Metamath Proof Explorer


Theorem mpteqb

Description: Bidirectional equality theorem for a mapping abstraction. Equivalent to eqfnfv . (Contributed by Mario Carneiro, 14-Nov-2014)

Ref Expression
Assertion mpteqb x A B V x A B = x A C x A B = C

Proof

Step Hyp Ref Expression
1 elex B V B V
2 1 ralimi x A B V x A B V
3 fneq1 x A B = x A C x A B Fn A x A C Fn A
4 eqid x A B = x A B
5 4 mptfng x A B V x A B Fn A
6 eqid x A C = x A C
7 6 mptfng x A C V x A C Fn A
8 3 5 7 3bitr4g x A B = x A C x A B V x A C V
9 8 biimpd x A B = x A C x A B V x A C V
10 r19.26 x A B V C V x A B V x A C V
11 nfmpt1 _ x x A B
12 nfmpt1 _ x x A C
13 11 12 nfeq x x A B = x A C
14 simpll x A B = x A C x A B V C V x A B = x A C
15 14 fveq1d x A B = x A C x A B V C V x A B x = x A C x
16 4 fvmpt2 x A B V x A B x = B
17 16 ad2ant2lr x A B = x A C x A B V C V x A B x = B
18 6 fvmpt2 x A C V x A C x = C
19 18 ad2ant2l x A B = x A C x A B V C V x A C x = C
20 15 17 19 3eqtr3d x A B = x A C x A B V C V B = C
21 20 exp31 x A B = x A C x A B V C V B = C
22 13 21 ralrimi x A B = x A C x A B V C V B = C
23 ralim x A B V C V B = C x A B V C V x A B = C
24 22 23 syl x A B = x A C x A B V C V x A B = C
25 10 24 syl5bir x A B = x A C x A B V x A C V x A B = C
26 25 expd x A B = x A C x A B V x A C V x A B = C
27 9 26 mpdd x A B = x A C x A B V x A B = C
28 27 com12 x A B V x A B = x A C x A B = C
29 eqid A = A
30 mpteq12 A = A x A B = C x A B = x A C
31 29 30 mpan x A B = C x A B = x A C
32 28 31 impbid1 x A B V x A B = x A C x A B = C
33 2 32 syl x A B V x A B = x A C x A B = C