Theorem ssnelpss 4068

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 2887	. . 3 ⊢ ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → ¬ 𝐵 = 𝐴)
2	1	neqcomd 2772	. 2 ⊢ ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → ¬ 𝐴 = 𝐵)
3		dfpss2 4041	. . 3 ⊢ (𝐴 ⊊ 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ ¬ 𝐴 = 𝐵))
4	3	baibr 545	. 2 ⊢ (𝐴 ⊆ 𝐵 → (¬ 𝐴 = 𝐵 ↔ 𝐴 ⊊ 𝐵))
5	2, 4	imbitrid 247	1 ⊢ (𝐴 ⊆ 𝐵 → ((𝐶 ∈ 𝐵 ∧ ¬ 𝐶 ∈ 𝐴) → 𝐴 ⊊ 𝐵))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 400   = wceq 1569   ∈ wcel 2142   ⊆ wss 3904   ⊊ wpss 3905

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-ex 1809  df-cleq 2754  df-clel 2837  df-ne 2958  df-pss 3924

This theorem is used by:  ssnelpssd  4069  ssexnelpss  4070  isfin4p1  10305  canthp1lem2  10644  nqpr  11005  uzindi  14025  nthruc  16314  nthruz  16315  vitali  25783  onpsstopbas  36969  nthrucw  47635