Metamath Proof Explorer


Theorem dfafn5a

Description: Representation of a function in terms of its values, analogous to dffn5 (only one direction of implication!). (Contributed by Alexander van der Vekens, 25-May-2017)

Ref Expression
Assertion dfafn5a ( 𝐹 Fn 𝐴𝐹 = ( 𝑥𝐴 ↦ ( 𝐹 ''' 𝑥 ) ) )

Proof

Step Hyp Ref Expression
1 fnrel ( 𝐹 Fn 𝐴 → Rel 𝐹 )
2 dfrel4v ( Rel 𝐹𝐹 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝑥 𝐹 𝑦 } )
3 1 2 sylib ( 𝐹 Fn 𝐴𝐹 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝑥 𝐹 𝑦 } )
4 fnbr ( ( 𝐹 Fn 𝐴𝑥 𝐹 𝑦 ) → 𝑥𝐴 )
5 4 ex ( 𝐹 Fn 𝐴 → ( 𝑥 𝐹 𝑦𝑥𝐴 ) )
6 5 pm4.71rd ( 𝐹 Fn 𝐴 → ( 𝑥 𝐹 𝑦 ↔ ( 𝑥𝐴𝑥 𝐹 𝑦 ) ) )
7 eqcom ( 𝑦 = ( 𝐹 ''' 𝑥 ) ↔ ( 𝐹 ''' 𝑥 ) = 𝑦 )
8 fnbrafvb ( ( 𝐹 Fn 𝐴𝑥𝐴 ) → ( ( 𝐹 ''' 𝑥 ) = 𝑦𝑥 𝐹 𝑦 ) )
9 7 8 syl5bb ( ( 𝐹 Fn 𝐴𝑥𝐴 ) → ( 𝑦 = ( 𝐹 ''' 𝑥 ) ↔ 𝑥 𝐹 𝑦 ) )
10 9 pm5.32da ( 𝐹 Fn 𝐴 → ( ( 𝑥𝐴𝑦 = ( 𝐹 ''' 𝑥 ) ) ↔ ( 𝑥𝐴𝑥 𝐹 𝑦 ) ) )
11 6 10 bitr4d ( 𝐹 Fn 𝐴 → ( 𝑥 𝐹 𝑦 ↔ ( 𝑥𝐴𝑦 = ( 𝐹 ''' 𝑥 ) ) ) )
12 11 opabbidv ( 𝐹 Fn 𝐴 → { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝑥 𝐹 𝑦 } = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥𝐴𝑦 = ( 𝐹 ''' 𝑥 ) ) } )
13 3 12 eqtrd ( 𝐹 Fn 𝐴𝐹 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥𝐴𝑦 = ( 𝐹 ''' 𝑥 ) ) } )
14 df-mpt ( 𝑥𝐴 ↦ ( 𝐹 ''' 𝑥 ) ) = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥𝐴𝑦 = ( 𝐹 ''' 𝑥 ) ) }
15 13 14 eqtr4di ( 𝐹 Fn 𝐴𝐹 = ( 𝑥𝐴 ↦ ( 𝐹 ''' 𝑥 ) ) )