Metamath Proof Explorer


Theorem xppreima2

Description: The preimage of a Cartesian product is the intersection of the preimages of each component function. (Contributed by Thierry Arnoux, 7-Jun-2017)

Ref Expression
Hypotheses xppreima2.1 φ F : A B
xppreima2.2 φ G : A C
xppreima2.3 H = x A F x G x
Assertion xppreima2 φ H -1 Y × Z = F -1 Y G -1 Z

Proof

Step Hyp Ref Expression
1 xppreima2.1 φ F : A B
2 xppreima2.2 φ G : A C
3 xppreima2.3 H = x A F x G x
4 3 funmpt2 Fun H
5 1 ffvelrnda φ x A F x B
6 2 ffvelrnda φ x A G x C
7 opelxp F x G x B × C F x B G x C
8 5 6 7 sylanbrc φ x A F x G x B × C
9 8 3 fmptd φ H : A B × C
10 9 frnd φ ran H B × C
11 xpss B × C V × V
12 10 11 sstrdi φ ran H V × V
13 xppreima Fun H ran H V × V H -1 Y × Z = 1 st H -1 Y 2 nd H -1 Z
14 4 12 13 sylancr φ H -1 Y × Z = 1 st H -1 Y 2 nd H -1 Z
15 fo1st 1 st : V onto V
16 fofn 1 st : V onto V 1 st Fn V
17 15 16 ax-mp 1 st Fn V
18 opex F x G x V
19 18 3 fnmpti H Fn A
20 ssv ran H V
21 fnco 1 st Fn V H Fn A ran H V 1 st H Fn A
22 17 19 20 21 mp3an 1 st H Fn A
23 22 a1i φ 1 st H Fn A
24 1 ffnd φ F Fn A
25 4 a1i φ x A Fun H
26 12 adantr φ x A ran H V × V
27 simpr φ x A x A
28 18 3 dmmpti dom H = A
29 27 28 eleqtrrdi φ x A x dom H
30 opfv Fun H ran H V × V x dom H H x = 1 st H x 2 nd H x
31 25 26 29 30 syl21anc φ x A H x = 1 st H x 2 nd H x
32 3 fvmpt2 x A F x G x B × C H x = F x G x
33 27 8 32 syl2anc φ x A H x = F x G x
34 31 33 eqtr3d φ x A 1 st H x 2 nd H x = F x G x
35 fvex 1 st H x V
36 fvex 2 nd H x V
37 35 36 opth 1 st H x 2 nd H x = F x G x 1 st H x = F x 2 nd H x = G x
38 34 37 sylib φ x A 1 st H x = F x 2 nd H x = G x
39 38 simpld φ x A 1 st H x = F x
40 23 24 39 eqfnfvd φ 1 st H = F
41 40 cnveqd φ 1 st H -1 = F -1
42 41 imaeq1d φ 1 st H -1 Y = F -1 Y
43 fo2nd 2 nd : V onto V
44 fofn 2 nd : V onto V 2 nd Fn V
45 43 44 ax-mp 2 nd Fn V
46 fnco 2 nd Fn V H Fn A ran H V 2 nd H Fn A
47 45 19 20 46 mp3an 2 nd H Fn A
48 47 a1i φ 2 nd H Fn A
49 2 ffnd φ G Fn A
50 38 simprd φ x A 2 nd H x = G x
51 48 49 50 eqfnfvd φ 2 nd H = G
52 51 cnveqd φ 2 nd H -1 = G -1
53 52 imaeq1d φ 2 nd H -1 Z = G -1 Z
54 42 53 ineq12d φ 1 st H -1 Y 2 nd H -1 Z = F -1 Y G -1 Z
55 14 54 eqtrd φ H -1 Y × Z = F -1 Y G -1 Z