Metamath Proof Explorer


Theorem fvmptf

Description: Value of a function given by an ordered-pair class abstraction. This version of fvmptg uses bound-variable hypotheses instead of distinct variable conditions. (Contributed by NM, 8-Nov-2005) (Revised by Mario Carneiro, 15-Oct-2016)

Ref Expression
Hypotheses fvmptf.1 𝑥 𝐴
fvmptf.2 𝑥 𝐶
fvmptf.3 ( 𝑥 = 𝐴𝐵 = 𝐶 )
fvmptf.4 𝐹 = ( 𝑥𝐷𝐵 )
Assertion fvmptf ( ( 𝐴𝐷𝐶𝑉 ) → ( 𝐹𝐴 ) = 𝐶 )

Proof

Step Hyp Ref Expression
1 fvmptf.1 𝑥 𝐴
2 fvmptf.2 𝑥 𝐶
3 fvmptf.3 ( 𝑥 = 𝐴𝐵 = 𝐶 )
4 fvmptf.4 𝐹 = ( 𝑥𝐷𝐵 )
5 2 nfel1 𝑥 𝐶 ∈ V
6 nfmpt1 𝑥 ( 𝑥𝐷𝐵 )
7 4 6 nfcxfr 𝑥 𝐹
8 7 1 nffv 𝑥 ( 𝐹𝐴 )
9 8 2 nfeq 𝑥 ( 𝐹𝐴 ) = 𝐶
10 5 9 nfim 𝑥 ( 𝐶 ∈ V → ( 𝐹𝐴 ) = 𝐶 )
11 3 eleq1d ( 𝑥 = 𝐴 → ( 𝐵 ∈ V ↔ 𝐶 ∈ V ) )
12 fveq2 ( 𝑥 = 𝐴 → ( 𝐹𝑥 ) = ( 𝐹𝐴 ) )
13 12 3 eqeq12d ( 𝑥 = 𝐴 → ( ( 𝐹𝑥 ) = 𝐵 ↔ ( 𝐹𝐴 ) = 𝐶 ) )
14 11 13 imbi12d ( 𝑥 = 𝐴 → ( ( 𝐵 ∈ V → ( 𝐹𝑥 ) = 𝐵 ) ↔ ( 𝐶 ∈ V → ( 𝐹𝐴 ) = 𝐶 ) ) )
15 4 fvmpt2 ( ( 𝑥𝐷𝐵 ∈ V ) → ( 𝐹𝑥 ) = 𝐵 )
16 15 ex ( 𝑥𝐷 → ( 𝐵 ∈ V → ( 𝐹𝑥 ) = 𝐵 ) )
17 1 10 14 16 vtoclgaf ( 𝐴𝐷 → ( 𝐶 ∈ V → ( 𝐹𝐴 ) = 𝐶 ) )
18 elex ( 𝐶𝑉𝐶 ∈ V )
19 17 18 impel ( ( 𝐴𝐷𝐶𝑉 ) → ( 𝐹𝐴 ) = 𝐶 )