Theorem ssnelpss 3614

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 2452	. . 3 ⊢ ((C ∈ B ∧ ¬ C ∈ A) → ¬ B = A)
2		eqcom 2355	. . 3 ⊢ (B = A ↔ A = B)
3	1, 2	sylnib 295	. 2 ⊢ ((C ∈ B ∧ ¬ C ∈ A) → ¬ A = B)
4		dfpss2 3355	. . 3 ⊢ (A ⊊ B ↔ (A ⊆ B ∧ ¬ A = B))
5	4	baibr 872	. 2 ⊢ (A ⊆ B → (¬ A = B ↔ A ⊊ B))
6	3, 5	syl5ib 210	1 ⊢ (A ⊆ B → ((C ∈ B ∧ ¬ C ∈ A) → A ⊊ B))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 358   = wceq 1642   ∈ wcel 1710   ⊆ wss 3258   ⊊ wpss 3259

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-11 1746  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-an 360  df-ex 1542  df-cleq 2346  df-clel 2349  df-ne 2519  df-pss 3262

This theorem is used by:  ssnelpssd  3615