Theorem ssnelpss 3680

Description: A subclass missing a member is a proper subclass. (Contributed by NM, 12-Jan-2002.)

Assertion

Ref	Expression
ssnelpss	⊢ (𝐴 ⊆ 𝐵 → ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → 𝐴 ⊊ 𝐵))

Proof of Theorem ssnelpss

Step	Hyp	Ref	Expression
1		nelneq2 2713	. . 3 ⊢ ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → ¬ 𝐵 = 𝐴)
2		eqcom 2617	. . 3 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
3	1, 2	sylnib 317	. 2 ⊢ ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → ¬ 𝐴 = 𝐵)
4		dfpss2 3654	. . 3 ⊢ (𝐴 ⊊ 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ ¬ 𝐴 = 𝐵))
5	4	baibr 943	. 2 ⊢ (𝐴 ⊆ 𝐵 → (¬ 𝐴 = 𝐵 ↔ 𝐴 ⊊ 𝐵))
6	3, 5	syl5ib 233	1 ⊢ (𝐴 ⊆ 𝐵 → ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → 𝐴 ⊊ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 383   = wceq 1475   ∈ wcel 1977   ⊆ wss 3540   ⊊ wpss 3541

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-an 385  df-ex 1696  df-cleq 2603  df-clel 2606  df-ne 2782  df-pss 3556

This theorem is referenced by:  ssnelpssd  3681  ssexnelpss  3682  canthp1lem2  9354  nqpr  9715  uzindi  12643  nthruc  14819  nthruz  14820  vitali  23188  onpsstopbas  31599