Theorem mss 4211

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 2733	. . . . 5 ⊢ 𝑦 ∈ V
2	1	snss 3709	. . . 4 ⊢ (𝑦 ∈ 𝐴 ↔ {𝑦} ⊆ 𝐴)
3	1	snm 3703	. . . . 5 ⊢ ∃𝑤 𝑤 ∈ {𝑦}
4	1	snex 4171	. . . . . 6 ⊢ {𝑦} ∈ V
5		sseq1 3170	. . . . . . 7 ⊢ (𝑥 = {𝑦} → (𝑥 ⊆ 𝐴 ↔ {𝑦} ⊆ 𝐴))
6		eleq2 2234	. . . . . . . 8 ⊢ (𝑥 = {𝑦} → (𝑤 ∈ 𝑥 ↔ 𝑤 ∈ {𝑦}))
7	6	exbidv 1818	. . . . . . 7 ⊢ (𝑥 = {𝑦} → (∃𝑤 𝑤 ∈ 𝑥 ↔ ∃𝑤 𝑤 ∈ {𝑦}))
8	5, 7	anbi12d 470	. . . . . 6 ⊢ (𝑥 = {𝑦} → ((𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥) ↔ ({𝑦} ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ {𝑦})))
9	4, 8	spcev 2825	. . . . 5 ⊢ (({𝑦} ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ {𝑦}) → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
10	3, 9	mpan2 423	. . . 4 ⊢ ({𝑦} ⊆ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
11	2, 10	sylbi 120	. . 3 ⊢ (𝑦 ∈ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
12	11	exlimiv 1591	. 2 ⊢ (∃𝑦 𝑦 ∈ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
13		elequ1 2145	. . . . 5 ⊢ (𝑧 = 𝑤 → (𝑧 ∈ 𝑥 ↔ 𝑤 ∈ 𝑥))
14	13	cbvexv 1911	. . . 4 ⊢ (∃𝑧 𝑧 ∈ 𝑥 ↔ ∃𝑤 𝑤 ∈ 𝑥)
15	14	anbi2i 454	. . 3 ⊢ ((𝑥 ⊆ 𝐴 ∧ ∃𝑧 𝑧 ∈ 𝑥) ↔ (𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
16	15	exbii 1598	. 2 ⊢ (∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑧 𝑧 ∈ 𝑥) ↔ ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
17	12, 16	sylibr 133	1 ⊢ (∃𝑦 𝑦 ∈ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑧 𝑧 ∈ 𝑥))

Syntax hints:   → wi 4   ∧ wa 103   = wceq 1348  ∃wex 1485   ∈ wcel 2141   ⊆ wss 3121  {csn 3583

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-13 2143  ax-14 2144  ax-ext 2152  ax-sep 4107  ax-pow 4160

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-in 3127  df-ss 3134  df-pw 3568  df-sn 3589

This theorem is referenced by: (None)