Metamath Proof Explorer


Theorem subrfn

Description: Vector subtraction produces a function. (Contributed by Andrew Salmon, 27-Jan-2012)

Ref Expression
Assertion subrfn ( ( 𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷 ) → ( 𝐴 -𝑟 𝐵 ) Fn ℝ )

Proof

Step Hyp Ref Expression
1 ovex ⊢ ( ( 𝐴 ‘ 𝑥 ) − ( 𝐵 ‘ 𝑥 ) ) ∈ V
2 eqid ⊢ ( 𝑥 ∈ ℝ ↦ ( ( 𝐴 ‘ 𝑥 ) − ( 𝐵 ‘ 𝑥 ) ) ) = ( 𝑥 ∈ ℝ ↦ ( ( 𝐴 ‘ 𝑥 ) − ( 𝐵 ‘ 𝑥 ) ) )
3 1 2 fnmpti ⊢ ( 𝑥 ∈ ℝ ↦ ( ( 𝐴 ‘ 𝑥 ) − ( 𝐵 ‘ 𝑥 ) ) ) Fn ℝ
4 subrval ⊢ ( ( 𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷 ) → ( 𝐴 -𝑟 𝐵 ) = ( 𝑥 ∈ ℝ ↦ ( ( 𝐴 ‘ 𝑥 ) − ( 𝐵 ‘ 𝑥 ) ) ) )
5 4 fneq1d ⊢ ( ( 𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷 ) → ( ( 𝐴 -𝑟 𝐵 ) Fn ℝ ↔ ( 𝑥 ∈ ℝ ↦ ( ( 𝐴 ‘ 𝑥 ) − ( 𝐵 ‘ 𝑥 ) ) ) Fn ℝ ) )
6 3 5 mpbiri ⊢ ( ( 𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷 ) → ( 𝐴 -𝑟 𝐵 ) Fn ℝ )