Metamath Proof Explorer


Theorem isofrlem

Description: Lemma for isofr . (Contributed by NM, 29-Apr-2004) (Revised by Mario Carneiro, 18-Nov-2014)

Ref Expression
Hypotheses isofrlem.1 ⊢ φ → H Isom R , S A B
isofrlem.2 ⊢ φ → H x ∈ V
Assertion isofrlem ⊢ φ → S Fr B → R Fr A

Proof

Step Hyp Ref Expression
1 isofrlem.1 ⊢ φ → H Isom R , S A B
2 isofrlem.2 ⊢ φ → H x ∈ V
3 isof1o ⊢ H Isom R , S A B → H : A ⟶ 1-1 onto B
4 1 3 syl ⊢ φ → H : A ⟶ 1-1 onto B
5 f1ofn ⊢ H : A ⟶ 1-1 onto B → H Fn A
6 n0 ⊢ x ≠ ∅ ↔ ∃ y y ∈ x
7 fnfvima ⊢ H Fn A ∧ x ⊆ A ∧ y ∈ x → H ⁡ y ∈ H x
8 7 ne0d ⊢ H Fn A ∧ x ⊆ A ∧ y ∈ x → H x ≠ ∅
9 8 3expia ⊢ H Fn A ∧ x ⊆ A → y ∈ x → H x ≠ ∅
10 9 exlimdv ⊢ H Fn A ∧ x ⊆ A → ∃ y y ∈ x → H x ≠ ∅
11 6 10 biimtrid ⊢ H Fn A ∧ x ⊆ A → x ≠ ∅ → H x ≠ ∅
12 11 expimpd ⊢ H Fn A → x ⊆ A ∧ x ≠ ∅ → H x ≠ ∅
13 5 12 syl ⊢ H : A ⟶ 1-1 onto B → x ⊆ A ∧ x ≠ ∅ → H x ≠ ∅
14 f1ofo ⊢ H : A ⟶ 1-1 onto B → H : A ⟶ onto B
15 imassrn ⊢ H x ⊆ ran ⁡ H
16 forn ⊢ H : A ⟶ onto B → ran ⁡ H = B
17 15 16 sseqtrid ⊢ H : A ⟶ onto B → H x ⊆ B
18 14 17 syl ⊢ H : A ⟶ 1-1 onto B → H x ⊆ B
19 13 18 jctild ⊢ H : A ⟶ 1-1 onto B → x ⊆ A ∧ x ≠ ∅ → H x ⊆ B ∧ H x ≠ ∅
20 4 19 syl ⊢ φ → x ⊆ A ∧ x ≠ ∅ → H x ⊆ B ∧ H x ≠ ∅
21 dffr3 ⊢ S Fr B ↔ ∀ z z ⊆ B ∧ z ≠ ∅ → ∃ w ∈ z z ∩ S -1 w = ∅
22 sseq1 ⊢ z = H x → z ⊆ B ↔ H x ⊆ B
23 neeq1 ⊢ z = H x → z ≠ ∅ ↔ H x ≠ ∅
24 22 23 anbi12d ⊢ z = H x → z ⊆ B ∧ z ≠ ∅ ↔ H x ⊆ B ∧ H x ≠ ∅
25 ineq1 ⊢ z = H x → z ∩ S -1 w = H x ∩ S -1 w
26 25 eqeq1d ⊢ z = H x → z ∩ S -1 w = ∅ ↔ H x ∩ S -1 w = ∅
27 26 rexeqbi1dv ⊢ z = H x → ∃ w ∈ z z ∩ S -1 w = ∅ ↔ ∃ w ∈ H x H x ∩ S -1 w = ∅
28 24 27 imbi12d ⊢ z = H x → z ⊆ B ∧ z ≠ ∅ → ∃ w ∈ z z ∩ S -1 w = ∅ ↔ H x ⊆ B ∧ H x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅
29 28 spcgv ⊢ H x ∈ V → ∀ z z ⊆ B ∧ z ≠ ∅ → ∃ w ∈ z z ∩ S -1 w = ∅ → H x ⊆ B ∧ H x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅
30 2 29 syl ⊢ φ → ∀ z z ⊆ B ∧ z ≠ ∅ → ∃ w ∈ z z ∩ S -1 w = ∅ → H x ⊆ B ∧ H x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅
31 21 30 biimtrid ⊢ φ → S Fr B → H x ⊆ B ∧ H x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅
32 20 31 syl5d ⊢ φ → S Fr B → x ⊆ A ∧ x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅
33 4 adantr ⊢ φ ∧ x ⊆ A → H : A ⟶ 1-1 onto B
34 f1ofun ⊢ H : A ⟶ 1-1 onto B → Fun ⁡ H
35 33 34 syl ⊢ φ ∧ x ⊆ A → Fun ⁡ H
36 simpl ⊢ w ∈ H x ∧ H x ∩ S -1 w = ∅ → w ∈ H x
37 fvelima ⊢ Fun ⁡ H ∧ w ∈ H x → ∃ y ∈ x H ⁡ y = w
38 35 36 37 syl2an ⊢ φ ∧ x ⊆ A ∧ w ∈ H x ∧ H x ∩ S -1 w = ∅ → ∃ y ∈ x H ⁡ y = w
39 simpr ⊢ w ∈ H x ∧ H x ∩ S -1 w = ∅ → H x ∩ S -1 w = ∅
40 ssel ⊢ x ⊆ A → y ∈ x → y ∈ A
41 40 imdistani ⊢ x ⊆ A ∧ y ∈ x → x ⊆ A ∧ y ∈ A
42 isomin ⊢ H Isom R , S A B ∧ x ⊆ A ∧ y ∈ A → x ∩ R -1 y = ∅ ↔ H x ∩ S -1 H ⁡ y = ∅
43 1 41 42 syl2an ⊢ φ ∧ x ⊆ A ∧ y ∈ x → x ∩ R -1 y = ∅ ↔ H x ∩ S -1 H ⁡ y = ∅
44 sneq ⊢ H ⁡ y = w → H ⁡ y = w
45 44 imaeq2d ⊢ H ⁡ y = w → S -1 H ⁡ y = S -1 w
46 45 ineq2d ⊢ H ⁡ y = w → H x ∩ S -1 H ⁡ y = H x ∩ S -1 w
47 46 eqeq1d ⊢ H ⁡ y = w → H x ∩ S -1 H ⁡ y = ∅ ↔ H x ∩ S -1 w = ∅
48 43 47 sylan9bb ⊢ φ ∧ x ⊆ A ∧ y ∈ x ∧ H ⁡ y = w → x ∩ R -1 y = ∅ ↔ H x ∩ S -1 w = ∅
49 39 48 imbitrrid ⊢ φ ∧ x ⊆ A ∧ y ∈ x ∧ H ⁡ y = w → w ∈ H x ∧ H x ∩ S -1 w = ∅ → x ∩ R -1 y = ∅
50 49 exp42 ⊢ φ → x ⊆ A → y ∈ x → H ⁡ y = w → w ∈ H x ∧ H x ∩ S -1 w = ∅ → x ∩ R -1 y = ∅
51 50 imp ⊢ φ ∧ x ⊆ A → y ∈ x → H ⁡ y = w → w ∈ H x ∧ H x ∩ S -1 w = ∅ → x ∩ R -1 y = ∅
52 51 com3l ⊢ y ∈ x → H ⁡ y = w → φ ∧ x ⊆ A → w ∈ H x ∧ H x ∩ S -1 w = ∅ → x ∩ R -1 y = ∅
53 52 com4t ⊢ φ ∧ x ⊆ A → w ∈ H x ∧ H x ∩ S -1 w = ∅ → y ∈ x → H ⁡ y = w → x ∩ R -1 y = ∅
54 53 imp ⊢ φ ∧ x ⊆ A ∧ w ∈ H x ∧ H x ∩ S -1 w = ∅ → y ∈ x → H ⁡ y = w → x ∩ R -1 y = ∅
55 54 reximdvai ⊢ φ ∧ x ⊆ A ∧ w ∈ H x ∧ H x ∩ S -1 w = ∅ → ∃ y ∈ x H ⁡ y = w → ∃ y ∈ x x ∩ R -1 y = ∅
56 38 55 mpd ⊢ φ ∧ x ⊆ A ∧ w ∈ H x ∧ H x ∩ S -1 w = ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
57 56 rexlimdvaa ⊢ φ ∧ x ⊆ A → ∃ w ∈ H x H x ∩ S -1 w = ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
58 57 ex ⊢ φ → x ⊆ A → ∃ w ∈ H x H x ∩ S -1 w = ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
59 58 adantrd ⊢ φ → x ⊆ A ∧ x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
60 59 a2d ⊢ φ → x ⊆ A ∧ x ≠ ∅ → ∃ w ∈ H x H x ∩ S -1 w = ∅ → x ⊆ A ∧ x ≠ ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
61 32 60 syld ⊢ φ → S Fr B → x ⊆ A ∧ x ≠ ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
62 61 alrimdv ⊢ φ → S Fr B → ∀ x x ⊆ A ∧ x ≠ ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
63 dffr3 ⊢ R Fr A ↔ ∀ x x ⊆ A ∧ x ≠ ∅ → ∃ y ∈ x x ∩ R -1 y = ∅
64 62 63 imbitrrdi ⊢ φ → S Fr B → R Fr A