Metamath Proof Explorer


Theorem fresaunres2

Description: From the union of two functions that agree on the domain overlap, either component can be recovered by restriction. (Contributed by Stefan O'Rear, 9-Oct-2014)

Ref Expression
Assertion fresaunres2 F : A C G : B C F A B = G A B F G B = G

Proof

Step Hyp Ref Expression
1 ffn F : A C F Fn A
2 ffn G : B C G Fn B
3 id F A B = G A B F A B = G A B
4 resasplit F Fn A G Fn B F A B = G A B F G = F A B F A B G B A
5 1 2 3 4 syl3an F : A C G : B C F A B = G A B F G = F A B F A B G B A
6 5 reseq1d F : A C G : B C F A B = G A B F G B = F A B F A B G B A B
7 resundir F A B F A B G B A B = F A B B F A B G B A B
8 inss2 A B B
9 resabs2 A B B F A B B = F A B
10 8 9 ax-mp F A B B = F A B
11 resundir F A B G B A B = F A B B G B A B
12 10 11 uneq12i F A B B F A B G B A B = F A B F A B B G B A B
13 dmres dom F A B B = B dom F A B
14 dmres dom F A B = A B dom F
15 14 ineq2i B dom F A B = B A B dom F
16 disjdif B A B =
17 16 ineq1i B A B dom F = dom F
18 inass B A B dom F = B A B dom F
19 0in dom F =
20 17 18 19 3eqtr3i B A B dom F =
21 15 20 eqtri B dom F A B =
22 13 21 eqtri dom F A B B =
23 relres Rel F A B B
24 reldm0 Rel F A B B F A B B = dom F A B B =
25 23 24 ax-mp F A B B = dom F A B B =
26 22 25 mpbir F A B B =
27 difss B A B
28 resabs2 B A B G B A B = G B A
29 27 28 ax-mp G B A B = G B A
30 26 29 uneq12i F A B B G B A B = G B A
31 30 uneq2i F A B F A B B G B A B = F A B G B A
32 simp3 F : A C G : B C F A B = G A B F A B = G A B
33 32 uneq1d F : A C G : B C F A B = G A B F A B G B A = G A B G B A
34 uncom G B A = G B A
35 un0 G B A = G B A
36 34 35 eqtri G B A = G B A
37 36 uneq2i G A B G B A = G A B G B A
38 resundi G A B B A = G A B G B A
39 incom A B = B A
40 39 uneq1i A B B A = B A B A
41 inundif B A B A = B
42 40 41 eqtri A B B A = B
43 42 reseq2i G A B B A = G B
44 fnresdm G Fn B G B = G
45 2 44 syl G : B C G B = G
46 45 adantl F : A C G : B C G B = G
47 43 46 syl5eq F : A C G : B C G A B B A = G
48 38 47 eqtr3id F : A C G : B C G A B G B A = G
49 37 48 syl5eq F : A C G : B C G A B G B A = G
50 49 3adant3 F : A C G : B C F A B = G A B G A B G B A = G
51 33 50 eqtrd F : A C G : B C F A B = G A B F A B G B A = G
52 31 51 syl5eq F : A C G : B C F A B = G A B F A B F A B B G B A B = G
53 12 52 syl5eq F : A C G : B C F A B = G A B F A B B F A B G B A B = G
54 7 53 syl5eq F : A C G : B C F A B = G A B F A B F A B G B A B = G
55 6 54 eqtrd F : A C G : B C F A B = G A B F G B = G