Metamath Proof Explorer


Theorem imadifxp

Description: Image of the difference with a Cartesian product. (Contributed by Thierry Arnoux, 13-Dec-2017)

Ref Expression
Assertion imadifxp ⊢ C ⊆ A → R ∖ A × B C = R C ∖ B

Proof

Step Hyp Ref Expression
1 ima0 ⊢ R ∖ A × B ∅ = ∅
2 imaeq2 ⊢ C = ∅ → R ∖ A × B C = R ∖ A × B ∅
3 imaeq2 ⊢ C = ∅ → R C = R ∅
4 ima0 ⊢ R ∅ = ∅
5 3 4 eqtrdi ⊢ C = ∅ → R C = ∅
6 5 difeq1d ⊢ C = ∅ → R C ∖ B = ∅ ∖ B
7 0dif ⊢ ∅ ∖ B = ∅
8 6 7 eqtrdi ⊢ C = ∅ → R C ∖ B = ∅
9 1 2 8 3eqtr4a ⊢ C = ∅ → R ∖ A × B C = R C ∖ B
10 9 adantl ⊢ C ⊆ A ∧ C = ∅ → R ∖ A × B C = R C ∖ B
11 uncom ⊢ ∅ ∪ R ∖ A × B C = R ∖ A × B C ∪ ∅
12 un0 ⊢ R ∖ A × B C ∪ ∅ = R ∖ A × B C
13 11 12 eqtr2i ⊢ R ∖ A × B C = ∅ ∪ R ∖ A × B C
14 inundif ⊢ R ∩ A × B ∪ R ∖ A × B = R
15 14 imaeq1i ⊢ R ∩ A × B ∪ R ∖ A × B C = R C
16 imaundir ⊢ R ∩ A × B ∪ R ∖ A × B C = R ∩ A × B C ∪ R ∖ A × B C
17 15 16 eqtr3i ⊢ R C = R ∩ A × B C ∪ R ∖ A × B C
18 17 difeq1i ⊢ R C ∖ B = R ∩ A × B C ∪ R ∖ A × B C ∖ B
19 difundir ⊢ R ∩ A × B C ∪ R ∖ A × B C ∖ B = R ∩ A × B C ∖ B ∪ R ∖ A × B C ∖ B
20 18 19 eqtri ⊢ R C ∖ B = R ∩ A × B C ∖ B ∪ R ∖ A × B C ∖ B
21 inss2 ⊢ R ∩ A × B ⊆ A × B
22 imass1 ⊢ R ∩ A × B ⊆ A × B → R ∩ A × B C ⊆ A × B C
23 ssdif ⊢ R ∩ A × B C ⊆ A × B C → R ∩ A × B C ∖ B ⊆ A × B C ∖ B
24 21 22 23 mp2b ⊢ R ∩ A × B C ∖ B ⊆ A × B C ∖ B
25 xpima ⊢ A × B C = if A ∩ C = ∅ ∅ B
26 incom ⊢ C ∩ A = A ∩ C
27 dfss2 ⊢ C ⊆ A ↔ C ∩ A = C
28 27 biimpi ⊢ C ⊆ A → C ∩ A = C
29 26 28 eqtr3id ⊢ C ⊆ A → A ∩ C = C
30 29 adantl ⊢ C ≠ ∅ ∧ C ⊆ A → A ∩ C = C
31 simpl ⊢ C ≠ ∅ ∧ C ⊆ A → C ≠ ∅
32 30 31 eqnetrd ⊢ C ≠ ∅ ∧ C ⊆ A → A ∩ C ≠ ∅
33 neneq ⊢ A ∩ C ≠ ∅ → ¬ A ∩ C = ∅
34 iffalse ⊢ ¬ A ∩ C = ∅ → if A ∩ C = ∅ ∅ B = B
35 32 33 34 3syl ⊢ C ≠ ∅ ∧ C ⊆ A → if A ∩ C = ∅ ∅ B = B
36 25 35 eqtrid ⊢ C ≠ ∅ ∧ C ⊆ A → A × B C = B
37 36 difeq1d ⊢ C ≠ ∅ ∧ C ⊆ A → A × B C ∖ B = B ∖ B
38 difid ⊢ B ∖ B = ∅
39 37 38 eqtrdi ⊢ C ≠ ∅ ∧ C ⊆ A → A × B C ∖ B = ∅
40 24 39 sseqtrid ⊢ C ≠ ∅ ∧ C ⊆ A → R ∩ A × B C ∖ B ⊆ ∅
41 ss0 ⊢ R ∩ A × B C ∖ B ⊆ ∅ → R ∩ A × B C ∖ B = ∅
42 40 41 syl ⊢ C ≠ ∅ ∧ C ⊆ A → R ∩ A × B C ∖ B = ∅
43 df-ima ⊢ R ∖ A × B C = ran ⁡ R ∖ A × B ↾ C
44 df-res ⊢ R ∖ A × B ↾ C = R ∖ A × B ∩ C × V
45 44 rneqi ⊢ ran ⁡ R ∖ A × B ↾ C = ran ⁡ R ∖ A × B ∩ C × V
46 43 45 eqtri ⊢ R ∖ A × B C = ran ⁡ R ∖ A × B ∩ C × V
47 46 ineq1i ⊢ R ∖ A × B C ∩ B = ran ⁡ R ∖ A × B ∩ C × V ∩ B
48 xpss1 ⊢ C ⊆ A → C × V ⊆ A × V
49 sslin ⊢ C × V ⊆ A × V → R ∖ A × B ∩ C × V ⊆ R ∖ A × B ∩ A × V
50 rnss ⊢ R ∖ A × B ∩ C × V ⊆ R ∖ A × B ∩ A × V → ran ⁡ R ∖ A × B ∩ C × V ⊆ ran ⁡ R ∖ A × B ∩ A × V
51 48 49 50 3syl ⊢ C ⊆ A → ran ⁡ R ∖ A × B ∩ C × V ⊆ ran ⁡ R ∖ A × B ∩ A × V
52 ssn0 ⊢ C ⊆ A ∧ C ≠ ∅ → A ≠ ∅
53 52 ancoms ⊢ C ≠ ∅ ∧ C ⊆ A → A ≠ ∅
54 inss1 ⊢ A × V ∩ R ⊆ A × V
55 ssdif ⊢ A × V ∩ R ⊆ A × V → A × V ∩ R ∖ A × B ⊆ A × V ∖ A × B
56 54 55 ax-mp ⊢ A × V ∩ R ∖ A × B ⊆ A × V ∖ A × B
57 incom ⊢ A × V ∩ R ∖ A × B = R ∖ A × B ∩ A × V
58 indif2 ⊢ A × V ∩ R ∖ A × B = A × V ∩ R ∖ A × B
59 57 58 eqtr3i ⊢ R ∖ A × B ∩ A × V = A × V ∩ R ∖ A × B
60 difxp2 ⊢ A × V ∖ B = A × V ∖ A × B
61 56 59 60 3sstr4i ⊢ R ∖ A × B ∩ A × V ⊆ A × V ∖ B
62 rnss ⊢ R ∖ A × B ∩ A × V ⊆ A × V ∖ B → ran ⁡ R ∖ A × B ∩ A × V ⊆ ran ⁡ A × V ∖ B
63 61 62 mp1i ⊢ A ≠ ∅ → ran ⁡ R ∖ A × B ∩ A × V ⊆ ran ⁡ A × V ∖ B
64 rnxp ⊢ A ≠ ∅ → ran ⁡ A × V ∖ B = V ∖ B
65 63 64 sseqtrd ⊢ A ≠ ∅ → ran ⁡ R ∖ A × B ∩ A × V ⊆ V ∖ B
66 disj2 ⊢ ran ⁡ R ∖ A × B ∩ A × V ∩ B = ∅ ↔ ran ⁡ R ∖ A × B ∩ A × V ⊆ V ∖ B
67 65 66 sylibr ⊢ A ≠ ∅ → ran ⁡ R ∖ A × B ∩ A × V ∩ B = ∅
68 53 67 syl ⊢ C ≠ ∅ ∧ C ⊆ A → ran ⁡ R ∖ A × B ∩ A × V ∩ B = ∅
69 ssdisj ⊢ ran ⁡ R ∖ A × B ∩ C × V ⊆ ran ⁡ R ∖ A × B ∩ A × V ∧ ran ⁡ R ∖ A × B ∩ A × V ∩ B = ∅ → ran ⁡ R ∖ A × B ∩ C × V ∩ B = ∅
70 51 68 69 syl2an2 ⊢ C ≠ ∅ ∧ C ⊆ A → ran ⁡ R ∖ A × B ∩ C × V ∩ B = ∅
71 47 70 eqtrid ⊢ C ≠ ∅ ∧ C ⊆ A → R ∖ A × B C ∩ B = ∅
72 disj3 ⊢ R ∖ A × B C ∩ B = ∅ ↔ R ∖ A × B C = R ∖ A × B C ∖ B
73 71 72 sylib ⊢ C ≠ ∅ ∧ C ⊆ A → R ∖ A × B C = R ∖ A × B C ∖ B
74 73 eqcomd ⊢ C ≠ ∅ ∧ C ⊆ A → R ∖ A × B C ∖ B = R ∖ A × B C
75 42 74 uneq12d ⊢ C ≠ ∅ ∧ C ⊆ A → R ∩ A × B C ∖ B ∪ R ∖ A × B C ∖ B = ∅ ∪ R ∖ A × B C
76 20 75 eqtrid ⊢ C ≠ ∅ ∧ C ⊆ A → R C ∖ B = ∅ ∪ R ∖ A × B C
77 13 76 eqtr4id ⊢ C ≠ ∅ ∧ C ⊆ A → R ∖ A × B C = R C ∖ B
78 77 ancoms ⊢ C ⊆ A ∧ C ≠ ∅ → R ∖ A × B C = R C ∖ B
79 10 78 pm2.61dane ⊢ C ⊆ A → R ∖ A × B C = R C ∖ B