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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 ∘f 𝑅 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 ∘f 𝑅 𝐺 ) ‘ 𝑋 ) = ( ( 𝐹 ‘ 𝑋 ) 𝑅 ( 𝐺 ‘ 𝑋 ) ) )
8 7 anasss ⊢ ( ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) ∧ ( 𝐴 ∈ 𝑉 ∧ 𝑋 ∈ 𝐴 ) ) → ( ( 𝐹 ∘f 𝑅 𝐺 ) ‘ 𝑋 ) = ( ( 𝐹 ‘ 𝑋 ) 𝑅 ( 𝐺 ‘ 𝑋 ) ) )