vec ‘ 𝑈 )
3 eqid ⊢ ( 1st ‘ 𝑈 ) = ( 1st ‘ 𝑈 )
4 3 nvvc ⊢ ( 𝑈 ∈ NrmCVec → ( 1st ‘ 𝑈 ) ∈ CVecOLD )
5 eqid ⊢ ( +𝑣 ‘ 𝑈 ) = ( +𝑣 ‘ 𝑈 )
6 5 vafval ⊢ ( +𝑣 ‘ 𝑈 ) = ( 1st ‘ ( 1st ‘ 𝑈 ) )
7 1 smfval ⊢ 𝑆 = ( 2nd ‘ ( 1st ‘ 𝑈 ) )
8 eqid ⊢ ( BaseSet ‘ 𝑈 ) = ( BaseSet ‘ 𝑈 )
9 8 5 bafval ⊢ ( BaseSet ‘ 𝑈 ) = ran ( +𝑣 ‘ 𝑈 )
10 eqid ⊢ ( GId ‘ ( +𝑣 ‘ 𝑈 ) ) = ( GId ‘ ( +𝑣 ‘ 𝑈 ) )
11 6 7 9 10 vcz ⊢ ( ( ( 1st ‘ 𝑈 ) ∈ CVecOLD ∧ 𝐴 ∈ ℂ ) → ( 𝐴 𝑆 ( GId ‘ ( +𝑣 ‘ 𝑈 ) ) ) = ( GId ‘ ( +𝑣 ‘ 𝑈 ) ) )
12 4 11 sylan ⊢ ( ( 𝑈 ∈ NrmCVec ∧ 𝐴 ∈ ℂ ) → ( 𝐴 𝑆 ( GId ‘ ( +𝑣 ‘ 𝑈 ) ) ) = ( GId ‘ ( +𝑣 ‘ 𝑈 ) ) )
13 5 2 0vfval ⊢ ( 𝑈 ∈ NrmCVec → 𝑍 = ( GId ‘ ( +𝑣 ‘ 𝑈 ) ) )
14 13 adantr ⊢ ( ( 𝑈 ∈ NrmCVec ∧ 𝐴 ∈ ℂ ) → 𝑍 = ( GId ‘ ( +𝑣 ‘ 𝑈 ) ) )
15 14 oveq2d ⊢ ( ( 𝑈 ∈ NrmCVec ∧ 𝐴 ∈ ℂ ) → ( 𝐴 𝑆 𝑍 ) = ( 𝐴 𝑆 ( GId ‘ ( +𝑣 ‘ 𝑈 ) ) ) )
16 12 15 14 3eqtr4d ⊢ ( ( 𝑈 ∈ NrmCVec ∧ 𝐴 ∈ ℂ ) → ( 𝐴 𝑆 𝑍 ) = 𝑍 )