Metamath Proof Explorer


Theorem feqmptdf

Description: Deduction form of dffn5f . (Contributed by Mario Carneiro, 8-Jan-2015) (Revised by Thierry Arnoux, 10-May-2017)

Ref Expression
Hypotheses feqmptdf.1 𝑥 𝐴
feqmptdf.2 𝑥 𝐹
feqmptdf.3 ( 𝜑𝐹 : 𝐴𝐵 )
Assertion feqmptdf ( 𝜑𝐹 = ( 𝑥𝐴 ↦ ( 𝐹𝑥 ) ) )

Proof

Step Hyp Ref Expression
1 feqmptdf.1 𝑥 𝐴
2 feqmptdf.2 𝑥 𝐹
3 feqmptdf.3 ( 𝜑𝐹 : 𝐴𝐵 )
4 ffn ( 𝐹 : 𝐴𝐵𝐹 Fn 𝐴 )
5 fnrel ( 𝐹 Fn 𝐴 → Rel 𝐹 )
6 nfcv 𝑦 𝐹
7 2 6 dfrel4 ( Rel 𝐹𝐹 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝑥 𝐹 𝑦 } )
8 5 7 sylib ( 𝐹 Fn 𝐴𝐹 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝑥 𝐹 𝑦 } )
9 2 1 nffn 𝑥 𝐹 Fn 𝐴
10 nfv 𝑦 𝐹 Fn 𝐴
11 fnbr ( ( 𝐹 Fn 𝐴𝑥 𝐹 𝑦 ) → 𝑥𝐴 )
12 11 ex ( 𝐹 Fn 𝐴 → ( 𝑥 𝐹 𝑦𝑥𝐴 ) )
13 12 pm4.71rd ( 𝐹 Fn 𝐴 → ( 𝑥 𝐹 𝑦 ↔ ( 𝑥𝐴𝑥 𝐹 𝑦 ) ) )
14 eqcom ( 𝑦 = ( 𝐹𝑥 ) ↔ ( 𝐹𝑥 ) = 𝑦 )
15 fnbrfvb ( ( 𝐹 Fn 𝐴𝑥𝐴 ) → ( ( 𝐹𝑥 ) = 𝑦𝑥 𝐹 𝑦 ) )
16 14 15 bitrid ( ( 𝐹 Fn 𝐴𝑥𝐴 ) → ( 𝑦 = ( 𝐹𝑥 ) ↔ 𝑥 𝐹 𝑦 ) )
17 16 pm5.32da ( 𝐹 Fn 𝐴 → ( ( 𝑥𝐴𝑦 = ( 𝐹𝑥 ) ) ↔ ( 𝑥𝐴𝑥 𝐹 𝑦 ) ) )
18 13 17 bitr4d ( 𝐹 Fn 𝐴 → ( 𝑥 𝐹 𝑦 ↔ ( 𝑥𝐴𝑦 = ( 𝐹𝑥 ) ) ) )
19 9 10 18 opabbid ( 𝐹 Fn 𝐴 → { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝑥 𝐹 𝑦 } = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥𝐴𝑦 = ( 𝐹𝑥 ) ) } )
20 8 19 eqtrd ( 𝐹 Fn 𝐴𝐹 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥𝐴𝑦 = ( 𝐹𝑥 ) ) } )
21 df-mpt ( 𝑥𝐴 ↦ ( 𝐹𝑥 ) ) = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥𝐴𝑦 = ( 𝐹𝑥 ) ) }
22 20 21 eqtr4di ( 𝐹 Fn 𝐴𝐹 = ( 𝑥𝐴 ↦ ( 𝐹𝑥 ) ) )
23 3 4 22 3syl ( 𝜑𝐹 = ( 𝑥𝐴 ↦ ( 𝐹𝑥 ) ) )