Theorem ssnelpss 4090

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 2940	. . 3 ⊢ ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → ¬ 𝐵 = 𝐴)
2		eqcom 2830	. . 3 ⊢ (𝐵 = 𝐴 ↔ 𝐴 = 𝐵)
3	1, 2	sylnib 330	. 2 ⊢ ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → ¬ 𝐴 = 𝐵)
4		dfpss2 4064	. . 3 ⊢ (𝐴 ⊊ 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ ¬ 𝐴 = 𝐵))
5	4	baibr 539	. 2 ⊢ (𝐴 ⊆ 𝐵 → (¬ 𝐴 = 𝐵 ↔ 𝐴 ⊊ 𝐵))
6	3, 5	syl5ib 246	1 ⊢ (𝐴 ⊆ 𝐵 → ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → 𝐴 ⊊ 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 398   = wceq 1537   ∈ wcel 2114   ⊆ wss 3938   ⊊ wpss 3939

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-ext 2795

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1781  df-cleq 2816  df-clel 2895  df-ne 3019  df-pss 3956

This theorem is referenced by:  ssnelpssd  4091  ssexnelpss  4092  isfin4p1  9739  canthp1lem2  10077  nqpr  10438  uzindi  13353  nthruc  15607  nthruz  15608  vitali  24216  onpsstopbas  33780