Metamath Proof Explorer


Theorem ipfeq

Description: If the inner product operation is already a function, the functionalization of it is equal to the original operation. (Contributed by Mario Carneiro, 14-Aug-2015)

Ref Expression
Hypotheses ipffval.1 ⊢ 𝑉 = ( Base ‘ 𝑊 )
ipffval.2 ⊢ , = ( ·𝑖 ‘ 𝑊 )
ipffval.3 ⊢ · = ( ·if ‘ 𝑊 )
Assertion ipfeq ( , Fn ( 𝑉 × 𝑉 ) → · = , )

Proof

Step Hyp Ref Expression
1 ipffval.1 ⊢ 𝑉 = ( Base ‘ 𝑊 )
2 ipffval.2 ⊢ , = ( ·𝑖 ‘ 𝑊 )
3 ipffval.3 ⊢ · = ( ·if ‘ 𝑊 )
4 1 2 3 ipffval ⊢ · = ( 𝑥 ∈ 𝑉 , 𝑦 ∈ 𝑉 ↦ ( 𝑥 , 𝑦 ) )
5 fnov ⊢ ( , Fn ( 𝑉 × 𝑉 ) ↔ , = ( 𝑥 ∈ 𝑉 , 𝑦 ∈ 𝑉 ↦ ( 𝑥 , 𝑦 ) ) )
6 5 biimpi ⊢ ( , Fn ( 𝑉 × 𝑉 ) → , = ( 𝑥 ∈ 𝑉 , 𝑦 ∈ 𝑉 ↦ ( 𝑥 , 𝑦 ) ) )
7 4 6 eqtr4id ⊢ ( , Fn ( 𝑉 × 𝑉 ) → · = , )