Theorem ssnelpss 4137

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 2869	. . 3 ⊢ ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → ¬ 𝐵 = 𝐴)
2		eqcom 2747	. . 3 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
3	1, 2	sylnib 328	. 2 ⊢ ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → ¬ 𝐴 = 𝐵)
4		dfpss2 4111	. . 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 2108   ⊆ wss 3976   ⊊ wpss 3977

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1778  df-cleq 2732  df-clel 2819  df-ne 2947  df-pss 3996

This theorem is referenced by:  ssnelpssd  4138  ssexnelpss  4139  isfin4p1  10384  canthp1lem2  10722  nqpr  11083  uzindi  14033  nthruc  16300  nthruz  16301  vitali  25667  onpsstopbas  36396