Theorem mss 3981

Description: An inhabited class (even if proper) has an inhabited subset. (Contributed by Jim Kingdon, 17-Sep-2018.)

Assertion

Ref	Expression
mss	⊢ (∃𝑦 𝑦 ∈ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑧 𝑧 ∈ 𝑥))

Distinct variable groups:   𝑥,𝑦   𝑥,𝑧   𝑥,𝐴,𝑦

Allowed substitution hint:   𝐴(𝑧)

Proof of Theorem mss

Dummy variable 𝑤 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		vex 2604	. . . . 5 ⊢ 𝑦 ∈ V
2	1	snss 3516	. . . 4 ⊢ (𝑦 ∈ 𝐴 ↔ {𝑦} ⊆ 𝐴)
3	1	snm 3510	. . . . 5 ⊢ ∃𝑤 𝑤 ∈ {𝑦}
4	1	snex 3957	. . . . . 6 ⊢ {𝑦} ∈ V
5		sseq1 3020	. . . . . . 7 ⊢ (𝑥 = {𝑦} → (𝑥 ⊆ 𝐴 ↔ {𝑦} ⊆ 𝐴))
6		eleq2 2142	. . . . . . . 8 ⊢ (𝑥 = {𝑦} → (𝑤 ∈ 𝑥 ↔ 𝑤 ∈ {𝑦}))
7	6	exbidv 1746	. . . . . . 7 ⊢ (𝑥 = {𝑦} → (∃𝑤 𝑤 ∈ 𝑥 ↔ ∃𝑤 𝑤 ∈ {𝑦}))
8	5, 7	anbi12d 456	. . . . . 6 ⊢ (𝑥 = {𝑦} → ((𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥) ↔ ({𝑦} ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ {𝑦})))
9	4, 8	spcev 2692	. . . . 5 ⊢ (({𝑦} ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ {𝑦}) → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
10	3, 9	mpan2 415	. . . 4 ⊢ ({𝑦} ⊆ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
11	2, 10	sylbi 119	. . 3 ⊢ (𝑦 ∈ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
12	11	exlimiv 1529	. 2 ⊢ (∃𝑦 𝑦 ∈ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
13		elequ1 1640	. . . . 5 ⊢ (𝑧 = 𝑤 → (𝑧 ∈ 𝑥 ↔ 𝑤 ∈ 𝑥))
14	13	cbvexv 1836	. . . 4 ⊢ (∃𝑧 𝑧 ∈ 𝑥 ↔ ∃𝑤 𝑤 ∈ 𝑥)
15	14	anbi2i 444	. . 3 ⊢ ((𝑥 ⊆ 𝐴 ∧ ∃𝑧 𝑧 ∈ 𝑥) ↔ (𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
16	15	exbii 1536	. 2 ⊢ (∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑧 𝑧 ∈ 𝑥) ↔ ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
17	12, 16	sylibr 132	1 ⊢ (∃𝑦 𝑦 ∈ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑧 𝑧 ∈ 𝑥))

Syntax hints:   → wi 4   ∧ wa 102   = wceq 1284  ∃wex 1421   ∈ wcel 1433   ⊆ wss 2973  {csn 3398

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-13 1444  ax-14 1445  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063  ax-sep 3896  ax-pow 3948

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-in 2979  df-ss 2986  df-pw 3384  df-sn 3404

This theorem is referenced by: (None)