Theorem ss1o0el1o 7010

Description: Reformulation of ss1o0el1 4241 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 6515	. . . 4 ⊢ 1o = {∅}
2	1	eqcomi 2209	. . 3 ⊢ {∅} = 1o
3	2	sseq2i 3220	. 2 ⊢ (𝐴 ⊆ {∅} ↔ 𝐴 ⊆ 1o)
4		ss1o0el1 4241	. . 3 ⊢ (𝐴 ⊆ {∅} → (∅ ∈ 𝐴 ↔ 𝐴 = {∅}))
5	2	eqeq2i 2216	. . 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 1373   ∈ wcel 2176   ⊆ wss 3166  ∅c0 3460  {csn 3633  1_oc1o 6495

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 711  ax-5 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-10 1528  ax-11 1529  ax-i12 1530  ax-bndl 1532  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-ext 2187  ax-nul 4170

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1484  df-sb 1786  df-clab 2192  df-cleq 2198  df-clel 2201  df-nfc 2337  df-v 2774  df-dif 3168  df-un 3170  df-in 3172  df-ss 3179  df-nul 3461  df-sn 3639  df-suc 4418  df-1o 6502

This theorem is referenced by:  pw1dc1  7011