Theorem ss1o0el1o 7009

Description: Reformulation of ss1o0el1 4240 using 1_o instead of {∅}. (Contributed by BJ, 9-Aug-2024.)

Assertion

Ref	Expression
ss1o0el1o	⊢ (𝐴 ⊆ 1o → (∅ ∈ 𝐴 ↔ 𝐴 = 1o))

Proof of Theorem ss1o0el1o

Step	Hyp	Ref	Expression
1		df1o2 6514	. . . 4 ⊢ 1o = {∅}
2	1	eqcomi 2208	. . 3 ⊢ {∅} = 1o
3	2	sseq2i 3219	. 2 ⊢ (𝐴 ⊆ {∅} ↔ 𝐴 ⊆ 1o)
4		ss1o0el1 4240	. . 3 ⊢ (𝐴 ⊆ {∅} → (∅ ∈ 𝐴 ↔ 𝐴 = {∅}))
5	2	eqeq2i 2215	. . 3 ⊢ (𝐴 = {∅} ↔ 𝐴 = 1o)
6	4, 5	bitrdi 196	. 2 ⊢ (𝐴 ⊆ {∅} → (∅ ∈ 𝐴 ↔ 𝐴 = 1o))
7	3, 6	sylbir 135	1 ⊢ (𝐴 ⊆ 1o → (∅ ∈ 𝐴 ↔ 𝐴 = 1o))

Syntax hints:   → wi 4   ↔ wb 105   = wceq 1372   ∈ wcel 2175   ⊆ wss 3165  ∅c0 3459  {csn 3632  1_oc1o 6494

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-in1 615  ax-in2 616  ax-io 710  ax-5 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-10 1527  ax-11 1528  ax-i12 1529  ax-bndl 1531  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186  ax-nul 4169

This theorem depends on definitions:  df-bi 117  df-tru 1375  df-nf 1483  df-sb 1785  df-clab 2191  df-cleq 2197  df-clel 2200  df-nfc 2336  df-v 2773  df-dif 3167  df-un 3169  df-in 3171  df-ss 3178  df-nul 3460  df-sn 3638  df-suc 4417  df-1o 6501

This theorem is referenced by:  pw1dc1  7010