Theorem difsnpss 4770

Description: (𝐵 ∖ {𝐴}) is a proper subclass of 𝐵 if and only if 𝐴 is a member of 𝐵. (Contributed by David Moews, 1-May-2017.)

Assertion

Ref	Expression
difsnpss	⊢ (𝐴 ∈ 𝐵 ↔ (𝐵 ∖ {𝐴}) ⊊ 𝐵)

Proof of Theorem difsnpss

Step	Hyp	Ref	Expression
1		notnotb 318	. 2 ⊢ (𝐴 ∈ 𝐵 ↔ ¬ ¬ 𝐴 ∈ 𝐵)
2		difss 4075	. . . 4 ⊢ (𝐵 ∖ {𝐴}) ⊆ 𝐵
3	2	biantrur 540	. . 3 ⊢ ((𝐵 ∖ {𝐴}) ≠ 𝐵 ↔ ((𝐵 ∖ {𝐴}) ⊆ 𝐵 ∧ (𝐵 ∖ {𝐴}) ≠ 𝐵))
4		difsnb 4769	. . . 4 ⊢ (¬ 𝐴 ∈ 𝐵 ↔ (𝐵 ∖ {𝐴}) = 𝐵)
5	4	necon3bbii 3006	. . 3 ⊢ (¬ ¬ 𝐴 ∈ 𝐵 ↔ (𝐵 ∖ {𝐴}) ≠ 𝐵)
6		df-pss 4017	. . 3 ⊢ ((𝐵 ∖ {𝐴}) ⊊ 𝐵 ↔ ((𝐵 ∖ {𝐴}) ⊆ 𝐵 ∧ (𝐵 ∖ {𝐴}) ≠ 𝐵))
7	3, 5, 6	3bitr4i 306	. 2 ⊢ (¬ ¬ 𝐴 ∈ 𝐵 ↔ (𝐵 ∖ {𝐴}) ⊊ 𝐵)
8	1, 7	bitri 278	1 ⊢ (𝐴 ∈ 𝐵 ↔ (𝐵 ∖ {𝐴}) ⊊ 𝐵)

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ∧ wa 401   ∈ wcel 2146   ≠ wne 2959   ⊆ wss 3900   ⊊ wpss 4016   ∖ cdif 4048  {csn 4588

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-v 3456  df-ss 3901  df-pss 4017  df-dif 4050  df-sn 4589

This theorem is used by:  marypha1lem  9438  infpss  10307  ominf4  10403  mrieqv2d  17828