Theorem npss0 4361

Description: No set is a proper subset of the empty set. Dual of nvpss 4363. (Contributed by NM, 17-Jun-1998.) (Proof shortened by Andrew Salmon, 26-Jun-2011.) (Proof shortened by JJ, 14-Jul-2021.)

Assertion

Ref	Expression
npss0	⊢ ¬ 𝐴 ⊊ ∅

Proof of Theorem npss0

Step	Hyp	Ref	Expression
1		0ss 4350	. 2 ⊢ ∅ ⊆ 𝐴
2		ssnpss 4055	. 2 ⊢ (∅ ⊆ 𝐴 → ¬ 𝐴 ⊊ ∅)
3	1, 2	ax-mp 5	1 ⊢ ¬ 𝐴 ⊊ ∅

This proof depends on syntax axioms:  ¬ wn 3   ⊆ wss 3899   ⊊ wpss 3900  ∅c0 4279

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-ne 2957  df-dif 3902  df-ss 3916  df-pss 3919  df-nul 4280

This theorem is used by:  pssnn  9168  pssn0  43249