Theorem ss1o0el1o 7148

Description: Reformulation of ss1o0el1 4293 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 6639	. . . 4 ⊢ 1o = {∅}
2	1	eqcomi 2235	. . 3 ⊢ {∅} = 1o
3	2	sseq2i 3255	. 2 ⊢ (𝐴 ⊆ {∅} ↔ 𝐴 ⊆ 1o)
4		ss1o0el1 4293	. . 3 ⊢ (𝐴 ⊆ {∅} → (∅ ∈ 𝐴 ↔ 𝐴 = {∅}))
5	2	eqeq2i 2242	. . 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 1398   ∈ wcel 2202   ⊆ wss 3201  ∅c0 3496  {csn 3673  1_oc1o 6618

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 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2213  ax-nul 4220

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2364  df-v 2805  df-dif 3203  df-un 3205  df-in 3207  df-ss 3214  df-nul 3497  df-sn 3679  df-suc 4474  df-1o 6625

This theorem is referenced by:  pw1dc1  7149