Theorem ss1o0el1o 7214

Description: Reformulation of ss1o0el1 4332 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 6695	. . . 4 ⊢ 1o = {∅}
2	1	eqcomi 2242	. . 3 ⊢ {∅} = 1o
3	2	sseq2i 3275	. 2 ⊢ (𝐴 ⊆ {∅} ↔ 𝐴 ⊆ 1o)
4		ss1o0el1 4332	. . 3 ⊢ (𝐴 ⊆ {∅} → (∅ ∈ 𝐴 ↔ 𝐴 = {∅}))
5	2	eqeq2i 2249	. . 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 1402   ∈ wcel 2209   ⊆ wss 3220  ∅c0 3520  {csn 3708  1_oc1o 6674

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 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220  ax-nul 4257

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-dif 3222  df-un 3224  df-in 3226  df-ss 3233  df-nul 3521  df-sn 3714  df-suc 4514  df-1o 6681

This theorem is referenced by:  pw1dc1  7215