Theorem mss 4220

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 2738	. . . . 5 ⊢ 𝑦 ∈ V
2	1	snss 3724	. . . 4 ⊢ (𝑦 ∈ 𝐴 ↔ {𝑦} ⊆ 𝐴)
3	1	snm 3709	. . . . 5 ⊢ ∃𝑤 𝑤 ∈ {𝑦}
4	1	snex 4180	. . . . . 6 ⊢ {𝑦} ∈ V
5		sseq1 3176	. . . . . . 7 ⊢ (𝑥 = {𝑦} → (𝑥 ⊆ 𝐴 ↔ {𝑦} ⊆ 𝐴))
6		eleq2 2239	. . . . . . . 8 ⊢ (𝑥 = {𝑦} → (𝑤 ∈ 𝑥 ↔ 𝑤 ∈ {𝑦}))
7	6	exbidv 1823	. . . . . . 7 ⊢ (𝑥 = {𝑦} → (∃𝑤 𝑤 ∈ 𝑥 ↔ ∃𝑤 𝑤 ∈ {𝑦}))
8	5, 7	anbi12d 473	. . . . . 6 ⊢ (𝑥 = {𝑦} → ((𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥) ↔ ({𝑦} ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ {𝑦})))
9	4, 8	spcev 2830	. . . . 5 ⊢ (({𝑦} ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ {𝑦}) → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
10	3, 9	mpan2 425	. . . 4 ⊢ ({𝑦} ⊆ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
11	2, 10	sylbi 121	. . 3 ⊢ (𝑦 ∈ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
12	11	exlimiv 1596	. 2 ⊢ (∃𝑦 𝑦 ∈ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
13		elequ1 2150	. . . . 5 ⊢ (𝑧 = 𝑤 → (𝑧 ∈ 𝑥 ↔ 𝑤 ∈ 𝑥))
14	13	cbvexv 1916	. . . 4 ⊢ (∃𝑧 𝑧 ∈ 𝑥 ↔ ∃𝑤 𝑤 ∈ 𝑥)
15	14	anbi2i 457	. . 3 ⊢ ((𝑥 ⊆ 𝐴 ∧ ∃𝑧 𝑧 ∈ 𝑥) ↔ (𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
16	15	exbii 1603	. 2 ⊢ (∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑧 𝑧 ∈ 𝑥) ↔ ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑤 𝑤 ∈ 𝑥))
17	12, 16	sylibr 134	1 ⊢ (∃𝑦 𝑦 ∈ 𝐴 → ∃𝑥(𝑥 ⊆ 𝐴 ∧ ∃𝑧 𝑧 ∈ 𝑥))

Syntax hints:   → wi 4   ∧ wa 104   = wceq 1353  ∃wex 1490   ∈ wcel 2146   ⊆ wss 3127  {csn 3589

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1445  ax-7 1446  ax-gen 1447  ax-ie1 1491  ax-ie2 1492  ax-8 1502  ax-10 1503  ax-11 1504  ax-i12 1505  ax-bndl 1507  ax-4 1508  ax-17 1524  ax-i9 1528  ax-ial 1532  ax-i5r 1533  ax-13 2148  ax-14 2149  ax-ext 2157  ax-sep 4116  ax-pow 4169

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1459  df-sb 1761  df-clab 2162  df-cleq 2168  df-clel 2171  df-nfc 2306  df-v 2737  df-in 3133  df-ss 3140  df-pw 3574  df-sn 3595

This theorem is referenced by: (None)