Metamath Proof Explorer


Theorem fvopabf4g

Description: Function value of an operator abstraction whose domain is a set of functions with given domain and range. (Contributed by Jeff Madsen, 1-Dec-2009) (Revised by Mario Carneiro, 12-Sep-2015)

Ref Expression
Hypotheses fvopabf4g.1 ⊢ 𝐶 ∈ V
fvopabf4g.2 ⊢ ( 𝑥 = 𝐴 → 𝐵 = 𝐶 )
fvopabf4g.3 ⊢ 𝐹 = ( 𝑥 ∈ ( 𝑅 ↑m 𝐷 ) ↦ 𝐵 )
Assertion fvopabf4g ( ( 𝐷 ∈ 𝑋 ∧ 𝑅 ∈ 𝑌 ∧ 𝐴 : 𝐷 ⟶ 𝑅 ) → ( 𝐹 ‘ 𝐴 ) = 𝐶 )

Proof

Step Hyp Ref Expression
1 fvopabf4g.1 ⊢ 𝐶 ∈ V
2 fvopabf4g.2 ⊢ ( 𝑥 = 𝐴 → 𝐵 = 𝐶 )
3 fvopabf4g.3 ⊢ 𝐹 = ( 𝑥 ∈ ( 𝑅 ↑m 𝐷 ) ↦ 𝐵 )
4 elmapg ⊢ ( ( 𝑅 ∈ 𝑌 ∧ 𝐷 ∈ 𝑋 ) → ( 𝐴 ∈ ( 𝑅 ↑m 𝐷 ) ↔ 𝐴 : 𝐷 ⟶ 𝑅 ) )
5 4 ancoms ⊢ ( ( 𝐷 ∈ 𝑋 ∧ 𝑅 ∈ 𝑌 ) → ( 𝐴 ∈ ( 𝑅 ↑m 𝐷 ) ↔ 𝐴 : 𝐷 ⟶ 𝑅 ) )
6 5 biimp3ar ⊢ ( ( 𝐷 ∈ 𝑋 ∧ 𝑅 ∈ 𝑌 ∧ 𝐴 : 𝐷 ⟶ 𝑅 ) → 𝐴 ∈ ( 𝑅 ↑m 𝐷 ) )
7 2 3 1 fvmpt ⊢ ( 𝐴 ∈ ( 𝑅 ↑m 𝐷 ) → ( 𝐹 ‘ 𝐴 ) = 𝐶 )
8 6 7 syl ⊢ ( ( 𝐷 ∈ 𝑋 ∧ 𝑅 ∈ 𝑌 ∧ 𝐴 : 𝐷 ⟶ 𝑅 ) → ( 𝐹 ‘ 𝐴 ) = 𝐶 )