Metamath Proof Explorer

Theorem kbval

Description: The value of the operator resulting from the outer product | A >. <. B | of two vectors. Equation 8.1 of Prugovecki p. 376. (Contributed by NM, 15-May-2006) (Revised by Mario Carneiro, 16-Nov-2013) (New usage is discouraged.)

		Ref	Expression
	Assertion	kbval	⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 ketbra 𝐵 ) ‘ 𝐶 ) = ( ( 𝐶 ·_ih 𝐵 ) ·_ℎ 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		kbfval	⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( 𝐴 ketbra 𝐵 ) = ( 𝑥 ∈ ℋ ↦ ( ( 𝑥 ·_ih 𝐵 ) ·_ℎ 𝐴 ) ) )
2	1	fveq1d	⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) → ( ( 𝐴 ketbra 𝐵 ) ‘ 𝐶 ) = ( ( 𝑥 ∈ ℋ ↦ ( ( 𝑥 ·_ih 𝐵 ) ·_ℎ 𝐴 ) ) ‘ 𝐶 ) )
3		oveq1	⊢ ( 𝑥 = 𝐶 → ( 𝑥 ·_ih 𝐵 ) = ( 𝐶 ·_ih 𝐵 ) )
4	3	oveq1d	⊢ ( 𝑥 = 𝐶 → ( ( 𝑥 ·_ih 𝐵 ) ·_ℎ 𝐴 ) = ( ( 𝐶 ·_ih 𝐵 ) ·_ℎ 𝐴 ) )
5		eqid	⊢ ( 𝑥 ∈ ℋ ↦ ( ( 𝑥 ·_ih 𝐵 ) ·_ℎ 𝐴 ) ) = ( 𝑥 ∈ ℋ ↦ ( ( 𝑥 ·_ih 𝐵 ) ·_ℎ 𝐴 ) )
6		ovex	⊢ ( ( 𝐶 ·_ih 𝐵 ) ·_ℎ 𝐴 ) ∈ V
7	4 5 6	fvmpt	⊢ ( 𝐶 ∈ ℋ → ( ( 𝑥 ∈ ℋ ↦ ( ( 𝑥 ·_ih 𝐵 ) ·_ℎ 𝐴 ) ) ‘ 𝐶 ) = ( ( 𝐶 ·_ih 𝐵 ) ·_ℎ 𝐴 ) )
8	2 7	sylan9eq	⊢ ( ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ) ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 ketbra 𝐵 ) ‘ 𝐶 ) = ( ( 𝐶 ·_ih 𝐵 ) ·_ℎ 𝐴 ) )
9	8	3impa	⊢ ( ( 𝐴 ∈ ℋ ∧ 𝐵 ∈ ℋ ∧ 𝐶 ∈ ℋ ) → ( ( 𝐴 ketbra 𝐵 ) ‘ 𝐶 ) = ( ( 𝐶 ·_ih 𝐵 ) ·_ℎ 𝐴 ) )