Metamath Proof Explorer


Theorem kmlem9

Description: Lemma for 5-quantifier AC of Kurt Maes, Th. 4, part of 3 => 4. (Contributed by NM, 25-Mar-2004)

Ref Expression
Hypothesis kmlem9.1 ⊢ A = u | ∃ t ∈ x u = t ∖ ⋃ x ∖ t
Assertion kmlem9 ⊢ ∀ z ∈ A ∀ w ∈ A z ≠ w → z ∩ w = ∅

Proof

Step Hyp Ref Expression
1 kmlem9.1 ⊢ A = u | ∃ t ∈ x u = t ∖ ⋃ x ∖ t
2 vex ⊢ z ∈ V
3 eqeq1 ⊢ u = z → u = t ∖ ⋃ x ∖ t ↔ z = t ∖ ⋃ x ∖ t
4 3 rexbidv ⊢ u = z → ∃ t ∈ x u = t ∖ ⋃ x ∖ t ↔ ∃ t ∈ x z = t ∖ ⋃ x ∖ t
5 2 4 1 elab2 ⊢ z ∈ A ↔ ∃ t ∈ x z = t ∖ ⋃ x ∖ t
6 vex ⊢ w ∈ V
7 eqeq1 ⊢ u = w → u = t ∖ ⋃ x ∖ t ↔ w = t ∖ ⋃ x ∖ t
8 7 rexbidv ⊢ u = w → ∃ t ∈ x u = t ∖ ⋃ x ∖ t ↔ ∃ t ∈ x w = t ∖ ⋃ x ∖ t
9 6 8 1 elab2 ⊢ w ∈ A ↔ ∃ t ∈ x w = t ∖ ⋃ x ∖ t
10 difeq1 ⊢ t = h → t ∖ ⋃ x ∖ t = h ∖ ⋃ x ∖ t
11 sneq ⊢ t = h → t = h
12 11 difeq2d ⊢ t = h → x ∖ t = x ∖ h
13 12 unieqd ⊢ t = h → ⋃ x ∖ t = ⋃ x ∖ h
14 13 difeq2d ⊢ t = h → h ∖ ⋃ x ∖ t = h ∖ ⋃ x ∖ h
15 10 14 eqtrd ⊢ t = h → t ∖ ⋃ x ∖ t = h ∖ ⋃ x ∖ h
16 15 eqeq2d ⊢ t = h → w = t ∖ ⋃ x ∖ t ↔ w = h ∖ ⋃ x ∖ h
17 16 cbvrexvw ⊢ ∃ t ∈ x w = t ∖ ⋃ x ∖ t ↔ ∃ h ∈ x w = h ∖ ⋃ x ∖ h
18 9 17 bitri ⊢ w ∈ A ↔ ∃ h ∈ x w = h ∖ ⋃ x ∖ h
19 reeanv ⊢ ∃ t ∈ x ∃ h ∈ x z = t ∖ ⋃ x ∖ t ∧ w = h ∖ ⋃ x ∖ h ↔ ∃ t ∈ x z = t ∖ ⋃ x ∖ t ∧ ∃ h ∈ x w = h ∖ ⋃ x ∖ h
20 eqeq12 ⊢ z = t ∖ ⋃ x ∖ t ∧ w = h ∖ ⋃ x ∖ h → z = w ↔ t ∖ ⋃ x ∖ t = h ∖ ⋃ x ∖ h
21 15 20 imbitrrid ⊢ z = t ∖ ⋃ x ∖ t ∧ w = h ∖ ⋃ x ∖ h → t = h → z = w
22 21 necon3d ⊢ z = t ∖ ⋃ x ∖ t ∧ w = h ∖ ⋃ x ∖ h → z ≠ w → t ≠ h
23 kmlem5 ⊢ h ∈ x ∧ t ≠ h → t ∖ ⋃ x ∖ t ∩ h ∖ ⋃ x ∖ h = ∅
24 ineq12 ⊢ z = t ∖ ⋃ x ∖ t ∧ w = h ∖ ⋃ x ∖ h → z ∩ w = t ∖ ⋃ x ∖ t ∩ h ∖ ⋃ x ∖ h
25 24 eqeq1d ⊢ z = t ∖ ⋃ x ∖ t ∧ w = h ∖ ⋃ x ∖ h → z ∩ w = ∅ ↔ t ∖ ⋃ x ∖ t ∩ h ∖ ⋃ x ∖ h = ∅
26 23 25 imbitrrid ⊢ z = t ∖ ⋃ x ∖ t ∧ w = h ∖ ⋃ x ∖ h → h ∈ x ∧ t ≠ h → z ∩ w = ∅
27 26 expd ⊢ z = t ∖ ⋃ x ∖ t ∧ w = h ∖ ⋃ x ∖ h → h ∈ x → t ≠ h → z ∩ w = ∅
28 22 27 syl5d ⊢ z = t ∖ ⋃ x ∖ t ∧ w = h ∖ ⋃ x ∖ h → h ∈ x → z ≠ w → z ∩ w = ∅
29 28 com12 ⊢ h ∈ x → z = t ∖ ⋃ x ∖ t ∧ w = h ∖ ⋃ x ∖ h → z ≠ w → z ∩ w = ∅
30 29 adantl ⊢ t ∈ x ∧ h ∈ x → z = t ∖ ⋃ x ∖ t ∧ w = h ∖ ⋃ x ∖ h → z ≠ w → z ∩ w = ∅
31 30 rexlimivv ⊢ ∃ t ∈ x ∃ h ∈ x z = t ∖ ⋃ x ∖ t ∧ w = h ∖ ⋃ x ∖ h → z ≠ w → z ∩ w = ∅
32 19 31 sylbir ⊢ ∃ t ∈ x z = t ∖ ⋃ x ∖ t ∧ ∃ h ∈ x w = h ∖ ⋃ x ∖ h → z ≠ w → z ∩ w = ∅
33 5 18 32 syl2anb ⊢ z ∈ A ∧ w ∈ A → z ≠ w → z ∩ w = ∅
34 33 rgen2 ⊢ ∀ z ∈ A ∀ w ∈ A z ≠ w → z ∩ w = ∅