Metamath Proof Explorer


Theorem fnfvof

Description: Function value of a pointwise composition. (Contributed by Stefan O'Rear, 5-Oct-2014) (Proof shortened by Mario Carneiro, 5-Jun-2015)

Ref Expression
Assertion fnfvof ⊢ F Fn A ∧ G Fn A ∧ A ∈ V ∧ X ∈ A → F R f G ⁡ X = F ⁡ X R G ⁡ X

Proof

Step Hyp Ref Expression
1 simpll ⊢ F Fn A ∧ G Fn A ∧ A ∈ V → F Fn A
2 simplr ⊢ F Fn A ∧ G Fn A ∧ A ∈ V → G Fn A
3 simpr ⊢ F Fn A ∧ G Fn A ∧ A ∈ V → A ∈ V
4 inidm ⊢ A ∩ A = A
5 eqidd ⊢ F Fn A ∧ G Fn A ∧ A ∈ V ∧ X ∈ A → F ⁡ X = F ⁡ X
6 eqidd ⊢ F Fn A ∧ G Fn A ∧ A ∈ V ∧ X ∈ A → G ⁡ X = G ⁡ X
7 1 2 3 3 4 5 6 ofval ⊢ F Fn A ∧ G Fn A ∧ A ∈ V ∧ X ∈ A → F R f G ⁡ X = F ⁡ X R G ⁡ X
8 7 anasss ⊢ F Fn A ∧ G Fn A ∧ A ∈ V ∧ X ∈ A → F R f G ⁡ X = F ⁡ X R G ⁡ X