Theorem sseq0b 4359

Description: The only subclass of the empty class is itself. (Contributed by NM, 7-Mar-2007.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) Strengthen sseq0 4360 to a biconditional. (Revised by BJ, 19-Jul-2026.)

Assertion

Ref	Expression
sseq0b	⊢ (𝐴 = ∅ → (𝐵 ⊆ 𝐴 ↔ 𝐵 = ∅))

Proof of Theorem sseq0b

Step	Hyp	Ref	Expression
1		sseq2 3962	. 2 ⊢ (𝐴 = ∅ → (𝐵 ⊆ 𝐴 ↔ 𝐵 ⊆ ∅))
2		ss0b 4357	. 2 ⊢ (𝐵 ⊆ ∅ ↔ 𝐵 = ∅)
3	1, 2	bitrdi 290	1 ⊢ (𝐴 = ∅ → (𝐵 ⊆ 𝐴 ↔ 𝐵 = ∅))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569   ⊆ wss 3904  ∅c0 4285

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-fal 1582  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-dif 3907  df-ss 3921  df-nul 4286

This theorem is used by:  sseq0  4360