Metamath Proof Explorer


Theorem dfac5lem3

Description: Lemma for dfac5 . (Contributed by NM, 12-Apr-2004)

Ref Expression
Hypothesis dfac5lem.1 ⊢ A = u | u ≠ ∅ ∧ ∃ t ∈ h u = t × t
Assertion dfac5lem3 ⊢ w × w ∈ A ↔ w ≠ ∅ ∧ w ∈ h

Proof

Step Hyp Ref Expression
1 dfac5lem.1 ⊢ A = u | u ≠ ∅ ∧ ∃ t ∈ h u = t × t
2 vsnex ⊢ w ∈ V
3 vex ⊢ w ∈ V
4 2 3 xpex ⊢ w × w ∈ V
5 neeq1 ⊢ u = w × w → u ≠ ∅ ↔ w × w ≠ ∅
6 eqeq1 ⊢ u = w × w → u = t × t ↔ w × w = t × t
7 6 rexbidv ⊢ u = w × w → ∃ t ∈ h u = t × t ↔ ∃ t ∈ h w × w = t × t
8 5 7 anbi12d ⊢ u = w × w → u ≠ ∅ ∧ ∃ t ∈ h u = t × t ↔ w × w ≠ ∅ ∧ ∃ t ∈ h w × w = t × t
9 4 8 elab ⊢ w × w ∈ u | u ≠ ∅ ∧ ∃ t ∈ h u = t × t ↔ w × w ≠ ∅ ∧ ∃ t ∈ h w × w = t × t
10 1 eleq2i ⊢ w × w ∈ A ↔ w × w ∈ u | u ≠ ∅ ∧ ∃ t ∈ h u = t × t
11 xpeq2 ⊢ w = ∅ → w × w = w × ∅
12 xp0 ⊢ w × ∅ = ∅
13 11 12 eqtrdi ⊢ w = ∅ → w × w = ∅
14 rneq ⊢ w × w = ∅ → ran ⁡ w × w = ran ⁡ ∅
15 3 snnz ⊢ w ≠ ∅
16 rnxp ⊢ w ≠ ∅ → ran ⁡ w × w = w
17 15 16 ax-mp ⊢ ran ⁡ w × w = w
18 rn0 ⊢ ran ⁡ ∅ = ∅
19 14 17 18 3eqtr3g ⊢ w × w = ∅ → w = ∅
20 13 19 impbii ⊢ w = ∅ ↔ w × w = ∅
21 20 necon3bii ⊢ w ≠ ∅ ↔ w × w ≠ ∅
22 df-rex ⊢ ∃ t ∈ h w × w = t × t ↔ ∃ t t ∈ h ∧ w × w = t × t
23 rneq ⊢ w × w = t × t → ran ⁡ w × w = ran ⁡ t × t
24 vex ⊢ t ∈ V
25 24 snnz ⊢ t ≠ ∅
26 rnxp ⊢ t ≠ ∅ → ran ⁡ t × t = t
27 25 26 ax-mp ⊢ ran ⁡ t × t = t
28 23 17 27 3eqtr3g ⊢ w × w = t × t → w = t
29 sneq ⊢ w = t → w = t
30 29 xpeq1d ⊢ w = t → w × w = t × w
31 xpeq2 ⊢ w = t → t × w = t × t
32 30 31 eqtrd ⊢ w = t → w × w = t × t
33 28 32 impbii ⊢ w × w = t × t ↔ w = t
34 equcom ⊢ w = t ↔ t = w
35 33 34 bitri ⊢ w × w = t × t ↔ t = w
36 35 anbi1ci ⊢ t ∈ h ∧ w × w = t × t ↔ t = w ∧ t ∈ h
37 36 exbii ⊢ ∃ t t ∈ h ∧ w × w = t × t ↔ ∃ t t = w ∧ t ∈ h
38 elequ1 ⊢ t = w → t ∈ h ↔ w ∈ h
39 38 equsexvw ⊢ ∃ t t = w ∧ t ∈ h ↔ w ∈ h
40 22 37 39 3bitrri ⊢ w ∈ h ↔ ∃ t ∈ h w × w = t × t
41 21 40 anbi12i ⊢ w ≠ ∅ ∧ w ∈ h ↔ w × w ≠ ∅ ∧ ∃ t ∈ h w × w = t × t
42 9 10 41 3bitr4i ⊢ w × w ∈ A ↔ w ≠ ∅ ∧ w ∈ h