Theorem ssnelpss 4124

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 2864	. . 3 ⊢ ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → ¬ 𝐵 = 𝐴)
2		eqcom 2742	. . 3 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
3	1, 2	sylnib 328	. 2 ⊢ ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → ¬ 𝐴 = 𝐵)
4		dfpss2 4098	. . 3 ⊢ (𝐴 ⊊ 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ ¬ 𝐴 = 𝐵))
5	4	baibr 536	. 2 ⊢ (𝐴 ⊆ 𝐵 → (¬ 𝐴 = 𝐵 ↔ 𝐴 ⊊ 𝐵))
6	3, 5	imbitrid 244	1 ⊢ (𝐴 ⊆ 𝐵 → ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → 𝐴 ⊊ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395   = wceq 1537   ∈ wcel 2106   ⊆ wss 3963   ⊊ wpss 3964

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1777  df-cleq 2727  df-clel 2814  df-ne 2939  df-pss 3983

This theorem is referenced by:  ssnelpssd  4125  ssexnelpss  4126  isfin4p1  10353  canthp1lem2  10691  nqpr  11052  uzindi  14020  nthruc  16285  nthruz  16286  vitali  25662  onpsstopbas  36413